2024 IB DP Biology Practice Paper with Answers

At \(0.4\text{ mol dm}^{-3}\), the plant tissue does not gain or lose mass. This indicates that the tissue is in an isotonic state relative to the surrounding sucrose solution. In this state, the water potential inside the plant cells is equal to the water potential of the sucrose solution outside. Although individual water molecules continue to diffuse across the semi-permeable cell membrane in both directions, the net movement of water is zero. Thus, there is no net change in mass.

Award 1 mark for the correct option B. Option A is incorrect because water molecules continue to move (dynamic equilibrium). Option C is incorrect because sucrose is not actively transported in significant quantities to balance water movement in this context. Option D is incorrect because full turgidity would correspond to being in a hypotonic solution, which would cause mass gain.

To find the energy incorporated into the biomass of tertiary consumers, we track the transfer through the trophic levels: Primary producers = \(12,000\text{ kJ m}^{-2}\text{ yr}^{-1}\); Primary consumers (second trophic level) = \(12,000 \times 0.10 = 1,200\text{ kJ m}^{-2}\text{ yr}^{-1}\); Secondary consumers (third trophic level) = \(1,200 \times 0.10 = 120\text{ kJ m}^{-2}\text{ yr}^{-1}\); Tertiary consumers (fourth trophic level) = \(120 \times 0.10 = 12\text{ kJ m}^{-2}\text{ yr}^{-1}\).

Award 1 mark for Option C. Option A represents primary consumers. Option B represents secondary consumers. Option D represents quaternary consumers.

Transcription synthesizes an RNA strand complementary and antiparallel to the DNA template. Because the template runs from \(3'\) to \(5'\), the mRNA is synthesized from \(5'\) to \(3'\). Following complementary base-pairing rules (where T pairs with A, A pairs with U, C pairs with G, and G pairs with C), the template sequence \(3'\text{-T A C G G C A A T C A G-}'5\) transcribed yields the mRNA sequence \(5'\text{-A U G C C G U U A G U C-}'3\).

Award 1 mark for Option A. Option B is incorrect because it contains thymine (T) instead of uracil (U). Option C is incorrect because the directionality is inverted. Option D is incorrect because it is a direct copy of the template using U instead of T, failing to reflect complementary base pairing.

Natural selection acts on existing genetic variation. When an antibiotic is introduced, it acts as a selective pressure, killing the non-resistant bacteria. Pre-existing resistant individuals survive, reproduce, and pass their resistance-conferring alleles to their offspring, thereby increasing the allele frequency in the population.

Award 1 mark for Option C. Option A is incorrect because antibiotics do not cause mutations; mutations are random and precede exposure. Option B is incorrect because resistance does not necessarily confer an advantage in environments lacking antibiotics. Option D is incorrect because it describes Lamarckian inheritance of acquired characteristics.

Type I pneumocytes are extremely thin, flattened cells specialized for efficient gas exchange by minimizing the diffusion distance. Type II pneumocytes are cuboidal cells that secrete pulmonary surfactant, a lipoprotein mixture that reduces alveolar surface tension, preventing alveoli from collapsing during exhalation.

Award 1 mark for Option A. Option B incorrectly reverses the functions of Type I and Type II pneumocytes. Option C describes the function of alveolar macrophages and goblet cells, not pneumocytes. Option D is incorrect because gas exchange is a passive process of diffusion, not active transport.

The medulla oblongata detects elevated carbon dioxide levels and sends signals via sympathetic nerves to the SA node to increase heart rate. Simultaneously, the adrenal glands are stimulated to secrete epinephrine (adrenaline) into the blood, reinforcing and sustaining this elevated heart rate.

Award 1 mark for Option B. Option A is incorrect because the parasympathetic nervous system releases acetylcholine to decrease heart rate. Option C is incorrect because somatic nerves control skeletal muscle, not cardiac muscle. Option D is incorrect because carbon dioxide is detected by chemoreceptors, and the response is mediated via the autonomic nervous system.

In competitive inhibition, the inhibitor binds to the active site, physically blocking substrate binding. Because they compete for the same site, the inhibition can be overcome by increasing substrate concentration. This leaves the overall \(V_{\text{max}}\) unchanged but shifts the curve to the right, increasing the apparent Michaelis constant (\(K_m\)).

Award 1 mark for Option B. Option A describes non-competitive inhibition, which decreases \(V_{\text{max}}\). Option C describes end-product inhibition, which is not defined by specific kinetic parameters. Option D describes irreversible inhibition, which permanently decreases \(V_{\text{max}}\).

Earth absorbs short-wave radiation from the Sun and re-emits it as long-wave infrared radiation. Greenhouse gases (such as carbon dioxide and methane) in the atmosphere absorb this outgoing long-wave radiation and re-radiate it in all directions, including back toward Earth, retaining heat within the atmosphere.

Award 1 mark for Option B. Option A is incorrect because reflecting solar radiation back into space would cool the Earth. Option C is incorrect because ozone depletion is a separate phenomenon from the greenhouse effect. Option D is incorrect because UV radiation is mainly absorbed by ozone in the stratosphere, not by lower-troposphere greenhouse gases.

Natural selection acts on pre-existing genetic variation within a population. Individual pests do not acquire immunity or undergo beneficial mutations in response to the pesticide. Instead, those individuals that already carried alleles conferring resistance survived the pesticide applications and passed these alleles to their offspring, increasing the frequency of resistance in subsequent generations.

Award 1 mark for the correct option B. Reject options suggesting acquired characteristics (D) or direct mutations induced by the pesticide (A and C).

At 0.3 mol dm\(^{-3}\), there is no net change in mass of the potato cylinders, indicating that the solution is isotonic to the cytoplasm of the potato cells. At this concentration, water enters and leaves the tissue at equal rates, meaning the osmolarity of the tissue is approximately equal to that of the external solution (0.3 mol dm\(^{-3}\)).

Award 1 mark for identifying D as the correct conclusion based on the isotonic point. Reject A because a higher tissue osmolarity would lead to water gain and a mass increase. Reject B because water entered the potato in the 0.1 mol dm\(^{-3}\) solution, meaning the solution had a higher water potential than the tissue.

The DNA template is 3'-TAC GGT CTA CTA-5'. The complementary mRNA strand transcribed from this template is 5'-AUG CCA GAU GAU-3'. The tRNA anticodons align antiparallel and complementary to the mRNA codons: mRNA 5'-AUG-3' pairs with tRNA 3'-UAC-5'; 5'-CCA-3' with 3'-GGU-5'; 5'-GAU-3' with 3'-CUA-5'; 5'-GAU-3' with 3'-CUA-5'. Thus, the sequence of tRNA anticodons written in the 3' to 5' direction is 3'-UAC GGU CUA CUA-5'.

Award 1 mark for the correct identification of the tRNA anticodon sequence in the proper 3'-to-5' orientation (C). All other options represent incorrect sequences or incorrect orientations.

Biomagnification is the process where chemical substances (like toxins) become more concentrated at each successive trophic level. Because toxins are not easily excreted or metabolized, they accumulate in the tissues of organisms. Large predatory fish, being at the top of this food chain, consume many small fish over their lifetime, resulting in the highest concentration of the toxin.

Award 1 mark for the correct choice D. Reject A, B, and C as biomagnification results in the highest concentrations at the highest trophic level, not at lower or intermediate ones.

During inhalation, the external intercostal muscles contract (pulling the ribcage upward and outward) and the diaphragm contracts (moving downward and flattening). This active contraction increases the volume of the thoracic cavity, which causes a decrease in thoracic pressure below atmospheric pressure, allowing air to flow into the lungs.

Award 1 mark for the correct combination of muscle state and pressure change (C). Reject options that state thoracic pressure increases (A, D) or that internal intercostals are responsible for inspiration (B, D).

When the maximum velocity (\(V_{max}\)) of an enzymatic reaction is unaffected by an inhibitor at high substrate concentrations, the inhibitor is competitive. The substrate outcompetes the inhibitor for the active site at high concentrations. Consequently, the apparent Michaelis constant (\(K_m\)), which is the substrate concentration at which the reaction rate is half-maximal, increases because more substrate is required to reach this rate.

Award 1 mark for the correct answer B. Reject non-competitive inhibition options (A, D) since non-competitive inhibitors decrease \(V_{max}\). Reject C because a competitive inhibitor increases, rather than decreases, the apparent \(K_m\).

Methane is a much more potent greenhouse gas than carbon dioxide on a per-molecule basis (higher global warming potential), but due to its significantly lower atmospheric concentration, its overall contribution is less than that of carbon dioxide. Greenhouse gases absorb longwave infrared radiation, not shortwave solar radiation (making A incorrect). Nitrogen and oxygen do not absorb infrared radiation and are not greenhouse gases (making D incorrect).

Award 1 mark for the correct choice B. Reject A because greenhouse gases absorb outgoing longwave radiation, not incoming shortwave radiation. Reject C because ozone is not the primary greenhouse gas, and ultraviolet radiation is shortwave. Reject D because nitrogen and oxygen do not contribute to the greenhouse effect.

The codon 5'-UAA-3' is a stop codon. Changing a codon that specifies an amino acid (such as Tyrosine, 5'-UAC-3') to a stop codon is classified as a nonsense mutation. This leads to the premature termination of translation during protein synthesis, producing a shorter (truncated) and usually non-functional polypeptide.

Award 1 mark for the correct answer C. Reject A because a missense mutation results in a different amino acid, not a stop codon. Reject B because a silent mutation does not alter the amino acid sequence. Reject D because a single nucleotide substitution does not shift the reading frame.

Using the Lincoln Index formula: \(N = \frac{n_1 \times n_2}{m_2}\), where \(n_1\) is the number of individuals marked in the first sample (120), \(n_2\) is the total number of individuals captured in the second sample (80), and \(m_2\) is the number of marked individuals in the second sample (15). Therefore, \(N = \frac{120 \times 80}{15} = 640\). A key assumption is that marked individuals mix randomly with the unmarked population so that the second sample is representative of the whole population.

Award [1] for the correct option (A). Option B is incorrect because non-random recapture violates assumptions. Options C and D contain incorrect calculations.

The binding of three intracellular sodium ions (\(Na^+\)) triggers the autophosphorylation of the pump by ATP. This phosphorylation (addition of a phosphate group) directly induces a major conformational change in the pump protein, shifting its opening to the extracellular side and releasing the sodium ions.

Award [1] for the correct option (B). Option A is incorrect because potassium binding occurs afterward. Option C is incorrect because cholesterol modulates membrane fluidity but does not drive this active transport mechanism. Option D is incorrect because dephosphorylation triggers the release of potassium, not sodium.

The mRNA codon is 5'-ACG-3'. The complementary anticodon must pair antiparallel to it. Thus, the 3' end of the anticodon aligns with the 5' end of the codon, and the 5' end of the anticodon aligns with the 3' end of the codon. This gives: Codon: 5'-ACG-3' and Anticodon: 3'-UGC-5'. Reversing this sequence to read from 5' to 3' yields 5'-CGU-3'.

Award [1] for the correct option (B). Option A is the complementary sequence in the 3' to 5' direction. Option C is incorrect because tRNA contains uracil (U) instead of thymine (T). Option D is incorrect because it is not complementary.

Natural selection operates on pre-existing genetic variation. Individuals with advantageous traits (darker fur in a dark-soiled habitat) have a higher probability of surviving predation and reproducing. They pass on the beneficial allele to their offspring, causing the frequency of this allele to increase in the population over time.

Award [1] for the correct option (B). Option A represents a Lamarckian misconception. Option C is incorrect because alleles are inherited genetically, not transmitted via physical contact. Option D represents a non-evolutionary behavioral explanation.

During inhalation, the external intercostal muscles contract (raising the ribcage) and the diaphragm contracts and flattens. This increases the volume of the thoracic cavity, causing a decrease in pressure inside the lungs below atmospheric pressure, which draws air into the respiratory system.

Award [1] for the correct option (A). Options B and D describe elements associated with exhalation. Option C is incorrect because relaxation of external intercostals and a decrease in volume characterize exhalation.

In competitive inhibition, the inhibitor structurally resembles the substrate and competes for binding at the enzyme's active site. If the substrate concentration is increased significantly, the probability of substrate binding instead of inhibitor binding becomes extremely high. Thus, the maximum velocity (\(V_{max}\)) of the reaction remains unchanged.

Award [1] for the correct option (B). Options A and C describe non-competitive inhibition, which lowers the overall \(V_{max}\). Option D is incorrect because competitive inhibitors bind directly to the active site, not to allosteric sites.

Greenhouse gases are largely transparent to incoming short-wave solar radiation. When this radiation hits the Earth's surface, it is absorbed and re-emitted as longer-wave infrared (heat) radiation. Greenhouse gases absorb these longer wavelengths and re-emit them in all directions, including back toward the Earth, trapping heat.

Award [1] for the correct option (B). Option A is incorrect because greenhouse gases do not absorb short-wave solar radiation. Option C describes ozone depletion, which is a different phenomenon. Option D is incorrect because greenhouse gases work by radiation absorption and emission, not physical convection trapping.

The cross is between a pink-flowered plant (\(C^R C^W\)) and a white-flowered plant (\(C^W C^W\)). A Punnett square yields: 50% \(C^R C^W\) (pink) and 50% \(C^W C^W\) (white). This results in a phenotypic ratio of 1 pink : 1 white.

Award [1] for the correct option (B). Option A is the result of crossing homozygous red with homozygous white. Option C represents an incorrect segregation ratio. Option D is the expected F2 ratio from crossing two pink-flowered plants.

The rate of uptake plateaus at high concentrations because the transport proteins (carriers) become saturated. This is characteristic of facilitated diffusion. Simple diffusion does not show saturation kinetics because it does not rely on a limited number of membrane proteins.

Award 1 mark for the correct option (B). No partial marks are available for multiple choice questions.

During elongation, incoming aminoacyl-tRNAs bind first to the aminoacyl (A) site. The ribosome then catalyzes peptide bond formation, and the tRNA moves to the peptidyl (P) site. Finally, the deacylated tRNA moves to the exit (E) site to be released. Thus, the sequence is A \(\rightarrow\) P \(\rightarrow\) E.

Award 1 mark for the correct option (A). No partial marks are available for multiple choice questions.

This scenario describes disruptive selection, where both extreme phenotypes (dark and light) have a selective advantage over the intermediate phenotype. This type of selection increases phenotypic and genetic variation in the population, which can lead to speciation.

Award 1 mark for the correct option (C). No partial marks are available for multiple choice questions.

During inhalation, the external intercostal muscles contract (raising the ribs) and the diaphragm contracts (moving down and flattening). These actions increase the volume of the thoracic cavity, causing a decrease in pressure relative to atmospheric pressure, which draws air into the lungs.

Award 1 mark for the correct option (C). No partial marks are available for multiple choice questions.

A competitive inhibitor binds to the active site of the enzyme, directly competing with the substrate. Increasing the substrate concentration increases the probability of substrate-enzyme binding over inhibitor-enzyme binding, allowing the reaction to reach its original maximum rate (\(V_{\max}\)).

Award 1 mark for the correct option (B). No partial marks are available for multiple choice questions.

A base substitution that changes a codon for an amino acid into a stop codon is a nonsense mutation. This leads to premature termination of translation, resulting in a truncated (shorter) and typically non-functional polypeptide.

Award 1 mark for the correct option (C). No partial marks are available for multiple choice questions.

(a) Before the removal of Semibalanus balanoides, Chthamalus stellatus is restricted only to the high shore zone (between 6 m and 10 m). After Semibalanus is removed, Chthamalus is able to occupy a much wider vertical zone, extending down to 2 m.
(b) Every species has a fundamental niche (the full range of conditions under which it can survive) and a realized niche (the actual conditions it occupies due to biotic interactions). The fundamental niche of Chthamalus is broad (2-10 m), but interspecific competition / competitive exclusion by Semibalanus restricts it to a narrower realized niche (6-10 m). Once Semibalanus is removed, competition is eliminated, allowing Chthamalus to expand into its fundamental niche.
(c) Abiotic factors on the upper shore include prolonged exposure to air during low tide, which causes desiccation (drying out) and temperature extremes (overheating from solar radiation) that Semibalanus cannot tolerate.

(a) Max [2 marks]
- 1 mark for describing distribution before: restricted to high shore / 6 to 10 m.
- 1 mark for describing distribution after: expanded downward / found from 2 to 10 m.

(b) Max [3 marks]
- 1 mark for mentioning that Chthamalus has a wide fundamental niche but a smaller realized niche.
- 1 mark for identifying Semibalanus as a competitor / competitive exclusion.
- 1 mark for explaining that removing competition allows Chthamalus to occupy its fundamental niche.

(c) Max [2 marks]
- 1 mark for desiccation / drying out.
- 1 mark for high temperature / solar radiation / lack of water coverage for filter-feeding.

(a) The independent variable is the factor manipulated by the experimenter, which is the concentration of the sucrose solution (measured in M). The dependent variable is the measured outcome, which is the percentage change in the mass of the potato tissue.
(b) The osmolarity of the tissue corresponds to the concentration of sucrose where there is no net movement of water, meaning 0% change in mass. By interpolating the data, the 0% change point lies between 0.2 M (positive change) and 0.4 M (negative change), around 0.36 M (accept any value in the range of 0.34 M to 0.38 M).
(c) In a hypertonic 0.8 M sucrose solution, the water potential of the external solution is lower than that of the cytoplasm. Water leaves the potato cells via osmosis down the water potential gradient. This loss of water causes the vacuole and cytoplasm to shrink, pulling the cell membrane away from the cell wall, which results in plasmolysis and causes the tissue to become flaccid/limp.

(a) Max [2 marks]
- 1 mark for identifying sucrose concentration as the independent variable.
- 1 mark for identifying percentage change in mass as the dependent variable.

(b) Max [2 marks]
- 1 mark for estimating the value within the range of 0.34 M to 0.38 M.
- 1 mark for the justification: the concentration where there is zero net movement of water / 0% mass change.

(c) Max [3 marks]
- 1 mark for explaining that water moves out of the cells by osmosis / down a water potential gradient.
- 1 mark for stating that the cells become plasmolysed / flaccid.
- 1 mark for describing the cell membrane pulling away from the cell wall / loss of turgor pressure.

(a) The resistance of the population to streptomycin (as measured by MIC) increased overall over the 100 generations. The rate of resistance acquisition was non-linear: it rose gradually from 2.0 to 15.5 \(\mu\text{g mL}^{-1}\) by generation 50, and then increased exponentially/rapidly to 120.0 \(\mu\text{g mL}^{-1}\) by generation 100.
(b) Random mutations in the E. coli DNA generated genetic variation, with some individuals possessing alleles that provided a degree of resistance to streptomycin. When exposed to streptomycin (the selective agent/pressure), sensitive bacteria were killed, whereas those with resistant alleles survived. These surviving bacteria reproduced asexually and passed their resistance alleles to their offspring. Over many generations, the frequency of the resistance allele in the population increased, leading to a higher overall MIC.
(c) Practices such as overprescribing antibiotics for viral or minor infections, or patients failing to complete their fully prescribed course of antibiotics, leave semi-resistant bacteria alive to multiply and evolve.

(a) Max [2 marks]
- 1 mark for stating that resistance / MIC increased overall over the 100 generations.
- 1 mark for noting the rate change (e.g., slower increase initially up to generation 50, followed by a rapid/exponential increase up to generation 100).

(b) Max [4 marks]
- 1 mark for identifying that genetic variation arises in the bacterial population due to random mutation.
- 1 mark for identifying streptomycin as the selective pressure / environmental agent.
- 1 mark for stating that resistant bacteria survive and reproduce while non-resistant bacteria die.
- 1 mark for explaining that survivors pass on the resistance allele to offspring, increasing its frequency in the population over generations.

(c) Max [1 mark]
- 1 mark for overprescribing / incomplete courses / misuse of antibiotics in agriculture. (Accept any reasonable human medical misuse).

(a) As temperature increases from 15 \(^{\circ}\text{C}\) to 37 \(^{\circ}\text{C}\), the rate of translation increases from 3.2 to 18.1 aa/s (due to higher kinetic energy). However, further increase in temperature to 45 \(^{\circ}\text{C}\) results in a severe drop in the rate of translation to 2.1 aa/s.
(b) At 45 \(^{\circ}\text{C}\), the thermal energy is high enough to disrupt hydrogen bonds and other weak interactions holding the tertiary structure of proteins. Key enzymes and ribosomal proteins involved in protein synthesis (such as aminoacyl-tRNA synthetase or peptidyl transferase) denature, losing their functional 3D conformation and disabling translation.
(c) During elongation, the ribosome moves along the mRNA molecule in a 5' to 3' direction. tRNA molecules, each carrying a specific amino acid, enter the ribosome's A site. The anticodon of the tRNA binds to the complementary codon on the mRNA. The ribosome (specifically rRNA/peptidyl transferase) catalyzes the formation of a peptide bond between the amino acid in the A site and the growing polypeptide chain in the P site. The empty tRNA then exits via the E site.

(a) Max [2 marks]
- 1 mark for noting the increase in rate up to 37 \(^{\circ}\text{C}\) / stating that 37 \(^{\circ}\text{C}\) is the optimum.
- 1 mark for noting the sharp/drastic decrease at 45 \(^{\circ}\text{C}\).

(b) Max [2 marks]
- 1 mark for mentioning denaturation of proteins / enzymes at 45 \(^{\circ}\text{C}\).
- 1 mark for explaining that denaturation alters tertiary structure / active sites, destroying function / disrupting ribosomal structure.

(c) Max [3 marks]
- 1 mark for stating that tRNA molecules carry specific amino acids to the ribosome.
- 1 mark for describing complementary base pairing between tRNA anticodons and mRNA codons.
- 1 mark for stating that ribosomes catalyze the peptide bond formation / move along mRNA to elongate the polypeptide chain.

(a) Minute ventilation = Tidal volume \(\times\) Ventilation rate.
- At rest: \(0.5\text{ L} \times 12\text{ breaths/min} = 6.0\text{ L min}^{-1}\) (or \(\text{dm}^3\text{ min}^{-1}\))
- During exercise: \(2.4\text{ L} \times 35\text{ breaths/min} = 84.0\text{ L min}^{-1}\) (or \(\text{dm}^3\text{ min}^{-1}\))
(b) Inspiration requires an increase in the volume of the thoracic cavity to lower the internal pressure. The external intercostal muscles contract (while the internal intercostal muscles relax) to pull the ribcage up and out. Simultaneously, the diaphragm contracts and flattens (while abdominal muscles relax). These coordinated movements expand the thoracic cavity, lowering pressure inside the lungs below atmospheric pressure, forcing air to rush in.
(c) Type II pneumocytes are specialized cells in the alveolar wall that secrete a fluid containing surfactant. This surfactant reduces surface tension within the alveoli, preventing the walls from sticking together and preventing alveolar collapse during expiration, while also keeping the respiratory membrane moist.

(a) Max [2 marks]
- 1 mark for correct calculation for rest: 6.0 L min⁻¹ (or dm³ min⁻¹) AND exercise: 84.0 L min⁻¹ (or dm³ min⁻¹).
- 1 mark for correct units (L min⁻¹, L/min, dm³ min⁻¹, or dm³/min) associated with the numbers.

(b) Max [3 marks]
- 1 mark for stating that external intercostal muscles contract and internal intercostal muscles relax (to move ribs up/out).
- 1 mark for stating that the diaphragm contracts/flattens (and abdominal muscles relax).
- 1 mark for explaining that these actions increase thoracic volume, decreasing pressure below atmospheric pressure to draw air into lungs.

(c) Max [2 marks]
- 1 mark for stating they secrete pulmonary surfactant.
- 1 mark for explaining that surfactant reduces surface tension / prevents the alveoli from collapsing.

a. Fluid Mosaic Model:
- Phospholipids consist of a hydrophilic (water-attracting) polar phosphate head and hydrophobic (water-repelling) non-polar fatty acid tails.
- In an aqueous environment, phospholipids spontaneously assemble into a bilayer. The hydrophilic heads face outwards toward the aqueous extra- and intracellular environments, while the hydrophobic tails face inwards, shielded from water.
- Membrane proteins are interspersed within this bilayer. Integral proteins have hydrophobic regions that span the fatty acid core and hydrophilic regions exposed to the aqueous environments. Peripheral proteins are hydrophilic and attach to the surface of the membrane.
- Cholesterol molecules are amphipathic (having both hydrophilic and hydrophobic regions) and are embedded between phospholipids in animal membranes, where they regulate membrane fluidity and permeability.
- The membrane is 'fluid' because individual phospholipids and proteins can move laterally, and 'mosaic' because of the diverse, scattered arrangement of proteins across the bilayer.

b. Transport Mechanisms:
- Facilitated diffusion is a passive transport process that does not require ATP/metabolic energy because substances move down their concentration gradient (from high to low concentration).
- It utilizes specific, selective membrane proteins, which can be channel proteins (forming a pore) or carrier proteins (undergoing conformational changes).
- Example: Potassium ( K\(^+ \) ) channels in axon membranes allow rapid efflux of potassium during action potential repolarization.
- Active transport is an energy-requiring process that moves substances against their concentration gradient (from low to high concentration) using ATP.
- It involves protein pumps (integral carrier proteins) that hydrolyze ATP to drive a conformational change, pumping the substance across the membrane.
- Example: The sodium-potassium ( Na\(^+\)/K\(^+\) ) pump in axon membranes actively pumps three sodium ions ( 3Na\(^+\) ) out of the cell and two potassium ions ( 2K\(^+\) ) into the cell.

c. Osmolarity Effects:
- Osmosis is the passive movement of water molecules from a region of lower solute concentration (hypotonic / higher water potential) to a region of higher solute concentration (hypertonic / lower water potential) across a selectively permeable membrane.
- In a hypotonic solution, water moves into both cell types via osmosis.
- Animal cells lack a cell wall, so they swell and will eventually burst (lysis/hemolysis).
- Plant cells have a rigid cell wall that resists expansion; as water enters, internal pressure increases, making the cell turgid (providing structural support).
- In a hypertonic solution, water moves out of both cell types.
- Animal cells lose water, shrivel, and shrink (crenation).
- Plant cells lose water from their vacuoles, causing the cytoplasm to shrink away from the rigid cell wall, a state known as plasmolysis.
- In an isotonic solution, there is no net movement of water. Animal cells remain stable (normal volume), whereas plant cells become flaccid due to lack of turgor pressure.

Part a: [Max 4 marks]
- Phospholipids consist of hydrophilic phosphate heads and hydrophobic fatty acid tails [1].
- Phospholipids form a bilayer with heads facing outwards (aqueous environment) and tails facing inwards (shielded from water) [1].
- Integral proteins have hydrophobic regions embedded in the core and hydrophilic regions facing the outer/inner aqueous environment [1].
- Cholesterol is amphipathic (hydrophilic and hydrophobic regions) and regulates membrane fluidity in animal cells [1].
- Membrane is 'fluid' because components can move laterally and 'mosaic' due to the scattered proteins [1].

Part b: [Max 5 marks]
- Facilitated diffusion is passive / does not require ATP, moving solutes down a concentration gradient [1].
- Facilitated diffusion utilizes selective/specific channel or carrier proteins [1].
- Specific example of facilitated diffusion: e.g., potassium voltage-gated channels in axons / aquaporins for water / glucose transporters [1].
- Active transport requires ATP to move solutes against a concentration gradient [1].
- Active transport involves protein pumps that undergo conformational changes powered by ATP hydrolysis [1].
- Specific example of active transport: e.g., sodium-potassium pump (moves 3 Na\(^+ \) out and 2 K\(^+ \) in) [1].

Part c: [Max 6 marks]
- Osmosis is net movement of water from hypotonic to hypertonic solution across a selectively permeable membrane [1].
- In a hypotonic solution, water enters cells [1].
- Animal cells swell and burst (lysis) due to lack of cell wall, whereas plant cells become turgid because the cell wall limits expansion [1].
- In a hypertonic solution, water leaves cells [1].
- Animal cells shrink/shrivel (crenation) [1].
- Plant cells undergo plasmolysis / plasma membrane pulls away from cell wall [1].
- In isotonic solution, no net water movement occurs; animal cells remain normal while plant cells become flaccid [1].

Quality of Construction: [1 mark]
- 1 mark is awarded if the candidate writes a clear, structured response addressing all three parts (a, b, and c) using appropriate, accurate biological terminology throughout.

(a) Blotted cylinders ensure that only the mass of the tissue itself is measured. Surface water adhering to the cylinders would add variable, inaccurate extra mass.

(b) Calculation:
\(\text{Percentage change} = \frac{\text{Final Mass} - \text{Initial Mass}}{\text{Initial Mass}} \times 100\)
\(\text{Percentage change} = \frac{3.91 - 3.95}{3.95} \times 100 = \frac{-0.04}{3.95} \times 100 = -1.01\%\) (accept \(-1.0\%\) or \(-1\%\)).

(c) If a graph of percentage change in mass against sucrose concentration is plotted, the isotonic concentration is the point where the line of best fit intersects the x-axis (where there is \(0\%\) change in mass). This indicates no net movement of water. Reading from the data, this occurs between \(0.2\text{ mol dm}^{-3}\) (which gained mass) and \(0.4\text{ mol dm}^{-3}\) (which lost mass), specifically around \(0.36\) to \(0.38\text{ mol dm}^{-3}\).

(a)
* Award [1] for stating that blotting removes excess/surface water/liquid that would add to the final mass.

(b)
* Award [1] for showing correct working: \(\frac{3.91 - 3.95}{3.95} \times 100\).
* Award [1] for correct final answer of \(-1.01\%\) or \(-1.0\%\). (Do not award the second mark if the negative sign is missing, unless the word 'loss' is explicitly stated alongside \(1\%\)).

(c)
* Award [1] for stating that the isotonic point is where there is no net movement of water / \(0\%\) change in mass.
* Award [1] for stating that this is determined by finding the x-intercept (where the curve/line crosses the x-axis / \(0\%\) line) of the graph, yielding a value between \(0.36\) and \(0.38\text{ mol dm}^{-3}\).

(a) The independent variable is the concentration of hydrogen peroxide (substrate concentration). The dependent variable is the volume of oxygen gas produced in 30 seconds (rate of reaction).

(b) At higher substrate concentrations (above \(2.0\%\)), all of the active sites on the yeast catalase molecules are fully occupied/saturated. This means that increasing the substrate concentration further has no effect on the rate of reaction because enzyme concentration has become the limiting factor.

(c) To show that the catalase enzyme in the yeast is responsible, a control experiment should be set up where the yeast extract is either boiled (which denatures the enzyme) or replaced entirely with an equal volume of distilled water. All other variables (substrate concentration, temperature, pH) must remain identical. If no gas is produced in this trial, it confirms the active yeast catalase was responsible.

(a)
* Award [1] for correctly identifying both: Independent variable = hydrogen peroxide / substrate concentration AND Dependent variable = volume of oxygen gas produced (per unit time) / rate of reaction.

(b)
* Award [1] for stating that the enzyme active sites are saturated/fully occupied.
* Award [1] for stating that enzyme concentration becomes the limiting factor.

(c)
* Award [1] for describing the control treatment: use of boiled/denatured yeast extract OR replacing the yeast extract with distilled water.
* Award [1] for stating that all other variables must be kept constant and the amount of gas produced is monitored (expected to be zero).

(a) Calculation:
\(R_f = \frac{\text{distance traveled by pigment}}{\text{distance traveled by solvent front}} = \frac{36\text{ mm}}{80\text{ mm}} = 0.45\)
Identification: The blue-green pigment in chloroplasts is Chlorophyll a.

(b) The separation of pigments depends on two physical properties:
1. Solubility in the mobile phase/solvent: pigments that are highly soluble in the running solvent travel faster and further.
2. Adsorption/affinity to the stationary phase/paper: pigments that have high affinity for the cellulose fibers in chromatography paper form stronger intermolecular bonds and travel slower and less distance.

(c) Ink from a pen contains soluble pigments that would dissolve in the chromatography solvent and run up the paper along with the leaf pigments, contaminating the chromatogram. Pencil graphite is insoluble in the running solvent and remains at the origin.

(a)
* Award [1] for the correct calculation: \(R_f = 0.45\) (accept \(0.45\) without units).
* Award [1] for correctly identifying the pigment as Chlorophyll a.

(b)
* Award [1] for stating solubility of the pigment in the chromatography solvent / mobile phase.
* Award [1] for stating affinity / adsorption of the pigment to the chromatography paper / stationary phase.

(c)
* Award [1] for explaining that ink is soluble in the solvent and would run/separate with the plant pigments, whereas pencil graphite is insoluble and does not interfere.

### Part (a)
* **In situ conservation** involves preserving a species within its natural habitat. *Example:* Establishing a Marine Protected Area (MPA) or marine reserve to protect the coral reef habitat of green sea turtles.
* **Ex situ conservation** involves preserving a species outside of its natural habitat. *Example:* Captive breeding programs for endangered sea otters in public aquariums, or storing genetic material in gene banks.

### Part (b)
* Persistent organic pollutants (POPs) are highly stable chemical compounds that are resistant to environmental biodegradation.
* Producers (e.g., phytoplankton) absorb these toxins from water/sediment, leading to bioaccumulation within their cells.
* Because these substances are lipophilic (fat-soluble), they accumulate in the lipid-rich tissues of organisms and are not easily excreted or metabolized.
* As energy transfers up the food chain, consumers at each successive trophic level must ingest large quantities of biomass from the trophic level below them.
* Consequently, the concentration of the toxin increases at each successive trophic level (biomagnification), culminating in toxic concentrations in top predators.

### Part (c)
* **(i)**
$$\text{Biomagnification Factor} = \frac{\text{Concentration in tertiary consumer}}{\text{Concentration in producer}}$$
$$\text{Biomagnification Factor} = \frac{12.50 \text{ mg kg}^{-1}}{0.05 \text{ mg kg}^{-1}} = 250$$
The methylmercury concentration increased by a factor of 250.
* **(ii)** Two characteristics:
1. Lipid-solubility (not easily excreted in water-based waste).
2. Persistence / high chemical stability (not easily metabolized or broken down by enzymes).

### Part (d)
* An **indicator species** is an organism whose presence, absence, or abundance reflects a specific environmental condition or pollution level.
* A **biotic index** assigns a numerical tolerance rating (or score) to different indicator organisms based on their sensitivity or tolerance to pollution.
* By sampling the aquatic ecosystem and counting the abundance of these indicator species, an overall mathematical score is calculated.
* A high biotic index score indicates a healthy, unpolluted ecosystem (characterized by sensitive species like stonefly larvae), whereas a low score indicates a degraded or polluted ecosystem (dominated by tolerant species like tubifex worms).

### Part (e)
* **Size:** Larger reserves are more effective because they support larger populations, greater species richness, and higher trophic levels with a lower extinction risk.
* **Shape / Edge Effect:** Circular reserves are preferred over long, narrow ones because they minimize the perimeter-to-area ratio. This reduces "edge effects," where the environmental conditions at the boundary differ from the core interior habitat.
* **Corridors:** Connecting fragmented reserves with wildlife corridors allows gene flow, migration, and re-colonization between populations.
* **Active Management:** Humans must actively manage reserves (e.g., controlling invasive species, restoring native flora, and restricting human access) to maintain successional stages and protect endangered species.

### Part (a) [Max 4 marks]
* **Award 1 mark** for defining *in situ* conservation as preserving species within their natural habitat.
* **Award 1 mark** for defining *ex situ* conservation as preserving species outside their natural habitat.
* **Award 1 mark** for a valid marine *in situ* example (e.g., marine protected areas, marine reserves, restricting fishing zones).
* **Award 1 mark** for a valid marine *ex situ* example (e.g., public aquariums, captive breeding of marine mammals/fish, cryopreservation of coral gametes).

### Part (b) [Max 4 marks]
* **Award 1 mark** for stating that POPs are highly stable / persistent / not easily biodegraded.
* **Award 1 mark** for explaining that POPs are fat-soluble/lipophilic and accumulate in fatty tissues (bioaccumulation).
* **Award 1 mark** for stating that organisms at higher trophic levels must consume large quantities of biomass from lower levels to meet energy needs.
* **Award 1 mark** for concluding that the toxin concentration increases progressively at each successive trophic level, reaching its highest levels in top predators (biomagnification).

### Part (c) [Max 4 marks]
* **Part (i) [2 marks]:**
* **Award 1 mark** for showing the correct working: \(\frac{12.50}{0.05}\).
* **Award 1 mark** for the correct value: \(250\) (or \(250\times\)).
* **Part (ii) [2 marks]:**
* **Award 1 mark** for chemical stability / resistance to breakdown / persistence.
* **Award 1 mark** for lipid-solubility / solubility in fats.

### Part (d) [Max 4 marks]
* **Award 1 mark** for stating that indicator species are sensitive or tolerant to specific environmental conditions/pollution levels.
* **Award 1 mark** for explaining that a biotic index assigns numerical ratings to different indicator organisms based on their tolerance.
* **Award 1 mark** for stating that the abundance/presence of these organisms is sampled to calculate an overall numerical score.
* **Award 1 mark** for explaining that high scores indicate low pollution/high health, while low scores indicate high pollution/poor health.

### Part (e) [Max 4 marks]
* **Award 1 mark** for stating that larger areas are better as they support larger populations and greater biodiversity.
* **Award 1 mark** for explaining that circular shapes are better to minimize edge effects (boundary disturbances).
* **Award 1 mark** for explaining that wildlife corridors connect isolated patches to allow gene flow and migration.
* **Award 1 mark** for explaining that active management (e.g., culling invasive predators, habitat restoration) is necessary to maintain ecological stability.