2024 IB DP Biology 模擬試題連答案詳解

During a fight-or-flight response, the sympathetic nervous system is activated, releasing noradrenaline (norepinephrine) to increase the heart rate. Simultaneously, the endocrine system responds as the adrenal medulla secretes epinephrine (adrenaline) into the bloodstream, which also acts to increase heart rate, preparing the body for action.

[1 mark] B is correct. Sympathetic activity and adrenal secretions both increase heart rate. A is incorrect because parasympathetic action decreases heart rate. C is incorrect because cardiac muscle is under autonomic, not somatic, control. D is incorrect as baroreceptors detect high blood pressure and trigger a parasympathetic response to decrease heart rate.

As carbon dioxide levels rise, it reacts with water to form carbonic acid, which dissociates into hydrogen ions and bicarbonate ions, lowering blood pH. Central chemoreceptors in the medulla oblongata and peripheral chemoreceptors in the carotid and aortic bodies detect this decrease in pH. They send nerve impulses to the ventilation center in the medulla oblongata, which coordinates an increase in the rate and depth of ventilation to expel carbon dioxide.

[1 mark] C is correct. Chemoreceptors detect the drop in pH due to carbonic acid and signal the medulla's respiratory center. A is incorrect because CO2 does not directly stimulate bronchiolar stretch receptors. B is incorrect because blood pH decreases (not rises) and the cerebellum coordinates movement, not ventilation. D is incorrect as the pituitary gland does not store or release oxygen.

Cyclins are proteins whose concentrations fluctuate throughout the cell cycle. They bind to and activate specific cyclin-dependent kinases (CDKs). Once activated, these CDKs phosphorylate other target proteins that trigger events specific to a phase of the cell cycle, such as DNA replication or spindle formation.

[1 mark] A is correct. Cyclins activate CDKs, which phosphorylate target proteins. B is incorrect because cyclins are regulatory proteins, not replication enzymes. C is incorrect because cyclins fluctuate while CDKs are generally constant in concentration (but inactive without cyclins). D is incorrect because spindle fibers are made of microtubules, not synthesized directly by cyclins.

All cells, whether prokaryotic or eukaryotic, are enclosed by a plasma membrane that regulates the transport of materials entering and exiting the cell. Eukaryotes have 80S ribosomes and membrane-bound organelles, while prokaryotes have 70S ribosomes and lack membrane-bound organelles. Prokaryotes have a nucleoid region, whereas eukaryotes have a membrane-bound nucleus.

[1 mark] B is correct. Both cell types share a plasma membrane. A is incorrect because prokaryotes have 70S ribosomes. C is incorrect because prokaryotes lack membrane-bound organelles. D is incorrect because eukaryotes have a nucleus rather than a nucleoid region.

Energy transfer is inefficient (typically around 10%) because much of the chemical energy ingested by organisms is used in cellular respiration and lost to the environment as heat. Additionally, some parts of organisms are not eaten (e.g., bones, roots) or are indigestible and excreted as feces, meaning this energy is not transferred to the next trophic level.

[1 mark] B is correct. Energy is lost as heat, unconsumed material, and feces. A is incorrect because decomposers break down dead organic matter, not sunlight. C and D are biologically incorrect/nonsensical.

A competitive inhibitor binds to the active site of the enzyme, competing directly with the substrate. Because the substrate can outcompete the inhibitor at very high concentrations, the maximum rate of reaction (\(V_{\text{max}}\)) can still be achieved. However, more substrate is required to achieve this, which increases the concentration needed to reach half of the \(V_{\text{max}}\) (raising the apparent \(K_m\)).

[1 mark] C is correct. Competitive inhibitors do not change \(V_{\text{max}}\) but require a higher substrate concentration to reach half-maximal velocity. A is incorrect because non-competitive inhibitors decrease \(V_{\text{max}}\). B is incorrect because irreversible inhibitors permanently reduce the active enzyme pool, reducing \(V_{\text{max}}\). D is incorrect because inhibitors decrease, not increase, enzymatic efficiency.

The sodium-potassium pump is an active transport mechanism (an integral membrane protein) that hydrolyzes one ATP molecule to pump 3 sodium ions (\(\text{Na}^+\)) out of the cell and 2 potassium ions (\(\text{K}^+\)) into the cell, both against their concentration gradients, to maintain the resting potential of the neuron.

[1 mark] B is correct. The pump actively transports 3 \(\text{Na}^+\) out and 2 \(\text{K}^+\) in, utilizing ATP. A is incorrect because it is active transport, not passive diffusion. C is incorrect because the directions of the ions are reversed. D is incorrect because it is active transport, not facilitated diffusion.

The water potential of the plant cell is calculated using the formula: \(\Psi_w = \Psi_s + \Psi_p = -0.6\text{ MPa} + 0.2\text{ MPa} = -0.4\text{ MPa}\). The surrounding solution has a water potential of \(-0.8\text{ MPa}\). Water moves from an area of higher water potential (\(-0.4\text{ MPa}\)) to an area of lower water potential (\(-0.8\text{ MPa}\)). Thus, water moves out of the cell. In plant cells, this loss of water causes the plasma membrane to pull away from the cell wall, a process known as plasmolysis. Plant cells do not burst from water loss.

[1 mark] C is correct. Water potential of the cell is \(-0.4\text{ MPa}\), which is higher than the solution's \(-0.8\text{ MPa}\), so water flows out, causing plasmolysis. A is incorrect because water moves out. B is incorrect because there is a water potential gradient. D is incorrect because plant cells do not burst when losing water, and their rigid cell walls prevent bursting during water uptake anyway.

Epinephrine (adrenaline) is a hormone secreted during stressful situations, including exercise. It increases cardiac output (heart rate and stroke volume) to ensure adequate oxygen delivery to active skeletal muscles, and it stimulates glycogenolysis (the breakdown of stored glycogen into glucose) in the liver to increase blood glucose levels for cellular respiration.

Award [1] for the correct answer (B). Correctly identifies that epinephrine increases cardiac output and increases liver glycogenolysis.

During intensive exercise, blood flow is redirected via vasoconstriction and vasodilation. Arterioles supplying non-essential organs (such as the digestive tract/small intestine, kidneys, and liver) constrict, decreasing blood flow to these regions. Concurrently, arterioles supplying skeletal muscles dilate, significantly increasing blood flow to meet the elevated metabolic demands of contracting muscles.

Award [1] for the correct answer (B). Recognizes that blood flow is shunted away from the digestive tract and directed toward skeletal muscles during exercise.

The mitotic index is calculated as the number of cells undergoing mitosis divided by the total number of cells observed.

1. Number of mitotic cells = \( 24 + 12 + 8 + 6 = 50 \) cells.
2. Total number of cells = \( 50 \text{ (mitotic)} + 150 \text{ (interphase)} = 200 \) cells.
3. Mitotic Index = \( \frac{50}{200} = 0.25 \).

Award [1] for the correct answer (A). Correctly calculates the mitotic index as 0.25 (or 25%).

Cyclins are regulatory proteins that control the progression of cells through the cell cycle. They function by binding to specific cyclin-dependent kinases (CDKs), forming active complexes. These active CDKs then phosphorylate target proteins that initiate key events in the cell cycle, such as DNA replication or spindle formation.

Award [1] for the correct answer (B). Correctly identifies the molecular mechanism of cyclin-dependent kinase activation and phosphorylation.

Pili are hair-like surface appendages on prokaryotic cells that allow them to attach to host cells or other surfaces, and are also used in conjugation (transfer of genetic material). Prokaryotes contain 70S ribosomes (not 80S), have a non-membrane-bound circular chromosome in the nucleoid region, and possess a cell wall made of peptidoglycan, not cellulose.

Award [1] for the correct answer (B). Identifies the correct function and presence of pili in prokaryotes.

The resolution of a microscope is limited by the wavelength of the radiation used to image the specimen. Electron beams have a much shorter wavelength than visible light, allowing electron microscopes to resolve much smaller, closely spaced structures.

Award [1] for the correct answer (A). Correctly attributes the higher resolution limit to the shorter wavelength of electron beams compared to light.

We must calculate the energy transfer sequentially through the trophic levels:
1. Trophic level 1: Primary producers = \( 24,000 \text{ kJ m}^{-2}\text{ yr}^{-1} \)
2. Trophic level 2: Primary consumers = \( 24,000 \times 0.10 = 2,400 \text{ kJ m}^{-2}\text{ yr}^{-1} \)
3. Trophic level 3: Secondary consumers = \( 2,400 \times 0.10 = 240 \text{ kJ m}^{-2}\text{ yr}^{-1} \)
4. Trophic level 4: Tertiary consumers = \( 240 \times 0.10 = 24 \text{ kJ m}^{-2}\text{ yr}^{-1} \).

Award [1] for the correct answer (C). Correctly determines the energy value at the fourth trophic level (tertiary consumer) using 10% efficiency transfers.

Cellular respiration by decomposers (as well as plants and animals) breaks down organic molecules using oxygen, releasing carbon dioxide gas (\( \text{CO}_2 \)) directly back into the atmosphere. Photosynthesis removes carbon from the atmosphere, while fossilization and lithification lock carbon into the geosphere over geological time scales.

Award [1] for the correct answer (B). Identifies respiration as a carbon source returning carbon to the atmosphere.

The resolution limit of a standard light microscope is about 200 nm. Structure size of ribosomes (approx. 25-30 nm) falls well below this limit, making them impossible to resolve with visible light. Mitochondria (approx. 1-10 micrometers), nuclei (approx. 5-10 micrometers), and chloroplasts (approx. 2-10 micrometers) are large enough to be resolved with a light microscope.

Award 1 mark for the correct answer: B. Reject other options because they can all be seen using a light microscope.

Increased carbon dioxide leads to a decrease in blood pH due to the formation of carbonic acid. Chemoreceptors detect this drop and send nerve impulses to the cardiovascular centre in the medulla oblongata. The medulla oblongata then increases heart rate by sending signals via the sympathetic nerve to the sinoatrial node.

Award 1 mark for the correct answer: A. Option B is incorrect because baroreceptors detect pressure, not carbon dioxide. Option C is incorrect because blood pH decreases and the control center is the medulla, not the cerebrum. Option D is incorrect because hormonal detection is not the primary neural pathway for this feedback loop.

Carbohydrates are digested and absorbed as glucose into the capillaries of the villi, which empty into the hepatic portal vein, raising its glucose concentration. High blood glucose levels trigger beta cells in the pancreas to secrete insulin, while alpha cells decrease the secretion of glucagon.

Award 1 mark for the correct answer: B. Reject options showing decreasing glucose levels or incorrect hormone secretion patterns (insulin decreases or glucagon increases after a meal).

During metaphase of mitosis, the DNA has already replicated, so each of the 12 chromosomes consists of two sister chromatids, resulting in 24 chromatids. Mitosis produces genetically identical daughter cells with the same diploid chromosome number, so each daughter cell will contain 12 chromosomes after telophase is complete.

Award 1 mark for the correct answer: B. Reject any answer with 6 chromosomes because mitosis does not reduce the chromosome number, and reject answers with 12 chromatids because metaphase chromosomes consist of pairs of sister chromatids.

Primary producers have \(80\,000\text{ kJ m}^{-2}\text{ yr}^{-1}\). Primary consumers (second trophic level) receive \(10\%\) of this energy: \(8\,000\text{ kJ m}^{-2}\text{ yr}^{-1}\). Secondary consumers (third trophic level) receive \(10\%\) of that: \(800\text{ kJ m}^{-2}\text{ yr}^{-1}\). Tertiary consumers (fourth trophic level) receive \(10\%\) of that: \(80\text{ kJ m}^{-2}\text{ yr}^{-1}\).

Award 1 mark for the correct answer: C. Incorrect calculations or misidentifying the trophic level sequence (e.g., considering primary consumers as tertiary) will lead to incorrect options A, B, or D.

Aquaporins are channel proteins that facilitate the passive diffusion of water molecules across a membrane down their concentration gradient (facilitated diffusion/osmosis) without using ATP. Simple diffusion of CO2 does not use a protein. The sodium-potassium pump and mineral ion uptake against a gradient both require active transport using ATP.

Award 1 mark for the correct answer: B. Reject other options because option A and D are active transport processes requiring ATP, and option C is simple diffusion which does not require a membrane protein.

Competitive inhibitors bind reversibly to the active site. Because they compete directly with substrate molecules, raising the substrate concentration to high levels increases the probability of the substrate binding instead of the inhibitor, allowing the reaction rate to reach the same maximum rate (\(V_{\text{max}}\)) as without the inhibitor.

Award 1 mark for the correct answer: C. Reject A and B because they describe non-competitive (allosteric) inhibition. Reject D because inhibitors decrease the rate of reaction.

According to Chargaff's rules of complementary base pairing, in double-stranded DNA, the percentage of cytosine (C) equals the percentage of guanine (G), which is \(34\%\). Together, C and G make up \(68\%\) (\(34\% + 34\%\)) of the bases. The remaining \(32\%\) (\(100\% - 68\%\)) must consist of adenine (A) and thymine (T) in equal proportions. Therefore, the percentage of adenine is \(32\% \div 2 = 16\%\).

Award 1 mark for the correct answer: A. Reject other options due to incorrect calculation or misunderstanding of complementary base-pairing rules.

The resolution limit of a standard light microscope is approximately 200 nm. Structures smaller than this cannot be resolved. Ribosomes are very small organelles (about 20-30 nm) and require an electron microscope (which has a resolution limit of about 0.2 nm) to be visualized. Mitochondria (0.5 to 10 \(\mu\)m), nuclei (approx 10 \(\mu\)m), and chloroplasts (approx 5 \(\mu\)m) are all large enough to be resolved by a light microscope.

Award [1] for the correct answer (A).

The mitotic index is the ratio of cells undergoing mitosis (prophase + metaphase + anaphase + telophase) to the total number of cells.
Number of mitotic cells = 12 + 8 + 5 + 15 = 40.
Total number of cells = 40 (mitotic) + 160 (interphase) = 200.
Mitotic index = 40 / 200 = 0.20.

Award [1] for the correct answer (B).

Glucose is absorbed by active transport and facilitated diffusion across the epithelium of the small intestine villi into the capillaries. These capillaries consolidate into the hepatic portal vein, which carries the nutrient-dense blood directly to the liver for storage and regulation before entering the wider systemic circulation. Therefore, immediately after a meal, the hepatic portal vein has the highest glucose concentration.

Award [1] for the correct answer (B).

Carbon dioxide produced by respiring tissues reacts with water in the blood to form carbonic acid, which dissociates and lowers the blood pH. This decrease in pH is detected by chemoreceptors in the carotid arteries and aorta. These receptors send nerve impulses to the medulla oblongata (cardiovascular center), which in turn sends nerve impulses via the sympathetic nerve to the sinoatrial (SA) node, increasing the rate of depolarization and speeding up the heart rate.

Award [1] for the correct answer (B).

The energy transfer through trophic levels is calculated as follows:
- Trophic Level 1 (Primary Producers): \( 1.2 \times 10^6 \text{ kJ m}^{-2}\text{ yr}^{-1} \)
- Trophic Level 2 (Primary Consumers): \( 1.2 \times 10^5 \text{ kJ m}^{-2}\text{ yr}^{-1} \) (10% of producer energy)
- Trophic Level 3 (Secondary Consumers): \( 1.2 \times 10^4 \text{ kJ m}^{-2}\text{ yr}^{-1} \) (10% of primary consumer energy)
- Trophic Level 4 (Tertiary Consumers): \( 1.2 \times 10^3 \text{ kJ m}^{-2}\text{ yr}^{-1} \) (10% of secondary consumer energy).

Award [1] for the correct answer (C).

Carbon dioxide is released back into the atmosphere via cell respiration (by plants, animals, and decomposers) and combustion of organic biomass (like forest fires or burning fossil fuels). Photosynthesis, on the other hand, absorbs carbon dioxide from the atmosphere.

Award [1] for the correct answer (B).

(a) Percentage increase = \(\frac{\text{Maximum} - \text{Rest}}{\text{Rest}} \times 100 = \frac{45 - 12}{12} \times 100 = \frac{33}{12} \times 100 = 275\%\).

(b) During exercise, higher rates of cellular respiration in muscles release more carbon dioxide into the blood, where it forms carbonic acid and decreases blood pH. This decrease in pH is detected by chemoreceptors in the medulla oblongata, as well as in the carotid and aortic bodies. The cardiovascular center in the medulla oblongata responds by increasing the frequency of nerve impulses sent via the sympathetic (cardiac accelerator) nerve to the sinoatrial (SA) node, resulting in an increased heart rate.

(c) The respiratory and circulatory systems cooperate as follows: Active muscle cells produce carbon dioxide as a waste product, which diffuses down its concentration gradient into the blood capillaries. The heart pumps this blood to the lungs more rapidly due to increased heart rate and stroke volume. At the lungs, the carbon dioxide diffuses across the thin alveolar membrane into the alveoli. The high rate and depth of ventilation (respiratory system) continuously expels this carbon dioxide from the lungs, maintaining a steep concentration gradient for efficient gas exchange.

(a) [1.6 marks]
- Award 1.6 marks for the correct answer: 275%.
- Award 0.8 marks for setting up the calculation correctly but with an arithmetic error.

(b) [2.5 marks]
- Award up to 2.5 marks (0.83 marks per point):
- Cellular respiration during exercise produces carbon dioxide, lowering blood pH.
- Chemoreceptors in the medulla (or carotid/aortic bodies) detect the lower blood pH.
- Medulla oblongata (cardiac center) sends nerve impulses via the sympathetic/cardiac accelerator nerve.
- The neurotransmitter (noradrenaline) is released at the sinoatrial (SA) node, accelerating heart rate.

(c) [2.5 marks]
- Award up to 2.5 marks (0.83 marks per point):
- Carbon dioxide diffuses from active muscle fibers into capillary blood.
- The heart pumps blood containing carbon dioxide to the lungs at a higher rate (increased cardiac output).
- Carbon dioxide diffuses across the alveolar-capillary membrane from the blood into the alveolar air space.
- High ventilation rate removes carbon dioxide from the alveoli, maintaining a steep concentration gradient.

(a) The cephalic phase is an anticipatory phase. Sensory inputs such as the sight, smell, thought, or taste of food stimulate the cerebral cortex and medulla oblongata. The medulla sends parasympathetic nerve impulses down the vagus nerve directly to the stomach's enteric nervous system. This stimulates gastric parietal cells (to secrete acid) and chief cells (to secrete pepsinogen) before food is even swallowed.

(b) When food enters the stomach, its physical presence causes distension, and proteins trigger chemical receptors. These stimuli cause endocrine cells (G-cells) in the gastric antrum to secrete the hormone gastrin into the bloodstream. Gastrin travels back to stimulate parietal cells to produce hydrochloric acid (HCl). As the acid accumulates, the gastric pH drops. Once pH falls below 1.5 to 2.0, negative feedback is initiated: endocrine cells secrete hormones such as somatostatin, which directly inhibit gastrin release from G-cells, reducing acid secretion.

(c) The autonomic nervous system regulates the gut antagonistically: the parasympathetic nervous system (via the vagus nerve) stimulates digestive activity by accelerating gut motility (peristalsis) and increasing secretions; the sympathetic nervous system inhibits digestive activity by reducing motility, closing sphincters, and restricting blood flow to the digestive organs.

(a) [2.0 marks]
- Award 1.0 mark per point up to a maximum of 2.0 marks:
- Sensory stimuli (smell, sight, or taste of food) send signals to the medulla oblongata.
- The brain sends efferent impulses via the vagus nerve (parasympathetic system) to gastric gland cells, stimulating early secretion.

(b) [2.6 marks]
- Award up to 2.6 marks (0.87 marks per point):
- Peptides/distension in the stomach stimulate G-cells to release gastrin into the blood.
- Gastrin stimulates parietal cells to produce hydrochloric acid (HCl).
- Extremely low pH (highly acidic conditions) triggers the release of somatostatin / secretin.
- This hormone acts via negative feedback to inhibit gastrin release, decreasing acid production.

(c) [2.0 marks]
- Award 1.0 mark per point up to a maximum of 2.0 marks:
- Parasympathetic nervous system stimulates motility and secretions.
- Sympathetic nervous system inhibits motility and secretions (diverting resources away from digestion).

(a) Mitotic index ratio = \(0.24 / 0.08 = 3\). This indicates that Agent X-treated cells have a three-times higher proportion of cells locked in or undergoing mitosis than control cells.

(b) Spindle microtubules attach to centromeres of chromosomes during metaphase to pull sister chromatids apart in anaphase. If Agent X acts as a microtubule stabilizer (like taxol) or destabilizer (like colchicine), it prevents the dynamic remodeling of microtubules. Without correct tension/attachment, the spindle assembly checkpoint prevents progression to anaphase, locking cells in metaphase.

(c) Cyclins are proteins whose concentrations rise and fall during different stages of the cell cycle. They control cycle progression by binding to cyclin-dependent kinases (CDKs), activating them. Activated CDKs then phosphorylate target proteins that initiate key processes: Cyclin D triggers G1 to S phase transition; Cyclin E/A initiates DNA replication; Cyclin B activates mitosis entry.

(a) [1.6 marks]
- Award 0.8 marks for the correct ratio of 3 (or 3:1).
- Award 0.8 marks for stating that this means there is a higher proportion of cells in mitosis / arrested in division.

(b) [2.5 marks]
- Award up to 2.5 marks (0.83 marks per point):
- Microtubules must attach to kinetochores at the centromere of each chromosome.
- Microtubules must undergo dynamic assembly/disassembly (shortening) to align and eventually separate chromosomes.
- Agent X may prevent microtubule depolymerization / inhibit spindle fiber assembly.
- This failure triggers the spindle assembly checkpoint (M-checkpoint) which blocks progression to anaphase.

(c) [2.5 marks]
- Award up to 2.5 marks (0.83 marks per point):
- Cyclins accumulate and reach threshold levels at specific stages of the cell cycle.
- Cyclins bind to and activate Cyclin-Dependent Kinases (CDKs).
- Activated CDK-cyclin complexes phosphorylate target proteins that drive cell cycle events (e.g., DNA replication, nuclear envelope breakdown).
- Different cyclins (e.g., Cyclins D, E, A, B) regulate different phase transitions.

(a) Convert the measured image size to micrometres:
\(48\text{ mm} \times 1,000 = 48,000\text{ }\mu\text{m}\).
Using the formula \(\text{Actual Size} = \frac{\text{Image Size}}{\text{Magnification}}\):
\(\text{Actual Size} = \frac{48,000\text{ }\mu\text{m}}{8,000} = 6\text{ }\mu\text{m}\).

(b) Features supporting endosymbiosis include: presence of a double membrane (inner from original prokaryote, outer from host vesicle); circular DNA plasmid-like genome independent of nuclear DNA; 70S ribosomes (characteristic of prokaryotes, not eukaryotes).

(c) Compartmentalization isolates biochemical reactions within membrane-bound organelles. This enhances efficiency by:
1. Concentrating reactants/enzymes in a small volume, increasing the frequency of collisions.
2. Protecting the rest of the cell from self-digestion (e.g., keeping hydrolytic enzymes sequestered inside lysosomes).
3. Maintaining optimal microenvironments (such as the acidic pH of the lysosome or high proton concentration in the mitochondrial intermembrane space) without altering the neutral cytoplasm.

(a) [1.6 marks]
- Award 0.8 marks for converting 48 mm to 48,000 \(\mu\text{m}\) (or setting up equation: \(48 / 8,000 = 0.006\text{ mm}\)).
- Award 0.8 marks for the final correct answer of 6 \(\mu\text{m}\) with appropriate units.

(b) [2.0 marks]
- Award 1.0 mark per valid feature listed up to a maximum of 2.0 marks:
- Presence of circular / naked DNA.
- Presence of 70S ribosomes.
- Double membrane envelope.
- Replicates independently via binary fission-like division.

(c) [3.0 marks]
- Award 1.0 mark per valid explanation up to a maximum of 3.0 marks:
- Concentration of enzymes/substrates: limits space and speeds up reaction rates.
- Isolation of toxic/harmful substances: protects cell components (e.g., lysosome enzymes).
- Maintenance of localized chemical environments: such as maintaining a low pH for lysosomal enzymes, or proton gradients in the mitochondria for ATP synthesis.

(a) Percentage efficiency of energy transfer = \(\frac{\text{Energy in secondary consumers}}{\text{Energy in primary consumers}} \times 100\)
\(\text{Efficiency} = \frac{96\text{ kJ m}^{-2}\text{ yr}^{-1}}{1,200\text{ kJ m}^{-2}\text{ yr}^{-1}} \times 100 = 8\%\).

(b) Only a small percentage of energy is transferred because:
1. Organisms perform cellular respiration, converting chemical energy to kinetic energy/heat, which dissipates into the environment and cannot be reused.
2. Not all organic matter is ingested by consumers (e.g., bones, woody tissue, roots are left behind).
3. Some ingested material is indigestible (e.g., cellulose) and is lost in feces (egestion) or excreted as metabolic waste (e.g., urea).

(c) Energy enters ecosystems from an external source (sunlight), flows through food webs, is converted to heat, and eventually radiates into space. Thus, energy cannot be recycled and requires a continuous input. In contrast, chemical nutrients (nitrogen, phosphorus, carbon) exist in a finite pool and are cycled indefinitely between biotic (living organisms) and abiotic (soil, air, water) components via decomposers.

(a) [1.6 marks]
- Award 1.6 marks for the correct calculation showing 8%.
- Award 0.8 marks if the calculation is set up correctly (96 / 1200) but has a calculation error.

(b) [2.5 marks]
- Award up to 2.5 marks (0.83 marks per point):
- Energy is lost as heat during metabolic processes/cellular respiration.
- Not all parts of organisms are consumed by the next trophic level.
- Some material is indigestible and lost through feces/egestion.
- Some material is lost through excretion (urine/urea).

(c) [2.5 marks]
- Award up to 2.5 marks (0.83 marks per point):
- Energy enters from the sun, flows in a linear/one-way path, and is lost as heat (does not cycle).
- Nutrients cycle repeatedly between organic and inorganic forms (finite pool).
- Decomposers recycle nutrients back to autotrophs, but they do not recycle energy.

Part a: During physical exercise, active tissue respiration releases high amounts of carbon dioxide into the blood. This carbon dioxide reacts with water to form carbonic acid, which dissociates into hydrogen ions and hydrogencarbonate ions, causing blood pH to drop. Chemoreceptors located in the medulla oblongata, as well as in the carotid and aortic bodies, detect this decrease in pH. In response, the respiratory center of the medulla oblongata increases the frequency of nerve impulses sent via the phrenic and intercostal nerves to the diaphragm and external intercostal muscles, increasing ventilation rate and tidal volume. Concurrently, the cardiovascular center in the medulla oblongata sends sympathetic impulses via the cardiac nerve to the sinoatrial node to increase heart rate, accelerating the delivery of carbon dioxide to the lungs for excretion and restoring normal blood pH. Part b: Blood glucose regulation is an integrated feedback process. The pancreas acts as both a sensor and an endocrine gland; its islets of Langerhans contain alpha and beta cells. After a meal, high blood glucose triggers beta cells to secrete insulin, which travels via the blood to stimulate hepatocytes and muscle cells to absorb glucose and convert it to stored glycogen through glycogenesis. Conversely, when blood glucose levels fall (such as during fasting or prolonged exercise), alpha cells secrete glucagon, which signals hepatocytes to undergo glycogenolysis to break down glycogen into glucose and release it into the bloodstream. Superimposed on this endocrine loop, the sympathetic nervous system can stimulate the adrenal glands to release adrenaline during acute stress or intense exercise, which rapidly activates glycogenolysis in both liver and muscle cells to ensure a sufficient supply of circulating glucose. Part c: During vigorous physical activity, working skeletal muscles require rapid ATP production via aerobic respiration. To sustain this, the respiratory system increases gas exchange at the alveoli by increasing breathing rate and depth. Oxygenated blood is transported by the cardiovascular system, which increases cardiac output through elevated stroke volume and heart rate. Local metabolic factors (such as high carbon dioxide, heat, and lactic acid) alongside sympathetic nervous signals cause selective vasodilation of arterioles leading to working skeletal muscles, while arterioles supplying non-essential organs (like the digestive tract and kidneys) undergo vasoconstriction to shunt blood toward active muscles. Within these muscles, the increased temperature and acidity decrease hemoglobin's affinity for oxygen (the Bohr shift), maximizing oxygen unloading at the capillaries. Additionally, the mechanical contraction of skeletal muscles compresses local veins, operating as a skeletal muscle pump to increase venous return to the heart, which sustains high cardiac output.

Part a (Maximum [4] marks): - Active cellular respiration produces carbon dioxide, which dissolves to form carbonic acid, lowering blood pH. - Chemoreceptors in the medulla oblongata (or carotid/aortic bodies) detect the decrease in pH. - The respiratory center of the medulla oblongata sends nerve impulses to the diaphragm and intercostal muscles to increase ventilation rate/depth. - The cardiovascular center of the medulla oblongata sends sympathetic impulses (via the cardiac nerve) to the sinoatrial node to increase heart rate. - This dual response accelerates carbon dioxide elimination at the lungs to restore blood pH homeostasis. Part b (Maximum [5] marks): - The pancreas acts as the primary receptor and controller for blood glucose levels. - High blood glucose stimulates beta cells to release insulin, which promotes glucose uptake and glycogenesis (conversion of glucose to glycogen in liver/muscle cells). - Low blood glucose stimulates alpha cells to release glucagon, which promotes glycogenolysis (breakdown of glycogen to glucose in the liver). - The autonomic nervous system integrates with this hormonal control; sympathetic stimulation triggers adrenaline release from the adrenal medulla. - Adrenaline rapidly accelerates glycogen breakdown to increase circulating glucose levels for immediate metabolic demands. - This feedback system maintains glucose levels within narrow physiological limits. Part c (Maximum [6] marks): - Active skeletal muscles contract rapidly, requiring elevated rates of cellular respiration and ATP. - The respiratory system increases ventilation rate and depth to maximize oxygen absorption and carbon dioxide removal. - The circulatory system increases cardiac output by increasing both heart rate and stroke volume. - Arterioles supplying skeletal muscles undergo vasodilation, while arterioles supplying non-essential organs (e.g., gut, kidneys) undergo vasoconstriction (blood shunting). - Higher temperature and lower pH in active muscles cause the Bohr shift, reducing hemoglobin's affinity for oxygen and promoting oxygen release to muscle cells. - Rhythmic skeletal muscle contractions compress deep veins (skeletal muscle pump), facilitating venous return to the heart. Clarity (Maximum [1] mark): - Award [1] mark if the response is written in a clear, logical, and structured manner, using appropriate biological terminology, and clearly addresses all three parts of the question.

(a) 15 divisions of stage micrometer = 15 * 10 \(\mu\)m = 150 \(\mu\)m. Since 150 \(\mu\)m = 60 eyepiece divisions, 1 eyepiece division = 150 / 60 = 2.5 \(\mu\)m. (b) 3.5 eyepiece divisions * 2.5 \(\mu\)m/division = 8.75 \(\mu\)m. (c) Changing objective magnification alters the size of the specimen image projected onto the graticule, while the scale of the graticule itself does not change. Therefore, each division represents a larger real distance at lower magnification.

(a) [2 marks] 1 mark for showing correct working (e.g., 150 / 60 or 15 * 10 / 60), 1 mark for correct value (2.5 \(\mu\)m). (b) [1 mark] 1 mark for correct calculation of actual size (8.75 \(\mu\)m or 8.8 \(\mu\)m). (c) [2 marks] 1 mark for stating that magnification of the image changes but eyepiece scale remains fixed, 1 mark for stating that each graticule division represents a larger actual distance at lower magnification.

(a) KOH absorbs \(\text{CO}_2\) so that change in volume is solely due to oxygen uptake. (b) Radius r = diameter / 2 = 0.5 mm. Height h = 48 mm. Volume in 10 mins = \(3.14 * (0.5)^2 * 48 = 3.14 * 0.25 * 48 = 37.68\) \(\text{mm}^3\). Volume per minute = 37.68 / 10 = 3.768 \(\text{mm}^3\text{ min}^{-1}\) (or 3.77). (c) Enzyme-catalyzed metabolic reactions of respiration speed up at higher temperatures due to increased kinetic energy of molecules, leading to more frequent collisions.

(a) [1 mark] Absorbs carbon dioxide gas. (b) [2 marks] 1 mark for correct substitution with radius 0.5 mm (e.g., \(3.14 * 0.25 * 48 = 37.68\)), 1 mark for correct rate of 3.77 or 3.8 \(\text{mm}^3\text{ min}^{-1}\) (accept 3.768). (c) [2 marks] 1 mark for stating respiration rate is higher at 30 degrees Celsius because it is enzyme-controlled, 1 mark for stating that higher temperature increases molecular kinetic energy and collision rates.

(a) Mitotic index = (cells in mitosis / total cells) * 100. Control total = 120 + 40 = 160. Control mitotic index = (40 / 160) * 100 = 25%. Treated total = 150 + 10 = 160. Treated mitotic index = (10 / 160) * 100 = 6.25%. (b) Cut the actively dividing terminal 1-2 mm of root tips, place them in hydrochloric acid to hydrolyze cell walls/middle lamella, rinse and stain (e.g., aceto-orcein) to color chromosomes, squash gently under a coverslip. (c) Wear personal protective equipment (gloves/goggles) when handling concentrated acids or mutagens, and use caution with sharp blades.

(a) [2 marks] 1 mark for Control = 25%, 1 mark for Treated = 6.25%. (b) [2 marks] 1 mark for maceration/softening with acid (or heating), 1 mark for staining and squashing under a coverslip. (c) [1 mark] Correct safety precaution (e.g., wear safety goggles or gloves to protect against acidic solutions/stains, or cut away from body with scalpel).

Gastric juice secretion is highly regulated to ensure digestion occurs efficiently:
1. **Cephalic Phase (Nervous control):** The sight, smell, or taste of food triggers impulses in the brain. These travel down the vagus nerve (parasympathetic system) to the stomach, stimulating gastric glands to secrete acid and pepsinogen before food arrives.
2. **Gastric Phase (Nervous & Hormonal control):** As food enters the stomach, physical distension (stretch receptors) and the presence of proteins/peptides trigger the release of the hormone gastrin from G-cells in the stomach mucosa. Gastrin travels in the blood to stimulate parietal cells to secrete hydrochloric acid (HCl).
3. **Intestinal Phase (Hormonal control):** As acidic chyme enters the duodenum, hormones such as secretin and cholecystokinin (CCK) are released, which inhibit gastric acid secretion and slow down stomach emptying.

Award [1] mark for each of the following points up to a maximum of [4]:
* cephalic phase involves sensory stimuli (sight/smell) causing impulses via the vagus nerve to stimulate gastric juice secretion;
* gastric phase involves stomach distension / peptides triggering the release of the hormone gastrin from G-cells;
* gastrin travels through the blood to stimulate parietal cells to produce hydrochloric acid (HCl);
* intestinal phase involves acidic chyme entering the duodenum, triggering the release of inhibitory hormones (e.g., secretin/somatostatin/CCK) that reduce gastric juice secretion.

Erythrocyte recycling is a vital function of the liver:
1. Aged or damaged red blood cells (erythrocytes) are recognized and engulfed via phagocytosis by Kupffer cells, which are specialized macrophages located within the liver sinusoids.
2. Inside the Kupffer cells, hemoglobin is broken down into globin proteins and a heme group.
3. The globin chains are hydrolyzed into free amino acids, which are released into the blood and recycled for protein synthesis.
4. The heme group is further disassembled: the iron is removed and either stored in the liver (as ferritin) or sent to the bone marrow to produce new hemoglobin, while the remaining organic component is converted to bilirubin (bile pigment) and excreted into bile.

Award [1] mark for each of the following points up to a maximum of [4]:
* Kupffer cells (macrophages in liver sinusoids) phagocytose aged/damaged erythrocytes;
* hemoglobin is broken down into globin and heme;
* globin is hydrolyzed into amino acids which are reused for protein synthesis;
* iron is removed from heme and stored in the liver (as ferritin) / transported to bone marrow;
* the remaining heme component is converted into bilirubin (bile pigment) and secreted into bile.

An ECG records the electrical signals of the heart, which directly precede and coordinate the mechanical events of muscle contraction (systole) and relaxation (diastole):
1. **P wave (Atrial Depolarization):** This electrical signal spreads from the SA node across the atria, causing them to depolarize. This electrical event triggers the mechanical contraction of the atria (atrial systole), pumping blood into the ventricles.
2. **QRS complex (Ventricular Depolarization):** This large electrical spike represents the rapid depolarization of the ventricles. This electrical wave triggers the mechanical contraction of the ventricles (ventricular systole), causing the AV valves to close and pumping blood into the aorta and pulmonary artery.
3. **Atrial Repolarization:** Although atrial relaxation (diastole) occurs during the QRS interval, its electrical signal is masked by the much larger QRS complex.
4. **T wave (Ventricular Repolarization):** This wave represents the repolarization of the ventricles as they prepare for the next cycle. This leads to the mechanical relaxation of the ventricles (ventricular diastole).

Award [1] mark for each of the following points up to a maximum of [4]:
* P wave represents atrial depolarization, which triggers atrial contraction (atrial systole);
* QRS complex represents ventricular depolarization, which triggers ventricular contraction (ventricular systole);
* during the QRS complex, atrial repolarization/relaxation (atrial diastole) also occurs (but is hidden);
* T wave represents ventricular repolarization, which leads to ventricular relaxation (ventricular diastole).

During exercise, muscles undergo high rates of aerobic respiration, releasing carbon dioxide (\(\text{CO}_2\)).
1. The increased \(\text{CO}_2\) reacts with water to form carbonic acid (catalyzed by carbonic anhydrase), which dissociates into hydrogen ions (\(\text{H}^+\)) and hydrogen carbonate, lowering the pH of the blood/tissue fluid.
2. A lower pH and higher partial pressure of \(\text{CO}_2\) (\(p\text{CO}_2\)) alter the conformation of hemoglobin, decreasing its affinity for oxygen. This phenomenon is known as the **Bohr shift**.
3. On a graph, this causes the sigmoidal oxygen dissociation curve of hemoglobin to shift to the **right**.
4. The physiological benefit is that at any given partial pressure of oxygen (\(p\text{O}_2\)), hemoglobin releases/unloads oxygen more readily, ensuring that rapidly respiring muscles receive more oxygen to sustain aerobic respiration.

Award [1] mark for each of the following points up to a maximum of [4]:
* high carbon dioxide concentration lowers the pH of the blood/tissue environment (due to the production of carbonic acid/hydrogen ions);
* low pH/high carbon dioxide reduces hemoglobin's affinity for oxygen;
* this results in a shift of the oxygen-hemoglobin dissociation curve to the right (known as the Bohr shift);
* physiological benefit: hemoglobin releases/unloads oxygen more easily to the metabolically active/respiring tissues where it is needed.

When the body is dehydrated:
1. **Detection:** Osmoreceptors in the hypothalamus detect an increase in the solute concentration (osmolarity) of the blood (or a decrease in water potential).
2. **Hormone release:** This triggers the posterior pituitary gland to secrete antidiuretic hormone (ADH) into the bloodstream.
3. **Target tissue action:** ADH travels to the kidneys and binds to specific receptors on the cells of the collecting ducts.
4. **Mechanism:** This binding initiates a cellular cascade that causes vesicles containing water channels (aquaporins) to fuse with the apical membrane of the collecting duct cells, greatly increasing their permeability to water.
5. **Reabsorption:** Consequently, water is reabsorbed by osmosis out of the collecting duct and into the highly concentrated (hypertonic) interstitial fluid of the renal medulla, returning to the bloodstream and producing a low volume of highly concentrated urine.

Award [1] mark for each of the following points up to a maximum of [4]:
* osmoreceptors in the hypothalamus detect high blood osmolarity/solute concentration (or low water potential);
* the posterior pituitary gland releases ADH into the bloodstream;
* ADH binds to receptors on the collecting duct cells, causing the insertion of aquaporins (water channels) into their membranes;
* this increases the permeability of the collecting duct to water, allowing more water to be reabsorbed by osmosis (into the hypertonic medulla), resulting in concentrated/low volume urine.