2025 IB DP Biology 模擬試題連答案詳解

Using the Lincoln Index formula: \(N = \frac{n_1 \times n_2}{m_2}\), where \(n_1 = 80\) (first sample marked), \(n_2 = 120\) (second sample captured), and \(m_2 = 15\) (marked individuals recaptured). Calculating this gives \(N = \frac{80 \times 120}{15} = \frac{9600}{15} = 640\). For this estimate to be reliable, several assumptions must hold true, including that there is no change in population size due to births, deaths, immigration, or emigration between the two sampling periods (a closed population). If marked individuals suffered higher mortality, the estimate would be skewed. Therefore, the population is estimated at 640 beetles and requires no immigration or emigration to be reliable.

1 mark for the correct option (A).

The cleaner wrasse and the client fish have a mutualistic relationship (mutualism) because both species benefit: the wrasse gets food (ectoparasites), and the client fish gets rid of harmful parasites. The blenny and the client fish have a parasitic relationship (parasitism) because the blenny benefits by obtaining food (tissue from the client), while the client fish is harmed by losing tissue, though it is usually not killed directly.

1 mark for the correct option (A).

The initial scenario shows bottom-up control, where resource availability (nutrients) limits the growth of higher trophic levels. If a top predator (piscivorous fish) is introduced, it will prey on the planktivorous fish, reducing their numbers. Since planktivorous fish prey on zooplankton, a reduction in planktivorous fish will relieve predation pressure on zooplankton, leading to an increase in the zooplankton population via a trophic cascade.

1 mark for the correct option (A).

During forced exhalation, the process is active. The internal intercostal muscles contract, pulling the ribcage downward and inward. The abdominal muscles contract, pushing the organs up against the diaphragm. Simultaneously, the external intercostal muscles and the diaphragm relax. These actions decrease the thoracic volume, which increases pressure inside the lungs, forcing air out.

1 mark for the correct option (B).

Type I pneumocytes are extremely thin, flattened cells that form the majority of the alveolar wall, minimizing the diffusion distance for gas exchange. Type II pneumocytes are thicker, rounded cells that secrete pulmonary surfactant, which reduces surface tension and prevents the alveoli from collapsing.

1 mark for the correct option (C).

Because malonate is structurally similar to succinate, it competes with succinate for the active site of succinate dehydrogenase, making it a competitive inhibitor. This type of inhibition can be overcome by increasing the substrate (succinate) concentration, as this increases the probability that the active site binds a substrate molecule rather than an inhibitor molecule.

1 mark for the correct option (B).

Sickle cell anemia is caused by a base substitution mutation where a single base in the DNA template is changed, resulting in the transcribed mRNA codon changing from GAG to GUG. This alters the translation of the polypeptide, replacing the hydrophilic amino acid glutamic acid with the hydrophobic amino acid valine.

1 mark for the correct option (B).

During the transitional phase, the rate of population increase slows down as density-dependent factors, such as the depletion of nutrients and the accumulation of toxic waste products (like ethanol in yeast), increase the death rate and decrease the birth rate, moving the population towards its plateau phase.

1 mark for the correct option (B).

The calculated chi-squared value (4.12) is greater than the critical value (3.84) at the \(p = 0.05\) level with 1 degree of freedom. This means we reject the null hypothesis of independence. Therefore, there is a statistically significant association between the presence of *Plantago major* and *Taraxacum officinale*. A sample size of 100 quadrats is standard and sufficient for a chi-squared test of association.

Award [1] for the correct answer (B).
- Reject A because the reasoning is contradictory (a calculated value greater than the critical value indicates a significant association, not a non-significant one).
- Reject C because the inequality stated is incorrect.
- Reject D because 100 quadrats is a standard, valid sample size.

The Lincoln index formula is used to estimate population size:

\[N = \frac{n_1 \times n_2}{m_2}\]

where:
- \(n_1\) = number of individuals marked and released in the first sample = 120
- \(n_2\) = total number of individuals captured in the second sample = 150
- \(m_2\) = number of marked individuals recaptured in the second sample = 15

Substituting the values:

\[N = \frac{120 \times 150}{15} = 120 \times 10 = 1,200\]

Award [1] for the correct answer (B).
- Award [0] for any incorrect calculations or misapplications of the Lincoln index formula.

Bottom-up control occurs when the population size of a trophic level is limited by the availability of resources (such as food or nutrients) from lower trophic levels. In this scenario, the primary consumer (the insect) is directly regulated by its food source, the host plant (the producer).

Award [1] for the correct answer (B).
- Reject A because top-down control is regulation by predators or parasites.
- Reject C because resource availability is a density-dependent factor, and the scenario depicts trophic level dependency.
- Reject D because competitive exclusion refers to one species outcompeting another for the same niche, not resource-driven population regulation.

In the mutualistic relationship between corals and zooxanthellae:
- The coral animal provides the zooxanthellae with a protected environment (shelter), carbon dioxide (\(\text{CO}_2\) from respiration), and inorganic nitrogenous/phosphorus waste products.
- The zooxanthellae (photosynthetic dinoflagellates) use these raw materials to perform photosynthesis and supply the coral with organic molecules like glucose, amino acids, and glycerol, as well as oxygen (\(\text{O}_2\)).

Award [1] for the correct answer (B).
- Reject A because corals do not produce glucose for zooxanthellae.
- Reject C because zooxanthellae do not provide calcium carbonate to the corals; the corals deposit calcium carbonate themselves using energy from the nutrients provided.
- Reject D because corals provide shelter, not zooxanthellae.

Type II pneumocytes secrete a lipoprotein mixture called pulmonary surfactant. This surfactant forms a monolayer over the moisture lining the alveoli, which dramatically reduces the surface tension. By lowering surface tension, it prevents the alveoli from collapsing during exhalation and makes it much easier to inflate the lungs during inhalation.

Award [1] for the correct answer (B).
- Reject A because surfactant decreases (rather than increases) surface tension.
- Reject C because while some proteins have innate immune roles, the primary physical and physiological function of surfactant is reducing surface tension to prevent alveolar collapse.
- Reject D because surfactant does not bind bicarbonate or chemically accelerate CO2 diffusion.

During vigorous exercise, respiring muscles produce more carbon dioxide and lactic acid, which increases partial pressure of CO2 and lowers blood pH. This decreases hemoglobin's affinity for oxygen, shifting the curve to the right. This allows hemoglobin to release oxygen more readily to the actively respiring muscle tissues.

Award [1] for the correct answer (A).
- Reject B because decreased CO2 and increased pH would increase affinity, shifting the curve to the left.
- Reject C because increased pH does not occur during vigorous exercise, and increased affinity would not promote oxygen release to tissues.
- Reject D because decreased temperature shifts the curve to the left, and a rightward shift decreases affinity rather than increases overall transport capacity in a general sense.

This is an example of end-product inhibition, which is a form of non-competitive (allosteric) inhibition. The end product (isoleucine) binds to an allosteric site on the first enzyme (threonine deaminase), altering its active site conformation. Because the inhibitor does not compete with the substrate for the active site, the maximum rate of reaction (\(V_{\max}\)) is decreased and cannot be restored by increasing the concentration of the substrate threonine.

Award [1] for the correct answer (B).
- Reject A and C because this is non-competitive inhibition, not competitive inhibition.
- Reject D because the substrate is not chemically modified by the inhibitor; instead, the enzyme's active site conformation is changed.

In the CRISPR-Cas9 system, the single guide RNA (sgRNA) is engineered to contain a 20-nucleotide sequence that is complementary to the target genomic DNA, guiding the complex to the correct location. The Cas9 protein functions as an endonuclease that binds to the guide RNA and cleaves both strands of the double-stranded DNA at the target locus.

Award [1] for the correct answer (C).
- Reject A because sgRNA does not act as a template for homologous recombination, and Cas9 is an endonuclease, not a polymerase.
- Reject B because sgRNA targets specific loci via base pairing (not merely general promoters), and Cas9 cuts highly specific sites targeted by the guide RNA, not random sites.
- Reject D because Cas9 does cleave the phosphodiester sugar-phosphate backbone, and sgRNA is not a DNA ligase helper molecule.

An increase in primary producers (food resources) increases the resources available to the herbivorous beetle, which raises the carrying capacity \(K\) of the habitat. However, the introduction of a predator imposes top-down regulation, meaning the actual population density of the beetle will fluctuate or decrease in response to predation, rather than simply matching the carrying capacity.

Award 1 mark for identifying that carrying capacity increases due to resource abundance, while the actual population density fluctuates due to the introduction of a predator.

A keystone species has a disproportionately large effect on its community relative to its abundance. Removing the starfish (*Pisaster ochraceus*) allows the mussel population to grow unchecked, outcompeting other species and significantly reducing biodiversity. This is a classic example of a keystone species preventing competitive exclusion.

Award 1 mark for the correct identification of the keystone species concept.

The fundamental niche of a species is the full potential range of physical, chemical, and biological factors it can use if there is no competition. The realized niche is the actual range of conditions it occupies due to limiting factors like competition. When competitor Species Y is removed, Species X expands from its realized niche to its fundamental niche.

Award 1 mark for linking the expansion of the species' range and resource use after removal of a competitor to the difference between its fundamental and realized niche.

Increased cellular respiration elevates blood \(pCO_2\). Carbon dioxide reacts with water to form carbonic acid, which dissociates and lowers blood pH (making it more acidic). Chemoreceptors in the medulla oblongata and major blood vessels detect this decrease in pH, stimulating the respiratory center in the medulla oblongata to increase the ventilation rate.

Award 1 mark for identifying that high \(pCO_2\) lowers blood pH, which is detected by the medulla oblongata to increase ventilation rate.

Type II pneumocytes are cuboidal cells that secrete a fluid containing pulmonary surfactant. This surfactant reduces the surface tension of water in the alveolus, preventing the alveolar walls from sticking together and collapsing during exhalation. Type I pneumocytes, on the other hand, are extremely thin cells adapted for gas diffusion.

Award 1 mark for identifying that Type II pneumocytes secrete a surfactant that prevents alveolar collapse.

In competitive inhibition, the inhibitor binds to the active site and directly competes with the substrate. Increasing the substrate concentration raises the likelihood of substrate binding instead of the inhibitor, thereby overcoming the inhibition and allowing the reaction to reach the original maximum velocity (\(V_{max}\)).

Award 1 mark for identifying competitive inhibition and explaining that its effect is overcome at high substrate concentrations.

A nonsense mutation is a point mutation in a sequence of DNA that results in a premature stop codon in the transcribed mRNA, leading to a truncated, incomplete, and usually nonfunctional protein product.

Award 1 mark for identifying that a mutation introducing a premature stop codon is a nonsense mutation.

The primary producers have a net primary productivity (NPP) of \(22,000\text{ kJ m}^{-2}\text{ yr}^{-1}\). Primary consumers receive \(10\%\) of this energy: \(22,000 \times 0.10 = 2,200\text{ kJ m}^{-2}\text{ yr}^{-1}\). Secondary consumers receive \(10\%\) of the energy transferred to primary consumers: \(2,200 \times 0.10 = 220\text{ kJ m}^{-2}\text{ yr}^{-1}\).

Award 1 mark for correctly applying the 10% efficiency transfer twice to obtain \(220\text{ kJ m}^{-2}\text{ yr}^{-1}\).

The Lincoln index is calculated using the formula: N = (n1 * n2) / m2, where n1 is the number of individuals marked in the first sample (120), n2 is the total number of individuals in the second sample (150), and m2 is the number of marked individuals recaptured in the second sample (30). Thus, N = (120 * 150) / 30 = 600 beetles. A critical assumption of this method is that the marked individuals mix randomly with the rest of the population and do not experience altered survival rates or changes in behavior that make them more or less likely to be recaptured.

Award 1 mark for the correct calculation of 600 beetles combined with the correct assumption of random mixing and equal recapture probability. Reject options showing incorrect population calculations (300) or invalid ecological assumptions (increased catchability, or high immigration/emigration which violates the closed population assumption).

To find the expected frequency, we first construct the contingency table and calculate row and column totals: Total quadrats where Species X is present = 25 (both) + 15 (only X) = 40. Total quadrats where Species Y is present = 25 (both) + 35 (only Y) = 60. Grand total of quadrats = 100. Using the formula for expected frequency: Expected = (Row Total * Column Total) / Grand Total. For the cell representing the presence of both species: Expected = (40 * 60) / 100 = 24.

Award 1 mark for calculating the correct expected frequency of 24. Option A (15) is incorrect (it is the observed frequency of only Species X). Option C (25) is incorrect (it is the observed frequency of both species). Option D (40) is incorrect (it is the total number of quadrats containing Species X).

This is an example of a top-down trophic cascade: 1. A reduction in the population of large predatory fish (tertiary consumers) decreases the predation pressure on small planktivorous fish (secondary consumers), causing their population to increase. 2. The increased population of small planktivorous fish leads to higher predation pressure on zooplankton (primary consumers), causing their population to decrease. 3. The decline in zooplankton reduces grazing pressure on phytoplankton (primary producers), allowing the phytoplankton population to increase.

Award 1 mark for the correct pattern of population changes corresponding to a top-down cascade: small fish increase, zooplankton decrease, phytoplankton increase. All other combinations represent incorrect ecological logic.

Inspiration is an active process that requires muscular contraction to increase thoracic cavity volume: The external intercostal muscles contract while the internal intercostal muscles relax (acting antagonistically) to pull the ribcage upward and outward. The diaphragm contracts, moving downward and flattening. These combined actions increase the volume of the thoracic cavity, which decreases the pressure inside the alveoli below atmospheric pressure. Air flows down this pressure gradient into the lungs.

Award 1 mark for identifying that the external intercostal muscles contract, internal intercostals relax, diaphragm contracts, and alveolar pressure decreases. Reject options indicating internal intercostal contraction during inspiration or diaphragm relaxation.

Type I pneumocytes are squamous (thin and flat) cells that cover approximately 95% of the alveolar surface. Their thinness minimizes the diffusion distance for oxygen and carbon dioxide, facilitating rapid gas exchange. Type II pneumocytes are larger, cuboidal cells that secrete pulmonary surfactant, which reduces surface tension and prevents the alveoli from collapsing.

Award 1 mark for correctly identifying Type I pneumocytes as thin, flat, and responsible for gas exchange. Reject options that misidentify Type I as surfactant secretors or Type II as thin and flat.

Competitive inhibitors compete with the substrate for the enzyme's active site. At very high substrate concentrations, the substrate outcompetes the inhibitor, allowing the reaction to reach its original maximum velocity (Vmax), though a higher concentration of substrate is needed to reach half Vmax (increased Km). Non-competitive inhibitors bind to an allosteric site, inducing a conformational change in the enzyme that reduces its catalytic activity. This lowers the maximum rate of reaction (Vmax) regardless of the substrate concentration.

Award 1 mark for identifying Inhibitor A as competitive (active site) and Inhibitor B as non-competitive (allosteric site). Reject options that swap these descriptions or incorrectly identify binding locations.

Enzymes increase the rate of chemical reactions by lowering the activation energy barrier required for the transition state. However, they do not alter the chemical equilibrium or change the free energy of the reactants or products. Consequently, the net free energy change of the reaction remains completely unaffected.

Award 1 mark for the option showing decreased activation energy and unaltered net free energy change. Reject all other combinations.

In the CRISPR-Cas9 system, the guide RNA (gRNA) is engineered to contain a sequence of approximately 20 nucleotides that is complementary to the target genomic DNA sequence. By complementary base pairing, the gRNA directs the Cas9-gRNA complex to the precise location in the genome. The Cas9 protein is an endonuclease enzyme that introduces a double-strand break (DSB) at this specific target site, allowing for gene knockout or targeted gene insertion.

Award 1 mark for recognizing that gRNA guides via complementary base pairing and Cas9 acts as an endonuclease to cleave DNA. Reject options claiming gRNA acts as a cutting enzyme, template, or messenger regulator.

The Lincoln index formula is: \(N = \frac{n_1 \times n_2}{m_2}\) where: \(n_1\) = number of individuals marked in the first capture (80), \(n_2\) = total number of individuals in the second capture (60), and \(m_2\) = number of marked individuals in the second capture (15). Substituting the values: \(N = \frac{80 \times 60}{15} = \frac{4800}{15} = 320\).

Award [1] for the correct answer A. Other options represent incorrect applications of the formula or mathematical errors.

Allelopathy is a biological phenomenon where an organism produces biochemicals that influence the growth, survival, or reproduction of other organisms. The secretion of juglone by black walnut trees to inhibit other plants is a classic example of allelopathy.

Award [1] for the correct answer B. Reject other options: A (mutualism is a mutually beneficial interaction), C (parasitism involves one organism living on or in another to its detriment), and D (herbivory is the consumption of plants by animals).

The null hypothesis (\(H_0\)) states that there is no significant association between the two species. Because the calculated chi-squared value (4.12) is greater than the critical value (3.84), the null hypothesis is rejected, indicating that there is a significant association between the two species.

Award [1] for the correct answer B. Reject options A and C because they incorrectly accept the null hypothesis despite the calculated value exceeding the critical value. Reject D because closeness to the critical value does not change the binary decision of the significance test.

Type I pneumocytes are extremely thin cells adapted to carry out gas exchange by diffusion. Type II pneumocytes are thicker, rounded cells that secrete a surfactant to reduce surface tension inside the alveoli and prevent them from collapsing.

Award [1] for the correct answer A. Reject B because it reverses the functions of the two cell types. Reject C and D because they attribute incorrect functions (such as phagocytosis or active transport of gases) to pneumocytes.

During inhalation, both the diaphragm and the external intercostal muscles contract. The diaphragm moves downward, and the external intercostal muscles pull the ribcage upward and outward. This increases the volume of the thoracic cavity, which decreases the pressure inside the lungs below atmospheric pressure, forcing air to enter.

Award [1] for the correct answer A. Reject other options because they state incorrect muscle actions or describe incorrect relationships between volume and pressure (which violate Boyle's law).

In competitive inhibition, the inhibitor binds directly to the active site, competing with the substrate. Increasing the substrate concentration increases the probability of substrate molecules binding instead of the inhibitor, allowing the reaction to eventually reach the same maximum velocity (\(V_{\max}\)) as the uninhibited reaction.

Award [1] for the correct answer B. Reject A because non-competitive inhibition cannot be overcome by high substrate concentrations. Reject C and D because they describe mechanisms that prevent the enzyme from achieving its normal \(V_{\max}\).

Sickle-cell anemia is caused by a point mutation where the DNA template triplet CAC (corresponding to CTC on the coding strand) is changed to CTC (corresponding to CAC on the coding strand). This transcribes into a change on the mRNA codon from GAG to GUG, causing the polar amino acid glutamic acid to be replaced by the non-polar amino acid valine.

Award [1] for the correct answer A. Reject B because it reverses the direction of the mutation. Reject C and D because sickle-cell anemia is a point mutation (base substitution), not a frameshift or triplet expansion mutation.

Top-down control occurs when a predator at a higher trophic level controls the structure or population dynamics of the ecosystem. A trophic cascade is an ecological phenomenon triggered by the addition or removal of top predators, which propagates down the food web. In this case, removing the sea otters (top predator) causes a cascade that leads to the decimation of the kelp (producer).

Award [1] for the correct answer B. Reject A because bottom-up control is regulated by nutrients and primary productivity. Reject C and D because they refer to unrelated ecological phenomena.

a) Between Day 2 and Day 6, the population shows rapid, unrestricted growth, which is characteristic of the exponential (or log) phase.

b) Calculation:
- Population density at Day 2 = \(15\text{ cells/mL}\)
- Population density at Day 8 = \(240\text{ cells/mL}\)
- Increase = \(240 - 15 = 225\text{ cells/mL}\)
- Percentage increase = \(\frac{225}{15} \times 100\% = 1500\%\).

c) The carrying capacity \(K\) is approximately \(245\text{ to }250\text{ cells/mL}\). This is because the population growth curve levels off (plateaus) around this value on Days 10\u201312, showing that the system has reached the maximum population size that the culture environment can sustainably support.

d) Factors slowing growth and causing stabilization:
- Food availability: As the population of *Paramecium* increases, the yeast food supply becomes a limiting factor.
- Waste accumulation: In a closed flask, metabolic toxic wastes (e.g., ammonia or acids) accumulate, which increases the mortality rate of the cells.
- Space limitations: Physical space inside the flask becomes restricted, increasing crowding and physical interactions.
- Competition: Intraspecific competition for resources increases, leading to a decreased reproduction rate and increased death rate until birth rate equals death rate.

a) Award [1] for: Exponential phase / log phase.

b) Award [2] marks:
- [1] for correct working showing change divided by original, e.g., \(\frac{240 - 15}{15} \times 100\) or equivalent.
- [1] for correct answer: \(1500\%\).

c) Award [2] marks:
- [1] for identifying carrying capacity as \(245\), \(250\), or a value within the range \(245\text{ to }250\text{ cells/mL}\).
- [1] for explaining that the population curve plateaus / levels off, showing that mortality rate equals birth rate due to environmental resistance.

d) Award up to [3.75] marks for explanation points (up to four points, or three detailed points):
- [1] depletion of food / yeast resources.
- [1] accumulation of toxic metabolic waste products in a closed system.
- [1] increased intraspecific competition for resources.
- [0.75] limited physical space/crowding.

a) The null hypothesis (\(H_0\)) states that the distribution of *Aster alpinus* and *Gentiana verna* is independent, meaning there is no association between the presence of the two species.

b) Calculation of Expected Frequency:
- Formula: \(\text{Expected Frequency} = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}\)
- For the cell \"Both present\":
- Row Total (Gentiana present) = \(50\)
- Column Total (Aster present) = \(40\)
- Grand Total = \(100\)
- Calculation: \(\frac{50 \times 40}{100} = 20\).

c) Statistical Conclusion:
- Since the calculated chi-squared value (\(10.67\)) is greater than the critical value (\(3.84\)), the null hypothesis is rejected.
- There is a statistically significant association between the two species.
- Since the observed frequency of co-occurrence (28) is greater than the expected frequency (20), this indicates a positive ecological association (they tend to grow together).

d) Biological Explanation:
- Similar abiotic factors: Both species may have identical preferences for environmental conditions such as soil pH, moisture, sunlight exposure, or nutrient availability in the alpine meadow.
- Ecological facilitation / Mutualism: One species may shelter or alter the soil environment in a way that benefits the germination or survival of the other species (e.g., nitrogen fixation, soil stabilization).

a) Award [1] for stating that there is no significant association / the species are independent / distribute randomly relative to each other.

b) Award [3] marks for working:
- [1] for identifying the correct row total (50) and column total (40).
- [1] for showing the correct calculation structure: \(\frac{50 \times 40}{100}\).
- [1] for the correct expected value: \(20\).

c) Award [2.75] marks:
- [0.75] for noting that \(10.67 > 3.84\) (calculated > critical value).
- [1] for stating that the null hypothesis (\(H_0\)) is rejected (and/or alternative hypothesis is accepted).
- [1] for concluding there is a significant positive association (they co-occur more than expected by chance).

d) Award up to [2] marks (1 mark per distinct biological explanation):
- [1] sharing of similar abiotic niche requirements (e.g., water, soil pH, sunlight).
- [1] facilitation / mutualistic interaction (e.g., shelter from wind/grazers, nutrient replenishment, pollinator attraction).

a) Calculation:
- Formula: \(\text{Minute Ventilation} = \text{Ventilation Rate} \times \text{Tidal Volume}\)
- At rest: \(12\text{ breaths/min} \times 0.5\text{ L/breath} = 6.0\text{ L/min}\) (or \(\text{dm}^3\text{ min}^{-1}\))
- During heavy exercise: \(35\text{ breaths/min} \times 2.8\text{ L/breath} = 98.0\text{ L/min}\) (or \(\text{dm}^3\text{ min}^{-1}\))

b) Description:
- Both ventilation rate and tidal volume show a direct positive relationship with exercise intensity.
- Tidal volume increases from \(0.5\text{ L}\) to \(2.8\text{ L}\) (a 5.6-fold increase), while ventilation rate increases from \(12\) to \(35\text{ breaths/min}\) (nearly a 3-fold increase). Both factors cooperate to drastically elevate overall minute ventilation.

c) Control Mechanism:
- During exercise, active muscles increase the rate of aerobic cellular respiration to generate ATP, producing large amounts of carbon dioxide (\(\text{CO}_2\)) as a byproduct.
- Excess \(\text{CO}_2\) diffuses into blood plasma, where it reacts with water to form carbonic acid (\(\text{H}_2\text{CO}_3\)), dissociating into hydrogen ions (\(\text{H}^+\)) and bicarbonate ions (\(\text{HCO}_3^-\)), thereby lowering blood pH.
- Chemoreceptors in the medulla oblongata (central) and in the carotid and aortic bodies (peripheral) detect this drop in pH / rise in partial pressure of carbon dioxide (\(P_{\text{CO}_2}\)).
- These receptors send sensory nerve impulses to the ventilation (respiratory) center in the medulla oblongata.
- The medulla oblongata responds by sending increased motor impulses via the phrenic and intercostal nerves to the diaphragm and external intercostal muscles, increasing both the rate and depth of contractions.

d) Other physiological responses:
- An increase in heart rate / cardiac output to pump oxygenated blood faster.
- Selective vasodilation of arterioles feeding skeletal muscle tissues and vasoconstriction of arterioles feeding non-essential organs.
- The Bohr shift: high local \(P_{\text{CO}_2}\) and low pH shift the oxygen-hemoglobin dissociation curve to the right, causing hemoglobin to release oxygen more readily in respiring muscle tissue.

a) Award [2] marks:
- [1] for both calculations correct (\(6.0\) and \(98.0\)).
- [1] for correct units (\(\text{L/min}\) or \(\text{dm}^3\text{ min}^{-1}\) or \(\text{liters per minute}\)) included with both answers.

b) Award up to [2] marks:
- [1] for stating that both parameters increase as exercise intensity increases.
- [1] for describing a quantitative detail (e.g., tidal volume increases by a greater factor than ventilation rate, or citing specific values showing the scaling).

c) Award up to [3.75] marks for explanation details:
- [1] cellular respiration increases \(\text{CO}_2\) concentration in blood, lowering blood pH.
- [1] chemoreceptors in the medulla / aorta / carotid bodies detect decrease in pH / increase in \(\text{CO}_2\).
- [1] respiratory center in medulla oblongata is stimulated.
- [0.75] signals sent via nerves to diaphragm / intercostal muscles to contract faster and more strongly.

d) Award [1] mark for any of the following valid responses:
- increased heart rate / stroke volume / cardiac output.
- Bohr shift (hemoglobin releasing oxygen more easily under acidic conditions).
- vasodilation of skeletal muscle arterioles.

a) Identification and Reason:
- Inhibitor A is the competitive inhibitor.
- Reason: As substrate concentration [S] increases to high levels (20 mmol/L), the rate of reaction with Inhibitor A (\(58\ \mu\text{mol min}^{-1}\)) closely approaches the uninhibited control rate (\(60\ \mu\text{mol min}^{-1}\)).
- This indicates that high substrate concentrations outcompete the inhibitor, restoring the maximum rate of reaction, which is a hallmark of competitive inhibition.

b) Estimation:
- \(V_{\text{max}}\) is the maximum rate plateauing at high substrate concentrations. From the control data, \(V_{\text{max}}\) is estimated to be around \(60\ \mu\text{mol min}^{-1}\) (accept values between 60 and 65).
- The Michaelis constant \(K_m\) is the substrate concentration at which the reaction velocity is half of its maximum (\(\frac{1}{2} V_{\text{max}} = 30\ \mu\text{mol min}^{-1}\)).
- Looking at the control data, a velocity of \(30\) lies between substrate concentrations of \(2\text{ mmol/L}\) (V = 25) and \(5\text{ mmol/L}\) (V = 45). Thus, \(K_m\) is approximately \(2.5\text{ to }3.0\text{ mmol/L}\).

c) Explanation of Non-Competitive Inhibition:
- Non-competitive inhibitors bind to an allosteric site on the enzyme (a site other than the active site).
- This binding causes a conformational change in the enzyme's three-dimensional structure, which alters the shape of the active site.
- Consequently, the enzyme's catalytic efficiency is severely reduced or completely halted, regardless of whether the substrate binds.
- Therefore, the maximum velocity (\(V_{\text{max}}\)) is decreased because the population of active/functional enzyme molecules is reduced.
- However, the affinity of the remaining unaffected enzyme molecules for the substrate remains unchanged, which is why the Michaelis constant (\(K_m\)) remains unchanged.

d) Effect of extremely high substrate concentration:
- The rate of reaction will reach the maximum velocity (\(V_{\text{max}}\)) of the uninhibited enzyme, which is approximately \(60\ \mu\text{mol min}^{-1}\). This is because the competitive inhibitor (A) is completely outcompeted for binding to the active site by the abundance of substrate molecules.

a) Award [2.75] marks:
- [1] for identifying Inhibitor A as competitive.
- [1] for pointing out that at high [S] (20 mmol/L), the rate of reaction (58) approaches control levels (60).
- [0.75] for explaining that competitive inhibition is overcome by high substrate concentrations.

b) Award [2] marks:
- [1] for estimating \(V_{\text{max}}\) in range \(60\text{ to }65\ \mu\text{mol min}^{-1}\) (must include units).
- [1] for determining \(K_m\) in range \(2.5\text{ to }3.0\text{ mmol/L}\) (must include units).

c) Award up to [3] marks:
- [1] for explaining that non-competitive inhibitors bind to the allosteric site, altering active site conformation.
- [1] for explaining that \(V_{\text{max}}\) decreases because the catalytic speed of the enzymes is reduced (or enzymes are inactivated).
- [1] for explaining that \(K_m\) remains unchanged because substrate affinity of the unmodified enzymes is unaffected.

d) Award [1] mark for stating that the rate will reach the control maximum velocity (approx. \(60\ \mu\text{mol min}^{-1}\)).

(a) The initial rapid growth phase under abundant resource conditions is the exponential (or log) phase. (b) As the population density increases and approaches carrying capacity, resources such as space, water, and soil minerals become depleted. This triggers intense intraspecific competition. As a result, individuals have less energy for reproduction (lowering the birth rate) and may suffer from starvation or physiological stress (increasing the death rate), leading to a plateau. (c) Biotic limiting factors include herbivorous insects or disease transmission. Abiotic limiting factors include sunlight intensity, soil mineral concentration, or rainfall.

(a) 1 mark: Identify exponential or log phase. (b) 3 marks: 1 mark for noting resource depletion/scarcity; 1 mark for citing increased intraspecific competition; 1 mark for explaining that birth rate decreases and/or death rate increases until they equilibrate. (c) 1.85 marks: 0.9 marks for any valid biotic factor (e.g., predation, disease, herbivory, interspecific competition); 0.95 marks for any valid abiotic factor (e.g., temperature, water, nutrients, pH, light).

(a) The null hypothesis (\(H_0\)) states that the distribution of Species A is independent of the distribution of Species B (there is no association between them). (b) Since the calculated chi-squared value of 4.12 is greater than the critical value of 3.84, we reject the null hypothesis at the 5% significance level. This indicates there is a statistically significant association between the two plant species. (c) Limitations of quadrat sampling for animals include: 1. Animals are mobile, which can lead to overcounting if they move between quadrats during sampling, or undercounting if they leave the area. 2. Many animals exhibit avoidance behaviors (fleeing) or are highly camouflaged, making direct visual detection within a quadrat unreliable.

(a) 1 mark: State that there is no association or that the species are independent. (b) 2.85 marks: 1 mark for stating that the calculated value exceeds the critical value (4.12 > 3.84); 1 mark for stating the null hypothesis is rejected; 0.85 marks for concluding a statistically significant association. (c) 3 marks: 1.5 marks for explaining how mobility causes counting errors (double counting or exclusion); 1.5 marks for explaining how active behavior/avoidance causes observation bias.

(a) Transfer efficiency from Phytoplankton to Zooplankton: \((12,000 / 100,000) \times 100 = 12.0\%\). Transfer efficiency from Zooplankton to Planktivorous Fish: \((1,100 / 12,000) \times 100 = 9.17\%\) (accept 9.2%). (b) Energy is lost between trophic levels due to: 1. Cell respiration, which releases energy as metabolic heat that cannot be recycled. 2. Not all parts of the organism are consumed (e.g., bones, shells, roots). 3. Some consumed material is indigestible and is lost as feces (egestion). (c) A pyramid of energy must consist of stacked horizontal bars, with the lowest bar representing producers and subsequent bars representing consumers in ascending order. The width/area of each bar must be directly proportional to the energy content of that level.

(a) 2 marks: 1 mark for 12.0% and 1 mark for 9.17% (or 9.2%). (b) 3 marks: 1 mark for heat loss via cell respiration; 1 mark for unconsumed biomass; 1 mark for egested waste/feces. (c) 1.85 marks: 0.95 marks for describing the stacked, proportional stepped bars (not a simple triangle); 0.9 marks for indicating that the producer bar is at the base and is much wider than the upper consumer bars.

(a) Resting ventilation = \(12\text{ breaths min}^{-1} \times 0.5\text{ L} = 6.0\text{ L min}^{-1}\). Vigorous exercise ventilation = \(35\text{ breaths min}^{-1} \times 2.2\text{ L} = 77.0\text{ L min}^{-1}\). (b) During exercise, active muscles undergo higher rates of aerobic respiration, releasing carbon dioxide (\(\text{CO}_2\)) into the blood. This reacts with water to form carbonic acid, which dissociates and lowers the blood pH. This decrease in pH is detected by chemoreceptors in the carotid and aortic bodies, as well as in the medulla oblongata. These chemoreceptors send nerve signals to the ventilation center of the medulla oblongata, which responds by sending impulses to the diaphragm and intercostal muscles, increasing the rate and depth of ventilation. (c) During inhalation: 1. The diaphragm contracts and moves downwards (flattens), increasing the thoracic volume vertically. 2. The external intercostal muscles contract, pulling the ribcage upwards and outwards, increasing the lateral volume of the thoracic cavity.

(a) 2 marks: 1 mark for resting value (6.0 L min^-1) and 1 mark for vigorous exercise value (77.0 L min^-1). (b) 3 marks: 1 mark for linking exercise to increased CO2 production / decreased blood pH; 1 mark for identifying chemoreceptors detecting the pH/CO2 change; 1 mark for describing the brain signaling the respiratory muscles to increase ventilation rate/depth. (c) 1.85 marks: 0.9 marks for describing diaphragm contraction/flattening; 0.95 marks for describing external intercostal muscle contraction to move the ribcage up and out.

(a) Gas exchange occurs by passive diffusion down partial pressure gradients. Since the \(pO_2\) in the alveoli (104 mmHg) is higher than in the incoming deoxygenated capillary blood (40 mmHg), oxygen diffuses rapidly across the alveolar-capillary membrane into the blood. Conversely, the \(pCO_2\) in the deoxygenated blood (46 mmHg) is higher than in the alveolar air (40 mmHg), causing carbon dioxide to diffuse out of the blood and into the alveoli to be exhaled. (b) Alveolar adaptations: 1. Extremely thin walls (made of a single layer of Type I pneumocytes) to minimize the diffusion distance. 2. Surrounding capillary network is highly dense, maintaining a steep concentration gradient by carrying oxygenated blood away rapidly. 3. There are millions of alveoli in the lungs, creating a massive collective surface area for diffusion. (c) Type II pneumocytes secrete surfactant, a phospholipid-protein mixture that reduces surface tension. This prevents the moist walls of the alveoli from adhering to one another, preventing alveolar collapse during expiration and making it easier for the lungs to inflate.

(a) 2.85 marks: 1 mark for stating that diffusion occurs down partial pressure gradients; 0.9 marks for explaining oxygen movement from alveoli to blood (104 to 40 mmHg); 0.95 marks for explaining carbon dioxide movement from blood to alveoli (46 to 40 mmHg). (b) 3 marks: 1 mark for each of three clearly described adaptations (e.g., thin walls / one-cell thick, rich capillary network, large surface area, moist lining). (c) 1 mark for stating that surfactant reduces surface tension to prevent alveolar collapse or aid in expansion.

(a) Competitive inhibitors have a structure similar to the substrate and bind directly to the enzyme's active site, preventing substrate binding. They can be overcome by high substrate concentrations, so the maximum rate of reaction (\(V_{\max}\)) remains unchanged. Non-competitive inhibitors bind to an allosteric site (away from the active site), which alters the conformation of the active site so the substrate can no longer bind or be processed. They cannot be overcome by high substrate concentrations, resulting in a reduced \(V_{\max}\). (b) Inhibitor Y is the non-competitive inhibitor. At high substrate concentrations, the active site is saturated, which overcomes competitive inhibition (Inhibitor X) because the substrate outcompetes the inhibitor. However, because Inhibitor Y binds to an allosteric site, it disables the enzyme molecules regardless of substrate concentration, meaning \(V_{\max}\) cannot be reached. (c) End-product inhibition (or feedback inhibition) occurs when the final product of a multi-step metabolic pathway acts as a non-competitive inhibitor for the first enzyme in the pathway. When the end-product accumulates in excess, it binds to the allosteric site of the initial enzyme, halting the entire pathway and preventing unnecessary overproduction of metabolites.

(a) 3 marks: 1 mark for stating competitive binds to active site vs non-competitive to allosteric site; 1 mark for stating competitive does not alter Vmax vs non-competitive decreases Vmax; 1 mark for stating competitive can be overcome by high substrate concentration while non-competitive cannot. (b) 1.85 marks: 0.9 marks for identifying Inhibitor Y as non-competitive; 0.95 marks for explaining that its inhibition cannot be overcome by high substrate concentration because it binds to a different site, permanently reducing the number of active enzymes. (c) 2 marks: 1 mark for stating that the final/end product of the pathway binds to the first enzyme; 1 mark for stating this binding is allosteric/inhibitory, stopping the pathway when product concentration is high (negative feedback).

(a) 1. The mutation is a single base substitution in the HBB gene. 2. On the DNA template strand, the triplet CTC is mutated to CAC (on the coding/sense strand, GAG becomes GTG). 3. During transcription, this results in the mRNA codon GUG instead of the normal GAG codon. 4. During translation, this codon change leads to the insertion of the non-polar amino acid valine instead of the polar amino acid glutamic acid at the sixth position of the beta-globin polypeptide chain. (b) CRISPR-Cas9 gene-editing process: 1. A single guide RNA (sgRNA) is synthesized with a sequence complementary to the target region of the mutated beta-globin gene. 2. The Cas9 protein (an endonuclease) complexed with this guide RNA is introduced into target cells (e.g., hematopoietic stem cells). 3. The guide RNA directs the Cas9 enzyme to the matching target sequence, where Cas9 creates a double-strand DNA break. 4. A normal donor DNA template is co-delivered, enabling the cell's homology-directed repair mechanism to repair the break by replacing the mutated sequence with the correct sequence.

(a) 3.85 marks: 1 mark for identifying the mutation as a single base substitution; 1 mark for stating the DNA sequence change (GAG to GTG on sense/coding strand OR CTC to CAC on template strand); 0.95 marks for stating the mRNA codon change (GAG to GUG); 0.9 marks for stating the amino acid substitution (glutamic acid to valine). (b) 3 marks: 1 mark for describing the role of guide RNA in targeting the specific genomic locus via complementary base pairing; 1 mark for describing Cas9 making a double-stranded cut; 1 mark for explaining how a donor DNA template allows the cell's repair mechanism to correct the sequence.

This question tests the understanding of gas exchange adaptations, lung ventilation mechanisms, and physiological adaptation to challenging environmental conditions (high altitude). To achieve full marks, a candidate must structure their answer systematically: first addressing anatomical features of the alveoli, then the physical pressure changes driven by antagonistic muscle action during inspiration, and finally the transition from immediate hyperventilation to endocrine-driven acclimation (erythropoietin release and erythropoiesis) at high altitudes.

Part (a) [4 marks maximum]: Award 1 mark for each of the following points, up to 4 marks. - Thin wall: single layer of flattened Type I pneumocytes reducing diffusion distance. - Large surface area: high density of spherical alveoli. - Capillary network: dense surrounding blood vessels to maintain a steep concentration gradient. - Surfactant/Moisture: Type II pneumocytes secrete surfactant to dissolve gases and prevent alveolar collapse. Part (b) [5 marks maximum]: Award 1 mark for each of the following points, up to 5 marks. - Diaphragm contracts and flattens. - External intercostal muscles contract, pulling the ribcage up and out. - Internal intercostals relax (antagonistic action). - Thoracic volume increases. - Air pressure inside the lungs decreases (below atmospheric pressure). - Air flows down the pressure gradient into the lungs. Part (c) [6 marks maximum]: Award 1 mark for each of the following points, up to 6 marks. - Lower atmospheric pressure reduces the partial pressure of oxygen (\(pO_2\)). - Decreased diffusion gradient leads to lower hemoglobin saturation. - Short-term: ventilation rate and heart rate increase. - Mid/long-term: Kidneys detect low oxygen and secrete erythropoietin (EPO). - EPO stimulates red bone marrow to produce more red blood cells/hemoglobin. - Increased capillary density in muscles/tissues or increased myoglobin levels. - Shift in the oxygen dissociation curve to facilitate oxygen release at tissues. Clarity/Construction [1 mark]: Award 1 mark if the entire response is well-structured, coherent, and uses precise biological vocabulary correctly.

This question tests critical ecological concepts from Populations and Communities. Candidates must describe the classic S-shaped growth curve, distinguishing between its phases. Next, they must contrast density-dependent and density-independent regulatory factors, providing clear examples. Finally, they need to connect interspecific interactions (mutualism and competition) to the broader context of community structure, explaining how competitive exclusion leads to either local extinction or niche partitioning, which ultimately determines community biodiversity.

Part (a) [4 marks maximum]: Award 1 mark for each of the following points, up to 4 marks. - Exponential phase: abundant resources, rapid growth, birth rate far exceeds death rate. - Transitional phase: limiting factors begin to emerge, competition increases, growth rate slows. - Plateau phase: population stabilizes around carrying capacity (\(K\)). - Dynamic equilibrium: at plateau, birth rate + immigration equals death rate + emigration. Part (b) [5 marks maximum]: Award 1 mark for each of the following points, up to 5 marks. - Definition of density-dependent: effect increases as population density increases. - Examples of density-dependent: disease, food competition, territoriality, predation. - Definition of density-independent: effect is constant regardless of population density. - Examples of density-independent: forest fires, severe freezes, volcanic eruptions. - Regulation: density-dependent factors regulate population near carrying capacity, whereas density-independent factors cause sudden drops. Part (c) [6 marks maximum]: Award 1 mark for each of the following points, up to 6 marks. - Mutualism definition: symbiotic relationship where both species benefit. - Impact of mutualism: enhances survival, expands realized niches, and increases overall community biodiversity. - Competitive exclusion principle: two species cannot share the exact same niche indefinitely if resources are limiting. - Outcome 1: extinction/exclusion of the less competitive species. - Outcome 2: resource partitioning / niche differentiation (natural selection drives differences in resource use). - Community structure: these interactions restrict or expand species' realized niches, determining species richness and abundance. Clarity/Construction [1 mark]: Award 1 mark if the entire response is well-structured, coherent, and uses precise biological vocabulary correctly.