2023 IB DP Biology 模擬試題連答案詳解

Water molecules are polar and form intermolecular hydrogen bonds with one another. These bonds require a substantial amount of thermal energy to break, giving water a high latent heat of vaporization. In contrast, methane is non-polar and only experiences weak dispersion forces between its molecules, meaning it evaporates and boils at much lower temperatures.

[1 mark] C is correct. Award 1 mark for the correct option. Reject options A, B, and D as they contain incorrect physical or chemical claims about water and methane.

A competitive inhibitor competes with the substrate for the active site of the enzyme. At low substrate concentrations, the inhibitor reduces the reaction rate. However, because the binding is reversible, sufficiently high concentrations of substrate will outcompete the inhibitor, allowing the enzyme to reach its normal maximum rate (\(V_{\max}\)).

[1 mark] B is correct. Award 1 mark for the correct option. Reject other options because non-competitive inhibitors reduce the overall \(V_{\max}\) regardless of substrate concentration, and competitive inhibitors bind to the active site rather than allosteric sites.

The point where the line of best fit crosses the x-axis represents a 0% change in mass. This indicates that the potato tissue and the surrounding solution are in osmotic equilibrium (isotonic). Therefore, there is no net movement of water into or out of the cells, and the osmolarity of the potato tissue is equivalent to that of the \(0.35\text{ mol dm}^{-3}\) sucrose solution.

[1 mark] C is correct. Award 1 mark for the correct option. Reject other options because a mass change of zero corresponds directly to isotonicity, meaning the tissue osmolarity matches the external solution.

The genotype of the white-eyed female is \(X^r X^r\) and the genotype of the red-eyed male is \(X^R Y\). The female offspring will receive one \(X^r\) allele from the mother and one \(X^R\) allele from the father, resulting in the genotype \(X^R X^r\) (red eyes). The male offspring will receive one \(X^r\) allele from the mother and a \(Y\) chromosome from the father, resulting in the genotype \(X^r Y\) (white eyes). Thus, all females will have red eyes and all males will have white eyes.

[1 mark] B is correct. Award 1 mark for the correct option. Reject A, C, and D because they fail to correctly apply the rules of sex-linked inheritance for this cross.

In glycolysis, \(NAD^+\) is reduced to \(NADH\) while converting glucose to pyruvate. Under anaerobic conditions, yeast cells must regenerate \(NAD^+\) so that glycolysis can continue to produce ATP. This is achieved during ethanol fermentation, where pyruvate is reduced to ethanol, oxidizing \(NADH\) back to \(NAD^+\).

[1 mark] B is correct. Award 1 mark for the correct option. Reject A because mitochondria and the electron transport chain are not active in anaerobic respiration. Reject C and D as they are biochemically incorrect.

Greenhouse gases (such as carbon dioxide, water vapor, and methane) allow incoming short-wave solar radiation to pass through the atmosphere and heat the Earth's surface. The surface then re-emits this energy as longer-wave infrared radiation (heat). Greenhouse gases absorb this long-wave radiation and re-radiate it in all directions, trapping heat in the atmosphere.

[1 mark] C is correct. Award 1 mark for the correct option. Reject A and D because greenhouse gases interact primarily with outgoing long-wave radiation rather than incoming short-wave radiation. Reject B because it describes the mechanism of ozone depletion, not the greenhouse effect.

During exercise, cellular respiration increases, releasing more carbon dioxide (\(CO_2\)) into the blood. This decreases blood pH, which is detected by chemoreceptors. The chemoreceptors send nerve signals to the cardiovascular center in the medulla oblongata, which then sends impulses along the sympathetic nerve to the sinoatrial (SA) node to increase heart rate.

[1 mark] C is correct. Award 1 mark for the correct option. Reject A because parasympathetic nerves slow the heart rate down. Reject B because adrenaline is produced by the adrenal glands, not the pituitary. Reject D because pressure-based mechanisms do not directly bypass the brainstem to control the SA node in this manner.

Natural selection operates on pre-existing genetic variation. Within a bacterial population, random mutations occasionally produce alleles that confer resistance to a specific antibiotic. When the population is exposed to the antibiotic, susceptible bacteria are killed, while the resistant variants survive and reproduce, passing on their resistance genes. Over time, the frequency of resistance alleles increases.

[1 mark] C is correct. Award 1 mark for the correct option. Reject A because mutations occur randomly and are not induced adaptively by the presence of the antibiotic. Reject B and D because they contain scientifically inaccurate explanations of natural selection.

Water has a high specific heat capacity, meaning it requires a relatively large amount of heat energy to raise its temperature by one degree Celsius. This property allows large bodies of water to maintain relatively stable temperatures, buffering aquatic organisms from extreme fluctuations in air temperature.

Award 1 mark for the correct choice (B).
- Reject A: Latent heat of vaporization relates to cooling via evaporation (e.g., transpiration or sweating).
- Reject C: Cohesion is important for transport in xylem but doesn't directly regulate habitat temperature.
- Reject D: The lower density of ice allows aquatic life to survive under frozen surfaces, but specific heat capacity is what buffers general thermal stability against daily/seasonal air changes.

The surrounding solution in the beaker (0.5 M) is hypertonic relative to the cytoplasm of the artificial cell (0.2 M). Water will move out of the cell via osmosis, down its concentration gradient (from a region of higher water potential to lower water potential). This net movement of water out of the cell causes its volume to decrease, causing it to shrink.

Award 1 mark for the correct choice (B).
- Reject A: Water would move in if the cell were in a hypotonic solution, but here the environment is hypertonic.
- Reject C: Water movement out increases, rather than decreases, the internal sucrose concentration.
- Reject D: Sucrose is stated to be membrane-impermeable, meaning no sucrose diffusion occurs.

Competitive inhibitors bind to the active site and can be overcome by high substrate concentrations, so \(V_{\text{max}}\) remains unchanged, but they increase the apparent \(K_m\) (reducing substrate affinity). Non-competitive inhibitors bind to an allosteric site, decreasing the amount of active enzyme and lowering \(V_{\text{max}}\), but because they do not interfere with substrate binding to the active site, the \(K_m\) remains unchanged.

Award 1 mark for the correct choice (B).
- Reject A, C, D: These combinations incorrectly state the effects of competitive or non-competitive inhibitors on \(V_{\text{max}}\) and \(K_m\).

A test cross involves crossing a heterozygous individual (RrHh) with a homozygous recessive individual (rrhh). Because the genes are on different chromosomes, they assort independently. The gametes produced by the heterozygote are RH, Rh, rH, and rh in equal proportions (25% each). The homozygous recessive parent only produces rh gametes. Thus, the offspring will have the genotypes RrHh, Rrhh, rrHh, and rrhh in equal proportions, yielding a phenotypic ratio of 1:1:1:1.

Award 1 mark for the correct choice (B).
- Reject A: 9:3:3:1 is the expected phenotypic ratio for a dihybrid self-cross (RrHh x RrHh).
- Reject C: 3:1 is the expected phenotypic ratio for a monohybrid cross.
- Reject D: 9:7 is an epistatic ratio, not expected for simple independent assortment.

Energy transfer through trophic levels is calculated as follows:
- Trophic Level 1 (Primary Producers): \(10,000 \text{ kJ m}^{-2} \text{ yr}^{-1}\)
- Trophic Level 2 (Primary Consumers): \(10,000 \times 0.10 = 1,000 \text{ kJ m}^{-2} \text{ yr}^{-1}\)
- Trophic Level 3 (Secondary Consumers): \(1,000 \times 0.10 = 100 \text{ kJ m}^{-2} \text{ yr}^{-1}\)
- Trophic Level 4 (Tertiary Consumers): \(100 \times 0.10 = 10 \text{ kJ m}^{-2} \text{ yr}^{-1}\)

Award 1 mark for the correct choice (C).
- Reject A: This is the energy for primary consumers.
- Reject B: This is the energy for secondary consumers.
- Reject D: This is the energy for quaternary consumers (if present).

Carbon dioxide, water vapor, methane, and nitrous oxide are all atmospheric gases that absorb and re-emit infrared radiation (heat), acting as greenhouse gases. Nitrogen, oxygen, and argon do not absorb infrared radiation and are not greenhouse gases.

Award 1 mark for the correct choice (B).
- Reject A: Nitrogen, oxygen, and argon are not greenhouse gases.
- Reject C: Nitrogen and helium are not greenhouse gases.
- Reject D: Oxygen and sulfur dioxide are not greenhouse gases (sulfur dioxide is an aerosol precursor that has a cooling effect, not a direct greenhouse gas like the others).

Pepsin is produced as pepsinogen by the chief cells in the gastric glands of the stomach mucosa and active pepsin digests proteins into shorter polypeptides. Amylase acts on starch (not maltose). Lipase is secreted by the pancreas (bile emulsifies lipids but is not an enzyme). Trypsin is produced by the pancreas (not the small intestine mucosa) and activated in the duodenum.

Award 1 mark for the correct choice (B).
- Reject A: Amylase digests starch to maltose; it does not digest maltose (maltase does).
- Reject C: Lipase is produced by the pancreas, not the liver.
- Reject D: Trypsin is produced by the pancreas, not the small intestine mucosa.

A fall in blood glucose concentration is detected by the islets of Langerhans in the pancreas. In response, alpha (\(\alpha\)) cells secrete the hormone glucagon. Glucagon travels in the blood to the liver, where it binds to specific receptors and stimulates the breakdown of stored glycogen into glucose (glycogenolysis), which is then released into the blood to raise glucose levels.

Award 1 mark for the correct choice (B).
- Reject A: Insulin is secreted by beta cells when blood glucose levels are high, stimulating glycogenesis.
- Reject C: Beta cells secrete insulin, not glucagon.
- Reject D: Alpha cells secrete glucagon, not insulin.

The sodium-potassium pump (\( \text{Na}^+/\text{K}^+ \) ATPase) actively transports 3 sodium ions (\( \text{Na}^+ \)) out of the cell and 2 potassium ions (\( \text{K}^+ \)) into the cell per cycle. This process requires energy in the form of ATP hydrolysis to change the conformation of the transport protein against the concentration gradients of both ions.

Award 1 mark for selecting the correct option (A). [1 mark]

Competitive inhibitors bind to the active site and can be overcome by high substrate concentrations, so \( V_{\text{max}} \) remains unchanged but \( K_m \) increases (lower affinity). Non-competitive inhibitors bind to an allosteric site, decreasing the amount of active enzyme, which reduces \( V_{\text{max}} \) but does not affect the binding affinity of the active enzymes, leaving \( K_m \) unchanged.

Award 1 mark for selecting the correct option (B). [1 mark]

The woman must be heterozygous (\( I^A i \)) because her father had blood group O (\( ii \)). The man has the genotype \( I^A I^B \). Crossing \( I^A i \times I^A I^B \) results in the following possible offspring genotypes: \( I^A I^A \) (group A), \( I^A I^B \) (group AB), \( I^A i \) (group A), and \( I^B i \) (group B). There is a 1 in 4 (25%) chance of producing a child with blood group B.

Award 1 mark for selecting the correct option (B). [1 mark]

Glycolysis consists of an energy investment phase and an energy payoff phase. It consumes 2 ATP molecules and produces 4 ATP molecules, resulting in a net yield of 2 ATP. It also reduces 2 molecules of \( \text{NAD}^+ \) to produce 2 molecules of NADH.

Award 1 mark for selecting the correct option (A). [1 mark]

Increased atmospheric \( \text{CO}_2 \) dissolves in oceans to produce carbonic acid, which dissociates and increases the concentration of hydrogen ions (lowering the pH). These hydrogen ions react with carbonate ions (\( \text{CO}_3^{2-} \)) to form bicarbonate (\( \text{HCO}_3^- \)), thereby reducing the availability of carbonate ions needed by marine organisms to build their calcium carbonate (\( \text{CaCO}_3 \)) skeletons.

Award 1 mark for selecting the correct option (C). [1 mark]

The electrical impulse is initiated at the pacemaker (sinoatrial or SA node) and travels through the atria to the atrioventricular (AV) node. The AV node delays the signal, which then travels down the septum via the Bundle of His to the Purkinje fibers in the ventricle walls, causing ventricular contraction.

Award 1 mark for selecting the correct option (A). [1 mark]

Cyclins are regulatory proteins that control the cell cycle by binding to and activating enzymes called cyclin-dependent kinases (CDKs). Once activated, these kinases phosphorylate specific target proteins that trigger key events in the cell cycle.

Award 1 mark for selecting the correct option (B). [1 mark]

Energy transfer efficiency between trophic levels is typically about 10%. Using this rule: Primary Producers (Autotrophs) = \( 12,000\text{ kJ m}^{-2}\text{ yr}^{-1} \); Primary Consumers (Herbivores) = \( 1,200\text{ kJ m}^{-2}\text{ yr}^{-1} \); Secondary Consumers (Carnivores) = \( 120\text{ kJ m}^{-2}\text{ yr}^{-1} \); Tertiary Consumers (Top Carnivores) = \( 12\text{ kJ m}^{-2}\text{ yr}^{-1} \).

Award 1 mark for selecting the correct option (D). [1 mark]

Bacteria are prokaryotes characterized by a peptidoglycan cell wall and 70S ribosomes, and they lack membrane-bound organelles such as a nucleus. Archaea lack peptidoglycan in their cell walls. Plant and fungal cells are eukaryotic, possessing 80S ribosomes and a nuclear envelope.

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

Carbon dioxide reacts with water to form carbonic acid, which dissociates into hydrogen ions and hydrogencarbonate ions, lowering the pH of the blood. Chemoreceptors detect this decrease in pH (acidification) and transmit nerve impulses to the medulla oblongata. The respiratory center in the medulla oblongata then increases the rate and depth of ventilation to remove excess carbon dioxide.

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

Competitive inhibitors compete with the substrate for the active site of the enzyme. At very high substrate concentrations, the substrate molecules outnumber the inhibitor molecules, allowing the reaction to reach the same maximum velocity (\(V_{max}\)). However, a higher concentration of substrate is required to reach half \(V_{max}\), meaning the apparent \(K_m\) is increased.

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

The high latent heat of vaporization means that a substantial amount of heat energy is absorbed to break hydrogen bonds and evaporate water. This makes sweating or panting an exceptionally effective cooling mechanism, as evaporation of relatively small volumes of water removes a large quantity of body heat.

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

Using the Hardy-Weinberg equation, \(p^2 + 2pq + q^2 = 1\), where \(q^2\) is the frequency of the homozygous recessive genotype. Given \(q^2 = 1/10000 = 0.0001\), we find \(q = 0.01\). Since \(p + q = 1\), \(p = 1 - 0.01 = 0.99\). The frequency of carriers (heterozygotes) is \(2pq = 2 \times 0.99 \times 0.01 = 0.0198\).

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

The sodium-potassium pump actively transports three sodium ions (\(Na^+\)) out of the cell and two potassium ions (\(K^+\)) into the cell per cycle of ATP hydrolysis. Both ions are moved against their respective concentration gradients, which requires energy in the form of ATP.

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

Part (a):
- Compare: Both organs are supplied with oxygenated blood by the systemic arterial system and show changes regulated by the autonomic nervous system during exercise.
- Contrast: Blood flow to skeletal muscle increases dramatically during exercise (due to vasodilation of arterioles) to meet elevated oxygen demand. Conversely, blood flow to the kidneys decreases (due to vasoconstriction of renal arterioles) to divert blood to working muscles.

Part (b):
- At the onset of exercise, the cardiovascular control center in the medulla oblongata increases sympathetic nervous output and decreases parasympathetic output.
- Sympathetic fibers release noradrenaline, which increases the rate of depolarisation of the sinoatrial (SA) node and increases cardiac muscle contractility.
- The adrenal medulla releases adrenaline (epinephrine) into the blood, which circulates to the heart, binding to beta-receptors to further increase both heart rate and stroke volume (thereby increasing cardiac output).

Part (c):
- Resting Stroke Volume (SV) calculation:
\(\text{SV}_{\text{rest}} = \frac{\text{Cardiac Output}}{\text{Heart Rate}} = \frac{5.0\text{ L min}^{-1}}{50\text{ bpm}} = 0.1\text{ L} = 100\text{ mL}\)
- Maximal Stroke Volume (SV) calculation:
\(\text{SV}_{\text{max}} = \frac{24.0\text{ L min}^{-1}}{180\text{ bpm}} = 0.1333\text{ L} = 133.3\text{ mL}\)
- Percentage Increase calculation:
\(\text{Percentage increase} = \frac{133.3 - 100}{100} \times 100\% = 33.3\%\) (or \(33\%\))

Part (a): [Max 3 marks]
- Both receive oxygenated blood / are part of systemic circulation. [1 mark]
- Blood flow to skeletal muscle increases while blood flow to kidneys decreases during exercise. [1 mark]
- Vasodilation of arterioles occurs in skeletal muscle, while vasoconstriction of arterioles occurs in the kidneys. [1 mark]
- Muscle blood flow increases to meet high metabolic/oxygen demand, whereas renal blood flow is temporarily reduced to prioritize active muscles. [1 mark]

Part (b): [Max 3 marks]
- Sympathetic nervous system releases noradrenaline to increase heart rate / ventricular contraction strength. [1 mark]
- Parasympathetic system activity decreases / vagus nerve inhibition is reduced. [1 mark]
- Endocrine system secretes adrenaline (epinephrine) from the adrenal medulla. [1 mark]
- Adrenaline acts on the sinoatrial node (SA node) to further increase heart rate. [1 mark]

Part (c): [2.5 marks]
- Correct resting stroke volume: \(100\text{ mL}\) (or \(0.1\text{ L}\)). [0.5 marks]
- Correct maximal stroke volume: \(133.3\text{ mL}\) (or \(0.1333\text{ L}\)). [0.5 marks]
- Correct percentage increase formula or working: \(\frac{133.3 - 100}{100} \times 100\%\). [1 mark]
- Correct final answer: \(33.3\%\) or \(33\%\) (must include % sign). [0.5 marks]

Part (a):
- Compound D (the end-product) acts as an allosteric inhibitor. It binds to the allosteric site of the first enzyme, E1, which is structurally distinct from the active site.
- Binding of D causes a conformational (shape) change in E1's active site, preventing substrate A from binding to it.
- This acts as a negative feedback loop; when concentrations of D are high, the pathway is inhibited, preventing unnecessary accumulation of metabolic intermediates and conserving energy.

Part (b):
- Competitive inhibitors bind to the active site of the enzyme, blocking the substrate. They increase \(K_{\text{m}}\) (reducing affinity) but do not affect the maximum velocity (\(V_{\max}\)) because high substrate concentrations can outcompete the inhibitor.
- Non-competitive inhibitors bind to an allosteric site, altering the enzyme's structure. This decreases the overall rate of reaction (\(V_{\max}\) is reduced) but does not alter substrate binding affinity at uninhibited active sites (\(K_{\text{m}}\) remains unchanged).

Part (c):
- Deduction: D acts as a non-competitive inhibitor.
- Reason: Competitive inhibition can be fully overcome by very high substrate concentrations, reaching the normal \(V_{\max}\). Since the rate remains severely restricted (at 20% of uninhibited rate) despite extremely high substrate concentration, the substrate is unable to displace the inhibitor, which is characteristic of non-competitive inhibition.
- Effect on \(V_{\max}\): \(V_{\max}\) is significantly decreased/reduced (to 20% of the normal value).

Part (a): [Max 3 marks]
- End-product D binds to the allosteric site of the first enzyme E1. [1 mark]
- This causes a conformational/structural change in the active site of E1. [1 mark]
- Substrate A can no longer bind to E1 / enzyme-substrate complex cannot form. [1 mark]
- This prevents overproduction of the end-product/conserves energy/represents a negative feedback loop. [1 mark]

Part (b): [Max 3 marks]
- Competitive inhibitors bind to the active site; non-competitive inhibitors bind to an allosteric site. [1 mark]
- Competitive inhibition increases \(K_{\text{m}}\); non-competitive inhibition does not change \(K_{\text{m}}\). [1 mark]
- Competitive inhibition does not affect \(V_{\max}\); non-competitive inhibition decreases \(V_{\max}\). [1 mark]

Part (c): [2.5 marks]
- Deduction: Non-competitive (inhibitor/inhibition). [1 mark]
- Reason: High substrate concentration cannot overcome the inhibition / the reaction rate remains low (20%), indicating that substrate and inhibitor do not compete for the same site. [1 mark]
- Effect: \(V_{\max}\) is decreased / reduced to 20%. [0.5 marks]

Part (a):
- Definition: Net primary productivity (NPP) is the energy or organic matter accumulated by autotrophs (producers) that is stored as biomass and available to the next trophic level, after subtracting the energy lost to autotrophic respiration (\(R_{\text{a}}\)).
- Calculation:
\(\text{NPP} = \text{GPP} - R_{\text{a}} = 24,000 - 14,000 = 10,000\)
- Unit: \(\text{kJ m}^{-2} \text{ yr}^{-1}\)

Part (b):
- Percentage of NPP ingested:
\(\text{Percentage Ingested} = \frac{\text{Herbivore Ingestion}}{\text{NPP}} \times 100\% = \frac{2,000}{10,000} \times 100\% = 20\%\)
- Secondary Productivity calculation (the rate at which consumers convert organic chemical energy into their own biomass):
\(\text{Secondary Productivity} = \text{Ingestion} - \text{Fecal/Urinary Losses} - \text{Respiration}\)
\(\text{Secondary Productivity} = 2,000 - 700 - 900 = 400\text{ kJ m}^{-2} \text{ yr}^{-1}\)

Part (c):
- Pyramids of energy must always be upright because they depict energy flow over a defined period. The Second Law of Thermodynamics dictates that energy transformations are never 100% efficient, and energy is lost as heat at each trophic level (with only about 10% transferred to the next level).
- Pyramids of biomass represent the standing crop of dry organic matter at a single point in time. In aquatic systems, primary producers (phytoplankton) reproduce extremely rapidly and have a very high turnover rate. Even though their standing biomass at any one moment is small, their high productivity is sufficient to support a larger standing biomass of long-lived primary consumers (zooplankton/fish).

Part (a): [2.5 marks]
- Definition of NPP: Gross primary productivity minus plant respiration / organic material available to consumers. [1 mark]
- Calculation: \(10,000\) (accept \(24,000 - 14,000\)). [1 mark]
- Correct units: \(\text{kJ m}^{-2} \text{ yr}^{-1}\) (or \(\text{kJ/m}^{2}/\text{yr}\)). [0.5 marks]

Part (b): [3 marks]
- Percentage ingested calculation: \(\frac{2,000}{10,000} \times 100\%\) leading to \(20\%\). [1.5 marks: 1 mark for correct setting up/working, 0.5 marks for correct final answer with %]
- Secondary productivity calculation: \(2,000 - 700 - 900\) leading to \(400\text{ kJ m}^{-2} \text{ yr}^{-1}\). [1.5 marks: 1 mark for correct subtraction, 0.5 marks for correct final answer with correct units]

Part (c): [Max 3 marks]
- Pyramids of energy show energy flow over time / are restricted by the Second Law of Thermodynamics. [1 mark]
- Energy is lost as heat (respiration) at each trophic level / only about 10% transfers up, so energy can never increase. [1 mark]
- Pyramids of biomass represent standing crop at a single instant in time. [1 mark]
- Phytoplankton (aquatic producers) have high turnover rates / consume/reproduce rapidly, allowing low standing biomass to sustain higher consumer biomass. [1 mark]

Part (a):
- Deduction: Eukaryotic organism.
- Reason 1: The presence of a double-membrane-bound region containing highly folded DNA represents a true nucleus (prokaryotic cells have DNA floating freely in the nucleoid region with no enclosing membrane).
- Reason 2: The presence of 80S ribosomes, which are characteristic of eukaryotic cells (prokaryotic cells only contain smaller, 70S ribosomes).
- Reason 3: The presence of an internal, folded membrane system indicates compartmentalization/organelles, which are absent or extremely rudimentary in prokaryotes.

Part (b):
- Compartmentalization allows substrates and enzymes to be concentrated in specific areas, which greatly increases chemical reaction rates.
- It isolates damaging reactions, such as hydrolytic processes in lysosomes, preventing degradation of other cellular structures.
- It allows optimal local conditions (such as pH or ion concentrations) to be maintained for specific enzymatic pathways, independent of the surrounding cytoplasm.
- It makes the transport and movement of substances within the cell more efficient via vesicle-based targeting.

Part (c):
- Plant cell walls are composed of cellulose, a polymer of \(\beta\)-glucose monomers.
- Fungal cell walls are composed of chitin, a polymer of modified glucose monomers containing nitrogen (N-acetylglucosamine).
- Bacterial cell walls are composed of peptidoglycan (murein), which consists of repeating disaccharide units cross-linked by short amino acid chains (peptides).

Part (a): [3 marks]
- Deduction: Eukaryotic. [1 mark, must be correct to award reasons]
- Reasons (any two of the following for 1 mark each): [2 marks]
- Presence of a double-membrane-bound region / nucleus containing DNA.
- Presence of 80S ribosomes.
- Presence of internal membrane systems / compartmentalized organelles.

Part (b): [Max 3 marks]
- Increases rate of metabolic reactions by concentrating reactants/enzymes. [1 mark]
- Protects the cell from self-digestion / harmful materials (e.g. lysosomes). [1 mark]
- Maintains optimum pH/conditions for specific enzymes inside different compartments. [1 mark]
- Facilitates organized transport of substances via vesicles within the cytoplasm. [1 mark]

Part (c): [2.5 marks]
- Plant cell walls: Cellulose / \(\beta\)-glucose monomers. [1 mark]
- Fungal cell walls: Chitin / N-acetylglucosamine monomers. [1 mark]
- Bacterial cell walls: Peptidoglycan / amino acid-sugar complex. [0.5 marks]

**(a) Enzyme Action and Activation Energy**
Enzymes are biological catalysts that increase the rate of chemical reactions without being consumed in the process. They do this by lowering the activation energy (\(E_a\)), which is the minimum energy required for reactant molecules to collide with enough force and correct orientation to react. When a substrate binds to the specific active site of an enzyme, it forms an enzyme-substrate complex. This binding induces structural stress, stretching or weakening the chemical bonds within the substrate. Consequently, less thermal energy is required to reach the unstable transition state, speeding up the reaction. The net energy change of the reaction (\(\Delta G\)) remains unchanged.

**(b) Competitive vs. Non-Competitive Inhibition**
Enzyme inhibitors prevent substrates from binding or reacting. They differ significantly in their mechanism:
- **Structural Similarity:** Competitive inhibitors are structurally similar to the natural substrate, whereas non-competitive inhibitors have a completely different structure.
- **Binding Site:** Competitive inhibitors bind directly to the active site, blockading it. Non-competitive inhibitors bind to an alternative site, known as the allosteric site.
- **Mechanism of Inhibition:** Competitive inhibitors physically block substrate access. Non-competitive inhibitors alter the tertiary structure of the enzyme when they bind to the allosteric site, causing the active site to change shape and lose its functionality.
- **Effect of Substrate Concentration:** The effects of a competitive inhibitor can be overcome by significantly increasing the substrate concentration, allowing the reaction to reach its original maximum velocity (\(V_{max}\)). The effects of a non-competitive inhibitor cannot be overcome by increasing substrate concentration; the maximum rate of reaction (\(V_{max}\)) is permanently reduced.
- **Examples:** Malonate acts as a competitive inhibitor of succinate dehydrogenase, while cyanide binds non-competitively to cytochrome oxidase.

**(c) End-Product Inhibition and the Threonine-Isoleucine Pathway**
Metabolic pathways consist of chains or cycles of enzyme-catalyzed steps. To prevent the wasteful overproduction of substances, cells utilize end-product inhibition, a form of negative feedback control. In this process, the final product of a pathway acts as an allosteric inhibitor of the first enzyme in that same pathway.

In the conversion of the amino acid threonine to isoleucine:
1. The pathway consists of five distinct enzymatic steps.
2. The initial reactant is threonine, and the first enzyme in the pathway is threonine deaminase.
3. As threonine deaminase works, it converts threonine into intermediate products, which are sequentially modified by subsequent enzymes until the end-product, isoleucine, is produced.
4. When isoleucine levels build up and exceed the cell's immediate requirements, the excess isoleucine molecules bind to the allosteric site of threonine deaminase.
5. This allosteric binding alters the conformation of threonine deaminase's active site, preventing further threonine molecules from binding.
6. Consequently, the pathway is temporarily deactivated, halting the production of intermediates and the end-product.
7. When the cell consumes the existing pool of isoleucine, its concentration drops, causing isoleucine to detach from the allosteric sites of threonine deaminase.
8. The enzyme's active site returns to its active conformation, allowing the pathway to resume production. This precise feedback mechanism maintains homeostasis and conserves metabolic energy.

**Part (a): Lowering Activation Energy [Max 3 marks]**
* \(\bullet\) Enzymes function as biological catalysts to increase reaction rates. [1]
* \(\bullet\) Substrate binds to the enzyme's active site to form an enzyme-substrate complex. [1]
* \(\bullet\) Binding places physical stress/tension on chemical bonds in the substrate, destabilizing them. [1]
* \(\bullet\) This lowers the transition state energy / activation energy (\(E_a\)). [1]
* \(\bullet\) The overall free energy change of the reaction (\(\Delta G\)) is unaffected. [1]
*(Accept clear, annotated progress diagrams showing energy levels with and without enzymes for up to 2 marks)*

**Part (b): Competitive vs. Non-Competitive Inhibition [Max 5 marks]**
* Award 1 mark per row of clear distinction, up to 5 marks total:
* *Similarity:* Competitive inhibitor is structurally similar to substrate **vs.** non-competitive inhibitor is structurally different. [1]
* *Binding:* Competitive binds to the active site **vs.** non-competitive binds to an allosteric site. [1]
* *Active site change:* Competitive does not alter the active site shape **vs.** non-competitive changes the conformation of the active site. [1]
* *Substrate effect:* Competitive inhibition can be overcome by increasing substrate concentration **vs.** non-competitive inhibition cannot be overcome by adding more substrate. [1]
* *Kinetics:* Competitive allows the reaction to eventually reach normal maximum velocity (\(V_{max}\)) **vs.** non-competitive lowers the maximum velocity (\(V_{max}\)). [1]
* *Example:* Malonate as competitive inhibitor / Cyanide as non-competitive inhibitor (or any other valid specific biological example). [1]
*(Reject general terms like 'inhibitor blocks enzyme' without specifying active/allosteric site)*

**Part (c): End-Product Inhibition & Threonine-Isoleucine Pathway [Max 8 marks]**
* \(\bullet\) Metabolic pathways involve chains or cycles of enzyme-catalyzed reactions. [1]
* \(\bullet\) End-product inhibition is a form of negative feedback control. [1]
* \(\bullet\) The final product acts as an inhibitor of the first enzyme in the metabolic pathway. [1]
* \(\bullet\) The inhibitor binds to an allosteric site (rather than the active site). [1]
* \(\bullet\) This binding changes the shape of the enzyme's active site, preventing substrate binding. [1]
* \(\bullet\) In the specific pathway, threonine is the initial substrate and isoleucine is the final end-product. [1]
* \(\bullet\) The pathway consists of five steps / a series of intermediate steps. [1]
* \(\bullet\) The first enzyme in this pathway is threonine deaminase. [1]
* \(\bullet\) High levels of isoleucine lead to its binding to the allosteric site of threonine deaminase, shutting down the pathway. [1]
* \(\bullet\) As isoleucine concentration drops, fewer allosteric sites are bound, and the pathway reactivates. [1]
* \(\bullet\) This prevents the unnecessary accumulation of intermediates and the wasteful synthesis of excess isoleucine. [1]

a) The dependent variable is the rate of lactase activity / rate of reaction, which is determined by measuring the appearance of products (glucose or galactose) or disappearance of substrate (lactose) over time. Typically, glucose production is measured using a glucose biosensor or colorimetrically using a glucose assay kit over a fixed time interval.
b) Competitive inhibitors bind to the active site of the enzyme, temporarily blocking substrate binding. At low substrate concentrations, the rate of reaction is lower in the presence of the inhibitor. However, at very high substrate concentrations, the substrate outcompetes the inhibitor for the active site, meaning that the maximum rate of reaction (\(V_{\max}\)) remains the same for both setups. Graphically, the curve with the inhibitor is shifted to the right but reaches the same asymptote.
c) Standard variables that must be controlled in enzyme-catalyzed reactions include temperature (maintained using a water bath) and pH (maintained using a buffer solution of a specific pH) to ensure that the enzyme's three-dimensional structure and active site remain stable.

a) [2 marks max]
• rate of glucose/product production / initial rate of reaction; [1] (Accept: rate of lactose depletion)
• measured using a colorimeter / glucose meter / glucose test strips at regular time intervals; [1]

b) [2 marks max]
• at very high substrate concentrations, both curves reach the same maximum rate / plateau / \(V_{\max}\); [1]
• the curve with the competitive inhibitor has a lower initial slope / rate at lower substrate concentrations than the curve without inhibitor; [1]

c) [1 mark max]
• temperature (controlled using a thermostatically controlled water bath) OR pH (controlled using a buffer solution); [1] (Do not accept 'water bath' or 'buffer' alone without the variable)

a) Calculating percentage change in mass standardizes the data, compensating for any differences in the starting mass or size of individual potato cylinders.
b) Distilled water has an osmolarity of \(0.0\text{ Osmol L}^{-1}\), which is lower than the osmolarity of the cytoplasm inside the potato cells (the external environment is hypotonic). Water enters the cytoplasm and vacuole of the potato cells by osmosis (movement of water from lower solute concentration to higher solute concentration across a selectively permeable membrane), causing the tissue to gain mass and become turgid.
c) When a graph of percentage change in mass (y-axis) versus sucrose concentration (x-axis) is constructed, a line of best fit is drawn. The point where the line intersects the x-axis (y = 0) represents the concentration where there is no net movement of water into or out of the cells (isotonic point). At this concentration, the osmolarity of the sucrose solution matches the internal osmolarity of the potato tissue cells.

a) [1 mark max]
• compensates for differences in the initial mass / size of the potato cylinders; [1] (Accept: standardizes the data for comparison)

b) [2 marks max]
• distilled water is hypotonic to the cytoplasm / has a higher water potential than the potato cells; [1]
• water enters the cells by osmosis / down the water potential gradient; [1] (Reject: active transport of water)

c) [2 marks max]
• plot percentage change in mass on the y-axis and sucrose concentration on the x-axis and draw a line of best fit; [1]
• find the sucrose concentration where the line crosses the x-axis / where percentage change in mass is zero; [1] (Accept: this is where the solution is isotonic to the tissue)

a) During aerobic respiration, germinating seeds consume oxygen (\(\text{O}_2\)) and release carbon dioxide (\(\text{CO}_2\)) gas. The \(\text{KOH}\) solution reacts with and absorbs the carbon dioxide gas, converting it to solid potassium carbonate.
b) Since the \(\text{CO}_2\) produced is absorbed by the \(\text{KOH}\), any change in gas volume inside the sealed chamber is solely due to the uptake of \(\text{O}_2\) by the seeds. The net reduction in gas volume decreases the pressure within the chamber relative to atmospheric pressure, pulling the liquid in the capillary tube towards the chamber.
c) Respirometers are highly sensitive to external changes in temperature (which affects gas volume according to Charles's Law) and barometric pressure. A thermo-barometer (control) is set up with inactive materials (like boiled, sterilized seeds or glass beads of the same volume) so that any changes in fluid movement due to physical factors can be measured and used to correct the experimental readings.

a) [1 mark max]
• absorbs carbon dioxide (\(\text{CO}_2\)) gas produced by respiration; [1] (Accept: removes carbon dioxide)

b) [2 marks max]
• oxygen (\(\text{O}_2\)) is absorbed/consumed by the germinating seeds and the carbon dioxide (\(\text{CO}_2\)) produced is absorbed by the \(\text{KOH}\); [1]
• this leads to a decrease in gas volume / gas pressure inside the respirometer chamber, causing the liquid to move towards the chamber; [1]

c) [2 marks max]
• use a control tube containing non-living material / boiled (dead) seeds / glass beads of the same volume as the seeds; [1]
• place it in the same environment to monitor changes in volume/pressure caused by temperature/pressure fluctuations, and subtract/add this movement to correct the experimental results; [1]

Kupffer cells, located in the liver sinusoids, engulf worn-out erythrocytes by phagocytosis. Inside Kupffer cells, hemoglobin is broken down into globin and heme. Globin is hydrolyzed into amino acids and released into the blood. Heme is broken down into iron and bilirubin. Iron is stored in the liver as ferritin or transported to the bone marrow to make new red blood cells. Bilirubin is taken up by hepatocytes and excreted in bile.

Part (a): 1 mark for: Kupffer cells. Part (b): Up to 4 marks for: Kupffer cells engulf old erythrocytes via phagocytosis; Hemoglobin is split into globin and heme; Globin is digested to amino acids; Heme is broken down into iron and bilirubin; Iron is stored as ferritin or transported to bone marrow via transferrin; Bilirubin is absorbed by hepatocytes and excreted in bile.

The QRS complex in an ECG corresponds to the depolarization of the ventricles, triggering ventricular systole. The delay at the AV node (about 0.1 seconds) is caused by smaller fiber diameters, fewer gap junctions, and a more negative resting potential in nodal cells. This delay is crucial because it ensures atrial contraction is completed, allowing all blood to move from the atria into the ventricles before ventricular contraction starts, maximizing pumping efficiency.

Part (a): 1 mark for: Ventricular depolarization. Part (b): Up to 4 marks for: Arrival of signal from SA node to AV node; Signal delay (about 0.1 seconds); Delay due to smaller fiber diameter / fewer gap junctions / higher resistance; Allows atrial systole/contraction to finish; Ensures ventricles are fully filled with blood; Prevents ventricles contracting too early / simultaneous contraction.

In the cephalic phase, the sight, smell, taste, or thought of food stimulates the brain to send signals via the vagus nerve (parasympathetic nervous system) to the stomach, stimulating gastric glands to secrete acid and pepsinogen. Gastrin is secreted by G-cells in the stomach mucosa in response to the presence of peptides or stomach stretching. Gastrin travels in the bloodstream to stimulate parietal cells to secrete HCl. When stomach pH drops very low, negative feedback inhibits further gastrin secretion to prevent over-acidification.

Part (a): Up to 2 marks for: Sensory stimuli (sight/smell/taste/thought of food) trigger impulses; Impulses travel via the vagus nerve to the stomach; Stimulates gastric glands / parietal cells / chief cells to secrete gastric juice. Part (b): Up to 3 marks for: Gastrin is released by G-cells (stimulated by peptides/distension); Gastrin stimulates parietal cells to release hydrochloric acid (HCl); Low pH / high acid level inhibits gastrin secretion via negative feedback.

During intense exercise, the oxygen dissociation curve shifts to the right, which is known as the Bohr shift. Rapidly respiring tissues release carbon dioxide, which reacts with water to form carbonic acid, reducing the pH (increasing hydrogen ion concentration). The hydrogen ions bind to hemoglobin, causing a conformational change that lowers its affinity for oxygen. Consequently, hemoglobin releases its oxygen more easily, ensuring that highly active tissues receive the oxygen needed for cellular respiration.

Part (a): 1 mark for: Shifts to the right. Part (b): Up to 4 marks for: Respiring tissues produce carbon dioxide (\(CO_2\)); \(CO_2\) forms carbonic acid (\(H_2CO_3\)), which lowers pH / increases \(H^+\); Hydrogen ions bind to hemoglobin, decreasing its affinity for oxygen; This causes a rightward shift of the dissociation curve; Physiological advantage: oxygen is released more readily to tissues with high metabolic activity.