2025 IB DP Biology Practice Paper with Answers

The rapid depolarization phase of an action potential is caused by the opening of voltage-gated sodium channels, allowing sodium ions to rapidly enter the axon down their electrochemical gradient. Voltage-gated potassium channels open later, during the repolarization phase.

Award 1 mark for selecting B. Correctly identifies the opening of voltage-gated sodium channels as the cause of depolarization.

Energy transfer between trophic levels is inefficient, with only about 10% of the energy being passed on to the next level. Most of the energy is lost as metabolic heat during cell respiration, limiting the total biomass that can be supported at higher trophic levels.

Award 1 mark for selecting A. Correctly identifies energy loss via respiration as the key cause of reduced biomass at higher trophic levels.

The 0.5 M sucrose solution inside the bag is hypertonic compared to the 0.1 M solution in the beaker. Water moves by osmosis down its concentration gradient (from high water potential to low water potential) into the bag, causing it to swell.

Award 1 mark for selecting B. Correctly identifies that water moves into the hypertonic bag by osmosis.

Competitive inhibitors bind to the active site of the enzyme, competing with the substrate. By increasing the substrate concentration significantly, the substrate outcompetes the inhibitor, allowing the reaction to reach the same maximum velocity (\(V_{\max}\)) as the uninhibited reaction.

Award 1 mark for selecting B. Correctly identifies competitive inhibition as the type where \(V_{\max}\) remains unchanged at high substrate concentrations.

Phloem transport (translocation) occurs bidirectionally from source to sink through living sieve tube elements and requires metabolic energy (active transport for loading). Xylem transport is unidirectional (from roots to leaves) through dead tracheids and vessel elements, driven passively by transpiration pull (tension).

Award 1 mark for selecting C. Correctly distinguishes active, bidirectional phloem transport from passive, unidirectional xylem transport.

A plant root cell is eukaryotic (possesses mitochondria and 80S ribosomes), contains a cellulosic cell wall, and a large central vacuole, but lacks chloroplasts as it is not exposed to light and does not perform photosynthesis. Animal cells do not have cell walls, while prokaryotes (bacteria) do not have eukaryotic organelles like mitochondria.

Award 1 mark for selecting B. Correctly identifies plant root cells as eukaryotic cells with cell walls and vacuoles but no chloroplasts.

During the synthesis of a triglyceride, the hydroxyl group of glycerol reacts with the carboxyl group of a fatty acid to form an ester bond. Because this is a condensation reaction, one molecule of water is released as a co-product for each ester bond formed.

Award 1 mark for selecting D. Correctly identifies the ester bond and water as the co-product of condensation.

This is a classical example of a top-down trophic cascade. The introduction of a predator at a high trophic level reduces the population of the primary consumers (zooplankton), which in turn reduces grazing pressure on the primary producers (phytoplankton), leading to their increase.

Award 1 mark for selecting B. Correctly identifies the top-down control cascade.

Tetrodotoxin blocks voltage-gated sodium channels, which are responsible for the rapid influx of sodium ions (\(\text{Na}^+\)) into the axon during the depolarization phase of an action potential. Without the opening of these channels, depolarization cannot occur, and the membrane cannot reach the threshold or generate an action potential. Consequently, the axon remains at its resting membrane potential of approximately -70 mV despite the suprathreshold stimulus.

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

First, find the percentage of incident solar energy stored as biomass (NPP): \(\frac{\text{NPP}}{\text{Incident Solar Energy}} \times 100 = \frac{1.6 \times 10^4}{2.0 \times 10^6} \times 100 = 0.8\%\). Second, calculate the energy lost through respiration (R): \(R = \text{GPP} - \text{NPP} = 4.0 \times 10^4 - 1.6 \times 10^4 = 2.4 \times 10^4 \text{ kJ m}^{-2} \text{ yr}^{-1}\). Now, calculate the percentage of GPP lost as respiration: \(\frac{R}{\text{GPP}} \times 100 = \frac{2.4 \times 10^4}{4.0 \times 10^4} \times 100 = 60\%\). Therefore, the correct option is B.

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

The osmolarity of the tissue corresponds to the concentration of sucrose where there is no net movement of water, resulting in a 0% change in mass. Looking at the data, a 0% change in mass lies between 0.2 mol dm\(^{-3}\) (+6.8%) and 0.4 mol dm\(^{-3}\) (-2.4%), which is approximately 0.35 mol dm\(^{-3}\). In a 0.6 mol dm\(^{-3}\) sucrose solution, the solution is hypertonic relative to the potato cytoplasm, causing water to leave the cells by osmosis, making the cells plasmolysed.

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

In competitive inhibition, the inhibitor structurally resembles the substrate and binds to the active site. Because it occupies the active site, it prevents substrate binding, increasing the concentration of substrate required to reach half-maximal velocity (\(K_m\) increases). However, at extremely high substrate concentrations, the substrate virtually outcompetes the inhibitor for the active sites, allowing the reaction to reach the same maximum velocity (\(V_{\max}\)) as in the uninhibited reaction.

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

Cohesion is the attraction between water molecules due to hydrogen bonding, which allows them to withstand extremely high tension (negative pressure) without pulling apart (cavitation). Adhesion is the attraction between water molecules and the hydrophilic cellulose components of the xylem cell walls, which helps support the water column against gravity. Xylem vessels are dead, hollow tubes (no cytoplasm) reinforced with lignin to prevent collapse under tension.

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

Both photosynthetic prokaryotes and plant cells contain a cell wall (peptidoglycan in prokaryotes, cellulose in plants), a cell membrane, cytoplasm, and ribosomes (70S in prokaryotes, and both 80S and 70S in plant cells). Prokaryotes do not have membrane-bound organelles like a nucleus, chloroplasts, or mitochondria, ruling out options A, C, and D.

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

Cellulose is a polymer of \(\beta\)-glucose monomers linked by \(\beta\)-1,4-glycosidic bonds. This configuration results in straight, unbranched chains that form hydrogen bonds with adjacent chains to create strong microfibrils, providing structural support in plant cell walls. Glycogen and amylopectin are made of \(\alpha\)-glucose and have both 1,4 and 1,6 bonds; glycogen is for animals, amylopectin for plants. Amylose has only \(\alpha\)-1,4-glycosidic bonds and is unbranched.

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

The correct order of phases in a sigmoid growth curve is: Lag phase (slow initial growth as individuals adapt) \(\rightarrow\) Exponential phase (rapid growth with abundant resources) \(\rightarrow\) Transitional phase (growth slows as resources become limiting) \(\rightarrow\) Plateau phase (growth stops, population stabilizes at carrying capacity). The transition to the plateau phase is driven by density-dependent factors such as nutrient depletion, space limitation, or accumulation of toxic wastes, which increase the mortality rate and/or decrease the reproduction rate.

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

In myelinated neurones, the lipid-rich myelin sheath acts as an electrical insulator, preventing ion flow across the axon membrane. Voltage-gated sodium and potassium channels are highly concentrated at the unmyelinated gaps called nodes of Ranvier. Consequently, membrane depolarization and the action potential can only occur at these nodes, allowing the electrical signal to effectively "jump" from one node to the next. This saltatory conduction significantly increases the velocity of the nerve impulse compared to continuous propagation along an unmyelinated axon, where the entire membrane must depolarize sequentially.

Award [1] for the correct answer (C). Reject options A, B, and D because the myelin sheath itself does not generate action potentials, it increases electrical resistance across the membrane rather than lowering it, and voltage-gated channels are concentrated at the nodes of Ranvier, not within the myelin sheath.

Myelin is an electrical insulator that prevents ion flow across the axon membrane. Consequently, action potentials can only occur at the unmyelinated gaps called nodes of Ranvier, where voltage-gated channels are concentrated. This causes the action potential to 'jump' from node to node, a process known as saltatory conduction, which significantly increases the speed of nerve impulse propagation. Options A, C, and D are incorrect because myelin decreases membrane capacitance, reduces the need for continuous active transport along the entire axon length, and acts as an insulator rather than a conductor.

Award 1 mark for the correct option (B). Option A is incorrect because myelination decreases capacitance. Option C is incorrect because myelin reduces overall ion leakage, leading to less energy expenditure by sodium-potassium pumps. Option D is incorrect because myelin acts as an insulator, not a conductor.

The total energy captured by producers via photosynthesis is represented by the Gross Primary Productivity (GPP), which is \(3.0 \times 10^4 \text{ kJ m}^{-2} \text{ yr}^{-1}\). Net Primary Productivity (NPP) is the energy remaining after producer respiration (R) is subtracted: \(\text{NPP} = \text{GPP} - R\). Rearranging this formula: \(R = \text{GPP} - \text{NPP} = 3.0 \times 10^4 - 1.5 \times 10^4 = 1.5 \times 10^4 \text{ kJ m}^{-2} \text{ yr}^{-1}\). The percentage of captured energy (GPP) lost to respiration is \(\frac{1.5 \times 10^4}{3.0 \times 10^4} \times 100\% = 50.0\%\).

Award 1 mark for the correct option (C). Option A incorrectly calculates percentage of GPP relative to total solar radiation (1.5%). Options B and D are distractors arising from incorrect ratios.

Mechanism I does not require ATP but does require a specific membrane protein (such as a channel or carrier protein) to move substances down their concentration gradient; this is the definition of facilitated diffusion. Mechanism II requires ATP and a specific protein to move substances against their gradient, which is active transport. Mechanism III requires no ATP and no protein to move substances down their concentration gradient, which is simple diffusion.

Award 1 mark for the correct option (B). Option A incorrectly identifies simple and facilitated diffusion. Option C incorrectly identifies osmosis and active transport. Option D completely scrambles the mechanisms.

Competitive inhibitors compete with the substrate for binding to the active site. Because they are temporary and reversible, a high enough concentration of substrate will outcompete the inhibitor, allowing the reaction to reach its original maximum velocity (\(V_{\text{max}}\)). Therefore, Inhibitor Y is a competitive inhibitor. Non-competitive inhibitors bind to an allosteric site, altering the shape of the active site so that the enzyme cannot function effectively, which decreases the maximum velocity (\(V_{\text{max}}\)) and cannot be overcome by adding more substrate. Therefore, Inhibitor Z is a non-competitive inhibitor.

Award 1 mark for the correct option (C). Option A is incorrect because Y is competitive. Option B is incorrect because non-competitive inhibitors reduce \(V_{\text{max}}\), not competitive ones. Option D is incorrect because Z is a non-competitive inhibitor and binds to an allosteric site, not a competitive inhibitor.

Xylem vessels are dead, hollow tubes reinforced with lignin, which prevents them from collapsing under the extreme tension (negative pressure) generated by transpiration pull. Phloem sieve tube elements are living cells with non-lignified walls and a highly reduced cytoplasm (lacking a nucleus and most organelles) to allow for the unimpeded flow of organic sap under positive hydrostatic pressure. Sieve plates are found in phloem, not xylem.

Award 1 mark for the correct option (B). Option A is incorrect because xylem is continuous, whereas phloem has sieve plates. Option C is incorrect because xylem vessels are dead and phloem sieve tubes are living. Option D is incorrect because xylem transports water under tension, and phloem transports organic solutes under hydrostatic pressure.

The transitional phase of a sigmoid (S-shaped) population growth curve occurs when limiting factors (such as food shortage, competition, predation, and disease) begin to take effect as the population approaches carrying capacity. During this phase, the natality (birth) rate decreases and/or the mortality (death) rate increases. As a result, the rate of population growth slows down, though the population continues to increase (since natality still exceeds mortality) until it reaches the plateau phase.

Award 1 mark for the correct option (D). Option A is incorrect because during the exponential phase, limiting factors are negligible and growth rate is accelerating. Option B is incorrect because the lag phase is the initial period of slow growth with low numbers of individuals. Option C is incorrect because during the plateau phase, population growth is zero as natality plus immigration equals mortality plus emigration.

1. Percentage increase = \( \frac{\text{Velocity of A} - \text{Velocity of C}}{\text{Velocity of C}} \times 100 \) = \( \frac{60 - 5}{5} \times 100 = 1100\\% \).

2. In myelinated axons, myelin sheath prevents the movement of ions across the axon membrane except at the gaps called nodes of Ranvier. Therefore, local currents can flow much further down the axon, and the action potential 'jumps' from node to node (saltatory conduction). In unmyelinated fibers, channels must open sequentially along the entire length of the membrane, which is a much slower process.

3. If myelin is lost, the local currents cannot reach the next node of Ranvier with enough intensity to depolarize it to threshold, leading to blocked or severely delayed nerve impulses. This disrupts the precise timing required for neural integration, leading to symptoms like muscle weakness or loss of coordination.

1. [1.25 marks total]:
- Correct working: \( \frac{60 - 5}{5} \times 100 \) [0.5 marks]
- Correct final value of 1100% (unit required) [0.75 marks]

2. [3 marks total]:
- Myelin acts as an electrical insulator / prevents ion flow across the axon membrane [1 mark]
- Action potentials/depolarization can only occur at the nodes of Ranvier [1 mark]
- Nerve impulse jumps from node to node / saltatory conduction, which is much faster than continuous propagation [1 mark]

3. [2 marks total]:
- Conduction velocity is reduced / signal transmission is blocked / action potentials fail to propagate [1 mark]
- Disruption of signal synchronization / delayed arrival of impulses at the synapse prevents successful summation / signal integration [1 mark]

1. Percentage of solar energy converted to GPP = \( \frac{\text{GPP}}{\text{Incident Solar Energy}} \times 100 \) = \( \frac{30,000}{1,200,000} \times 100 = 2.5\\% \).

2. Net Primary Productivity is equal to Gross Primary Productivity minus Respiration (NPP = GPP - R). Therefore, R = GPP - NPP = 30,000 - 12,000 = 18,000 kJ m\u207b\u00b2 yr\u207b\u00b9.

3. The energy transfer efficiency between trophic levels is low (typically around 10%) because:
- Organisms lose a significant amount of energy as heat during metabolic reactions/cellular respiration.
- Some organic matter is not consumed (e.g., bones, claws, roots, or woody tissues are left behind).
- Some consumed material is indigestible and is excreted as feces (egestion) or urine (excretion), transferring energy to decomposers rather than the next consumer level.

1. [1.25 marks total]:
- Correct fraction: 30,000 / 1,200,000 [0.5 marks]
- Correct calculation: 2.5% (unit required) [0.75 marks]

2. [2 marks total]:
- Recalling formula: GPP - NPP or NPP = GPP - R [1 mark]
- Correct value: 18,000 kJ m⁲ yrⁱ (accept without units if already specified in the prompt, but reject if wrong units are given) [1 mark]

3. [3 marks total, award max 3 marks from the following points]:
- Energy lost as heat during cellular respiration / metabolic activity [1 mark]
- Not all parts of organisms are eaten/accessible to the consumers [1 mark]
- Some eaten matter is undigested / lost as feces/egested [1 mark]
- Energy lost through waste excretion (e.g., urea) [1 mark]

1. The independent variable is the factor being changed by the investigator, which is the sucrose concentration (measured in M). The dependent variable is the measured outcome, which is the percentage change in mass of the potato tissue.

2. To find the isotonic concentration, we must determine where the line of best fit crosses the x-axis (0% change in mass). Because the mass increases at 0.2 M (+8.5%) and decreases at 0.4 M (-2.0%), the isotonic point must lie between these two values. A linear interpolation gives: 0.2 + (0.2 * (8.5 / (8.5 - (-2.0)))) \\approx 0.36 M. Justification: Isotonic means no net movement of water, which results in zero net change in the mass of the potato cylinders.

3. The 0.8 M sucrose solution has a much higher solute concentration than the internal environment of the potato cells, making the solution hypertonic. The water potential of the solution is lower than that of the cytoplasm. Consequently, water molecules move out of the cells down the water potential gradient by osmosis across the selectively permeable plasma membrane, leading to cell plasmolysis and a decrease in mass.

1. [1.25 marks total]:
- Correctly identifies both variables: Independent variable = Sucrose concentration AND Dependent variable = Percentage change in mass. (Award 1.25 marks for both correct, 0.5 marks if only one is correct).

2. [2 marks total]:
- Correct estimate of isotonic point: any value between 0.34 M and 0.38 M [1 mark]
- Valid justification: It is the point where there is no net change in mass / no net osmosis occurs / water potentials are equal [1 mark]

3. [3 marks total]:
- Water moves out of the potato tissue/cells [1 mark]
- Movement occurs by osmosis down the water potential gradient / from high water potential to low water potential [1 mark]
- Surrounding 0.8 M solution is hypertonic relative to the cytoplasm / has a higher solute concentration [1 mark]

1. Inhibitor A displays competitive inhibition because \( V_{max} \) is unaffected but \( K_m \) is increased (meaning affinity is reduced, requiring more substrate to reach half-saturation). Inhibitor B displays non-competitive inhibition because the maximum capacity to catalyze the reaction is lowered (lower \( V_{max} \)) but substrate affinity of the remaining functional enzymes is unchanged (constant \( K_m \)).

2. Competitive inhibitors have a structure similar to the substrate and physically fit into the active site, blocking the substrate from entering. Non-competitive inhibitors do not bind to the active site; instead, they bind to an alternative (allosteric) site. This binding alters the tertiary structure of the enzyme, changing the shape of the active site so that the substrate either cannot bind or cannot be converted into products.

3. At high substrate concentrations, the likelihood of a substrate molecule binding to an active site is much higher than that of a competitive inhibitor, effectively nullifying the inhibitor's effect and allowing the reaction to reach the original maximum rate. In contrast, non-competitive inhibitors reduce the overall number of active, functional enzyme molecules. No matter how much substrate is added, it cannot displace the inhibitor from the allosteric site, so the maximum rate of reaction remains depressed.

1. [1.25 marks total]:
- Inhibitor A = competitive AND Inhibitor B = non-competitive (both correct needed for 1.25 marks, otherwise 0.5 marks for one correct).

2. [2 marks total]:
- Competitive inhibitor: binds to the active site / has a similar shape to the substrate [1 mark]
- Non-competitive inhibitor: binds to an allosteric site / changes the shape of the active site [1 mark]

3. [3 marks total]:
- Under competitive inhibition, high substrate concentration outcompetes the inhibitor for the active site [1 mark]
- This allows the reaction rate to reach the normal maximum rate (Vmax) [1 mark]
- Under non-competitive inhibition, the inhibitor remains bound to the allosteric site regardless of substrate concentration, meaning the maximum rate (Vmax) cannot be restored [1 mark]

(a) The relationship between gross primary productivity, net primary productivity, and respiration is given by the equation: \(NPP = GPP - R\).
(b) At 25 °C, GPP is 210 g m⁻² yr⁻¹ and NPP is 110 g m⁻² yr⁻¹. Rearranging the equation to solve for R gives: \(R = GPP - NPP = 210 - 110 = 100\text{ g m}^{-2}\text{ yr}^{-1}\).
(c) As the temperature rises from 25 °C to 35 °C, enzyme-controlled metabolic processes speed up. Autotrophic respiration increases significantly from 100 to 160 g m⁻² yr⁻¹ (a 60% increase), whereas GPP only increases from 210 to 250 g m⁻² yr⁻¹ (a 19% increase). Since more energy is lost as heat via respiration than is fixed via photosynthesis, NPP decreases.
(d) In a closed chamber containing a plant, carbon dioxide is taken up for photosynthesis and released by cellular respiration. Under light conditions, the rate of decrease in CO₂ concentration represents net photosynthesis (NPP) and can be tracked using a digital carbon dioxide gas probe.

a. NPP = GPP - R (or GPP = NPP + R) [1]
b. 100 g m⁻² yr⁻¹ (both correct value and units are required) [1]
c. NPP decreases because respiration (R) increases more than GPP [1]; supporting data: R increases by 60 g m⁻² yr⁻¹ while GPP only increases by 40 g m⁻² yr⁻¹ [1]
d. Use of CO₂ sensor/probe inside a sealed transparent container [1]; measure the rate of decline of CO₂ concentration in the chamber over time under constant light conditions [1]

(a) The resting membrane potential is the stable potential before the stimulus and after recovery, which is -70 mV.
(b) When comparing, note that both axons reach the threshold potential of -55 mV at 1 ms. However, the control axon shows a full action potential spike with depolarization reaching +35 mV at 2 ms, while the treated axon does not depolarize, dropping slightly to -65 mV instead.
(c) The threshold of -55 mV normally triggers the opening of voltage-gated sodium channels, causing an influx of Na⁺ that depolarizes the membrane to +35 mV. Since the treated axon fails to show any depolarization beyond the threshold, STX-B must act as a blocker of voltage-gated sodium channels.
(d) After the peak of the action potential, potassium ions flow out of the axon during repolarization. The slow closing of voltage-gated potassium channels causes the potential to temporarily drop below the resting potential (-80 mV), a phase known as hyperpolarization.

a. -70 mV [1]
b. Both axons reach -55 mV at 1 ms (threshold) [1]; control axon depolarizes to +35 mV at 2 ms, while the STX-B treated axon fails to depolarize / drops to -65 mV [1]
c. STX-B blocks voltage-gated sodium (Na⁺) channels [1]; preventing the influx of sodium ions required for the depolarization phase of an action potential [1]
d. Hyperpolarization / undershoot / refractory period [1]

(a) The independent variable is the variable manipulated by the experimenter, which is the substrate (lactose) concentration.
(b) Lactase activity shows a characteristic enzyme-substrate curve: it increases rapidly at first as more substrate molecules bind to unoccupied active sites, then plateaus above 40 mmol dm⁻³ as the active sites of lactase become saturated with lactose.
(c) Competitive inhibitors bind reversibly to the active site. At low lactose concentrations, galactose successfully competes for the active site, reducing activity. At high lactose concentrations, the excess lactose molecules outcompete galactose, allowing the rate of reaction to approach the uninhibited maximum (V_max).
(d) Non-competitive inhibitors bind to an allosteric site, altering the shape of the active site so that the substrate cannot be catalyzed. Consequently, increasing substrate concentration cannot overcome this type of inhibition, and the maximum rate of reaction (V_max) is permanently reduced.

a. Lactose concentration [1]
b. Lactase activity increases as lactose concentration increases [1]; the rate of increase plateaus / levels off at high concentrations (above 40 mmol dm⁻³) due to active site saturation [1]
c. At low substrate concentrations, the rate of reaction is significantly lower with the inhibitor than without it [1]; at high substrate concentrations, the activity with the inhibitor rises to nearly match the uninhibited rate / the inhibition is overcome by high substrate concentrations [1]
d. V_max would decrease / be reduced [1]

(a) Simple diffusion does not require membrane proteins, and its rate is directly proportional to the concentration gradient. The data shows a linear increase of 10 arbitrary units of uptake for every 2 mmol L⁻¹ increase in solute concentration, indicating simple diffusion.
(b) Substance Y is transported via facilitated diffusion, which relies on channel proteins. As the concentration of Y increases, the rate of uptake initially rises rapidly, but above 6 mmol L⁻¹ it begins to plateau (saturate) at 95 units. This is because all the channel proteins are operating at their maximum capacity (saturated).
(c) The composition of the lipid bilayer determines its physical properties. Unsaturated fatty acids have double bonds that create bends or kinks, preventing tight packaging of lipids, which increases fluidity and allows hydrophobic molecules to pass more easily. Longer hydrocarbon tails increase the distance molecules must diffuse, thereby lowering the diffusion rate.

a. Simple diffusion [1]; because the rate of uptake increases linearly / is directly proportional to the solute concentration [1]
b. The rate of uptake plateaus / levels off [1]; because the transport channel proteins become saturated / have reached their maximum capacity to transport solute [1]
c. Unsaturated fatty acids (kinks) increase fluidity, which increases the rate of simple diffusion [1]; longer fatty acid tails increase bilayer thickness / hydrophobic barrier, which decreases the rate of simple diffusion [1]

(a) The Lincoln Index is calculated as: \(N = \frac{n_1 \times n_2}{n_3}\), where \(n_1 = 48\) (first sample marked), \(n_2 = 50\) (second sample size), and \(n_3 = 12\) (recaptured marked snails). \(N = \frac{48 \times 50}{12} = \frac{2400}{12} = 200\).
(b) For the Lincoln Index to be accurate, the population must be closed (no births, deaths, immigration, or emigration). Additionally, the marking method must not affect survival (e.g., making them more visible to predators) and must not get lost.
(c) Emigration of marked snails decreases the numerator of the recaptured ratio (or \(n_3\)). Because \(n_3\) is in the denominator of the Lincoln Index formula, a smaller value of \(n_3\) leads to a mathematically larger, and thus overestimated, population size \(N\).
(d) A population is defined by a single species in a specific habitat, whereas a community contains multiple interacting species within that habitat.

a. 200 (snails) [1]
b. Award [1] for each correct assumption up to [2]:
- No immigration or emigration
- No births or deaths
- Marking does not affect survival / predation risk
- Marks do not wash off / disappear
- Marked individuals mix randomly with the population
c. Emigration of marked individuals reduces the number of recaptured marked snails / decreases the value of recaptured individuals (n3) [1]; this results in an overestimation of the population size [1]
d. Population refers to one species in an area, while community refers to multiple species interacting in that area [1]

(a) Analyzing the table reveals that higher GC content is generally associated with higher on-target DSB efficiency (e.g., 35% GC gives 42% efficiency, whereas 70% GC gives 92% efficiency).
(b) For therapeutics, there must be a balance between high clinical efficacy and high safety (low off-target effects). gRNA 4 has a strong on-target efficiency of 81% while keeping off-target mutations extremely low (5), making it the safest and most effective option compared to gRNA 3 (which has 92% efficiency but 29 off-target cuts, which could damage essential genes).
(c) The CRISPR-Cas9 system relies on two key components: the guide RNA (gRNA) provides specificity by hybridizing to the target genomic sequence via complementary base-pairing, and the Cas9 enzyme acts as molecular scissors to cut the DNA strands at that precise site.
(d) Non-homologous end-joining (NHEJ) is an error-prone repair mechanism used by cells to heal double-strand breaks. It frequently introduces small insertions or deletions (indels) at the cut junction, which disrupt the reading frame of the target gene, resulting in a knockout.

a. Positive correlation / higher GC content corresponds to higher on-target DSB efficiency [1]
b. gRNA 4 is the best choice [1]; because it provides high efficiency (81%) with minimal safety risk / lowest number of off-target mutations among high-efficiency options [1]
c. gRNA directs the Cas9 complex to the target DNA sequence through complementary base pairing [1]; Cas9 is an endonuclease that cuts the double-stranded DNA [1]
d. Insertion/deletion (indel) mutation (accept frameshift mutation) [1]

**Part a (4 marks max):**
- The sodium-potassium pump (\text{Na}^+/\text{K}^+-ATPase) actively transports \text{Na}^+ and \text{K}^+ ions against their concentration gradients.
- It pumps three sodium ions (\text{Na}^+) out of the neurone for every two potassium ions (\text{K}^+) pumped in.
- This activity requires ATP (active transport).
- It establishes a steep concentration gradient: high \text{Na}^+ outside, high \text{K}^+ inside.
- The resting membrane is much more permeable to \text{K}^+ than to \text{Na}^+ due to open potassium leak channels.
- Potassium ions diffuse out of the cell down their concentration gradient.
- Large, negatively charged proteins and organic anions remain trapped inside the cytoplasm.
- This results in a negative potential difference across the membrane, typically around -70 mV.

**Part b (5 marks max):**
- An action potential in one region causes an influx of \text{Na}^+ ions, depolarizing that part of the membrane.
- This creates local currents as \text{Na}^+ ions diffuse internally along the inside of the axon to adjacent, resting sections.
- The diffusion of \text{Na}^+ reduces the negative charge in the adjacent region, raising it toward the threshold potential (approx. -55 mV).
- This depolarization causes voltage-gated \text{Na}^+ channels in the adjacent region to open, generating a new action potential.
- In myelinated axons, the myelin sheath acts as an electrical insulator, preventing ion exchange through the membrane.
- Action potentials/depolarization can only occur at the uninsulated gaps called nodes of Ranvier.
- Consequently, the local currents and action potentials jump from one node to the next (saltatory conduction), which greatly increases the speed of transmission.

**Part c (6 marks max):**
- The arrival of an action potential depolarizes the pre-synaptic membrane.
- This depolarization causes voltage-gated calcium (\text{Ca}^{2+}) channels to open.
- Calcium ions (\text{Ca}^{2+}) diffuse into the pre-synaptic knob down their concentration gradient.
- High calcium concentration causes synaptic vesicles containing neurotransmitter (e.g., acetylcholine) to move to and fuse with the pre-synaptic membrane.
- The neurotransmitter is released into the synaptic cleft by exocytosis.
- Neurotransmitter molecules diffuse across the synaptic cleft.
- The neurotransmitter binds to specific receptor proteins (ligand-gated ion channels) on the post-synaptic membrane.
- This binding opens ligand-gated sodium channels, allowing \text{Na}^+ to diffuse into the post-synaptic cell.
- The influx of \text{Na}^+ causes depolarization of the post-synaptic membrane (generating an excitatory post-synaptic potential or EPSP).

**Part a: Max [4 marks]**
- active transport of sodium/potassium ions by sodium-potassium pump;
- 3 \text{Na}^+ are pumped out for every 2 \text{K}^+ pumped in (requires ATP);
- creates concentration gradients (high \text{Na}^+ outside, high \text{K}^+ inside);
- membrane is more permeable to \text{K}^+ than \text{Na}^+ (due to leak channels);
- potassium diffuses out of the cell down its concentration gradient;
- large negatively charged organic anions inside the cell cannot leave;
- results in a resting potential of approx. -70 mV (negative inside relative to outside);

**Part b: Max [5 marks]**
- influx of \text{Na}^+ during depolarization creates local currents/concentration differences;
- sodium ions diffuse internally along the axon to adjacent resting regions;
- this depolarizes the adjacent region toward the threshold potential (approx. -55 mV);
- triggers the opening of voltage-gated \text{Na}^+ channels in the adjacent region;
- myelin sheath (produced by Schwann cells) acts as an electrical insulator;
- prevents ion movement across the membrane except at nodes of Ranvier;
- action potentials/depolarization can only occur at nodes of Ranvier;
- nerve impulse jumps from node to node (saltatory conduction), increasing transmission speed;

**Part c: Max [6 marks]**
- action potential arrives at pre-synaptic membrane and depolarizes it;
- causes voltage-gated calcium channels to open;
- calcium ions (\text{Ca}^{2+}) diffuse into the pre-synaptic knob;
- calcium causes synaptic vesicles to fuse with the pre-synaptic membrane;
- neurotransmitter (e.g., acetylcholine) is released into the synaptic cleft by exocytosis;
- neurotransmitter diffuses across the synaptic cleft;
- neurotransmitter binds to specific receptors on the post-synaptic membrane;
- causes ligand-gated sodium channels to open;
- sodium ions (\text{Na}^+) diffuse into the post-synaptic neurone, causing depolarization;

**Quality of Communication: [1 mark]**
- Award 1 mark for structured, coherent, and sequential answers that clearly connect physiological events across all three parts with precise biological terminology.