2025 IB DP Biology Practice Paper with Answers

The parent plant is heterozygous for both traits (RrYy), exhibiting the dominant phenotypes for both flower color and seed shape. In a dihybrid cross of two heterozygous individuals (RrYy x RrYy) on different chromosomes, the expected phenotypic ratio of the offspring is 9:3:3:1. The proportion of offspring with the dominant phenotype for both traits (same as the parent) is 9/16. Therefore, the proportion of offspring with a different phenotype is 1 - 9/16 = 7/16.

Award 1 mark for the correct answer (C).

Glycolysis of one glucose molecule yields two molecules of pyruvate. In the link reaction, the two pyruvates are decarboxylated to form two acetyl-CoA, releasing 2 \(CO_2\) and 2 reduced NAD. In the Krebs cycle, the two acetyl-CoA molecules are completely oxidized, releasing 4 \(CO_2\) and 6 reduced NAD. Therefore, the combined total for the link reaction and the Krebs cycle is 6 \(CO_2\) (2 + 4) and 8 reduced NAD (2 + 6).

Award 1 mark for the correct answer (C).

Competitive inhibitors compete with the substrate for the active site. At very high substrate concentrations, the substrate outcompetes the inhibitor, allowing \(V_{max}\) to remain unchanged, but a higher concentration of substrate is needed to reach half-maximum velocity, which increases the \(K_m\) value (Inhibitor X). Non-competitive inhibitors bind to an allosteric site, decreasing the number of functional enzymes and lowering \(V_{max}\), while the remaining functional enzymes retain their normal affinity for the substrate, meaning \(K_m\) is unchanged (Inhibitor Y).

Award 1 mark for the correct answer (A).

Using the Lincoln index formula: \(N = \frac{n_1 \times n_2}{m_2}\), where \(n_1\) is the number of individuals initially marked (80), \(n_2\) is the total number of individuals captured in the second sample (100), and \(m_2\) is the number of marked individuals recaptured (20). Thus, \(N = \frac{80 \times 100}{20} = 400\).

Award 1 mark for the correct answer (B).

Because both species live in the same geographic region, their isolation is sympatric. The difference in mating seasons (early spring vs late summer) is a temporal barrier, and the difference in mating habitats (temporary pools vs permanent streams) is an ecological (habitat) barrier.

Award 1 mark for the correct answer (B).

In situ conservation (conservation of species in their natural habitats) allows organisms to continue interacting with their natural environment, maintaining ecological relationships and allowing them to undergo natural selection and evolutionary adaptation. Options A, C, and D describe benefits that are typically associated with ex situ conservation (conservation outside natural habitats, e.g., in zoos, seed banks, or captive breeding programs).

Award 1 mark for the correct answer (B).

The sodium-potassium pump is an active transport mechanism. In each cycle of active transport, the pump uses the energy from the hydrolysis of one ATP molecule to export 3 sodium ions (\(Na^+\)) out of the cell and import 2 potassium ions (\(K^+\)) into the cell against their concentration gradients.

Award 1 mark for the correct answer (A).

The mitotic index is defined as the ratio of the number of cells in mitosis (prophase + metaphase + anaphase + telophase) to the total number of cells observed. Number of cells in mitosis = 20 + 10 + 8 + 12 = 50. Total cells = 250. Mitotic Index = 50 / 250 = 0.20.

Award 1 mark for the correct answer (C).

The Lincoln index formula is given by: \( N = \frac{n_1 \times n_2}{m_2} \), 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 (150), and \( m_2 \) is the number of marked individuals recaptured in the second sample (30). Substituting the values: \( N = \frac{120 \times 150}{30} = 600 \).

Award 1 mark for the correct answer (B). No partial marks are awarded for multiple-choice questions.

In a dihybrid cross involving unlinked genes, the phenotypic ratio in the \( F_2 \) generation is \( 9:3:3:1 \). The parental phenotypes are yellow/round (dominant, ratio 9/16) and green/wrinkled (recessive, ratio 1/16). The recombinant phenotypes are yellow/wrinkled (3/16) and green/round (3/16). The expected number of yellow/wrinkled offspring is \( \frac{3}{16} \times 800 = 150 \).

Award 1 mark for the correct answer (B).

One molecule of glucose undergoes glycolysis to produce two molecules of pyruvate. In the link reaction, each pyruvate is decarboxylated to form acetyl-CoA, releasing 1 \( \text{CO}_2 \) molecule (total of 2 \( \text{CO}_2 \) for one glucose). In the Krebs cycle, each acetyl-CoA is completely oxidized, releasing 2 \( \text{CO}_2 \) molecules (total of 4 \( \text{CO}_2 \) for two acetyl-CoA). Combining these reactions, \( 2 + 4 = 6 \) molecules of \( \text{CO}_2 \) are produced.

Award 1 mark for the correct answer (C).

Non-competitive inhibitors bind to an allosteric site on the enzyme rather than the active site. This decreases the overall concentration of active enzymes, lowering the maximum velocity (\( V_{\max} \)). However, because the inhibitor does not compete with the substrate for the active site, the affinity of the substrate for the remaining active enzymes is unchanged, meaning the Michaelis constant (\( K_m \)) remains the same.

Award 1 mark for the correct answer (A).

Sympatric speciation occurs when a new species evolves from a single ancestral species while occupying the same geographic region. Polyploidy in plants like *Allium* results in reproductive isolation without geographic barriers, which is a classic example of sympatric speciation.

Award 1 mark for the correct answer (A).

Edge effects occur at the boundary of a habitat where conditions differ from the interior. To minimize edge effects, a reserve should have a low perimeter-to-area ratio. A large circular shape has the lowest possible perimeter for a given area compared to long, narrow, or highly fragmented shapes. A single large reserve is also preferred over several small ones as it prevents fragmentation of the interior habitat.

Award 1 mark for the correct answer (B).

The primary driving force of the transpirational pull is the evaporation of water from the wet cell walls of spongy mesophyll cells into the air spaces of the leaf, followed by diffusion out of the stomata. This loss of water creates a tension (negative pressure) that is transmitted down the xylem column due to the cohesive properties of water.

Award 1 mark for the correct answer (C).

The osmolarity of the plant tissue is equivalent to the concentration of sucrose where there is no net movement of water into or out of the cells (i.e., a \( 0\% \) change in mass). A \( 0\% \) change in mass lies between the sucrose concentrations of \( 0.2\text{ mol dm}^{-3} \) (where mass increased by \( 4\% \)) and \( 0.4\text{ mol dm}^{-3} \) (where mass decreased by \( 3\% \)). Interposing these values gives an estimated osmolarity of approximately \( 0.3\text{ mol dm}^{-3} \).

Award 1 mark for the correct answer (B).

When ATP synthase is selectively blocked, protons (\(H^+\)) can no longer flow back down their electrochemical gradient into the mitochondrial matrix. Because the electron transport chain (ETC) continues to pump protons out of the matrix into the intermembrane space, the proton gradient across the inner membrane increases and becomes extremely steep. Eventually, the electrochemical gradient becomes so high that the free energy released by the electron transport chain is insufficient to pump more protons against this steep gradient. As a result, the ETC slows down dramatically or stops, causing a corresponding decrease in the consumption of oxygen, which serves as the terminal electron acceptor.

[1 mark] Award for option A.
- Reject B, C, and D as blocking ATP synthase will prevent proton dissipation (increasing the gradient) and consequently feedback-inhibit the electron transport chain (decreasing oxygen consumption).

To find the recombination frequency, we first identify the recombinant offspring phenotypes, which are those that differ from the parental combinations (Purple-cut and Green-potato). These are Purple-potato (104) and Green-cut (96).

1. Sum the recombinant individuals: \(104 + 96 = 200\).
2. Calculate the total number of offspring: \(412 + 388 + 104 + 96 = 1000\).
3. Calculate the recombination frequency: \(\frac{200}{1000} \times 100 = 20\%\).

Since the recombination frequency is significantly less than \(50\%\), it indicates that the two gene loci are linked on the same chromosome.

[1 mark] Award for option A.
- Reject B, C, and D because the correct calculation of the recombination frequency is \(20\%\), and any frequency less than \(50\%\) represents linkage on the same chromosome rather than independent assortment.

Temporal isolation occurs when two species reproduce at different times of the day, seasons, or years, preventing them from interbreeding. Because the two species of lettuce flower during different seasons, they cannot cross-pollinate. This is a pre-zygotic barrier because it prevents the formation of a zygote in the first place.

[1 mark] Award for option A.
- Reject B, C, and D. Ecological (habitat) isolation refers to living in different habitats, whereas differences in reproductive timing are temporal. Pre-zygotic barriers occur before fertilization, ruling out post-zygotic barriers.

A competitive inhibitor structurally resembles the substrate and competes with it for binding at the active site of the enzyme. At very high substrate concentrations, the substrate outcompetes the inhibitor, allowing the reaction to reach its original maximum velocity (\(V_{\max}\) remains unchanged). However, because of the competition, more substrate is required to achieve half-maximal velocity, which increases the Michaelis constant (\(K_m\)).

[1 mark] Award for option A.
- Reject B and D because non-competitive inhibitors decrease \(V_{\max}\).
- Reject C because competitive inhibitors bind directly to the active site, not an allosteric site.

The estimated population size is calculated using the Lincoln index formula: \(N = \frac{n_1 \times n_2}{m_2}\)
- Where \(n_1 = 120\) (number of individuals marked and released)
- \(n_2 = 150\) (total caught in second sample)
- \(m_2 = 30\) (number of marked individuals recaptured)

\(N = \frac{120 \times 150}{30} = 600\).

For the Lincoln index to be accurate, we must assume a closed population (no significant immigration, emigration, births, or deaths between the release and recapture events).

[1 mark] Award for option A.
- Reject B because the marked beetles being easier to spot violates the assumption of equal catchability.
- Reject C and D because the calculation is incorrect (yields 450 instead of 600) and clumping violates the assumption of random mixing.

Forest fragmentation increases the ratio of edge habitat to interior habitat. Edge habitats have different abiotic and biotic conditions (e.g., more wind, lower humidity, different light levels, and higher predation) compared to the forest interior. Consequently, interior specialist species, which require stable deep-forest conditions, decline, while edge species may benefit or remain stable.

[1 mark] Award for option A.
- Reject B because the total edge perimeter increases relative to interior area during fragmentation.
- Reject C because fragmentation leads to isolation and localized extinctions, reducing species richness.
- Reject D because smaller isolated populations are highly vulnerable to genetic drift, which reduces genotypic diversity.

Cholesterol functions as a bidirectional regulator of membrane fluidity. At high temperatures (such as normal physiological body temperature, \(37^\circ\text{C}\)), cholesterol stabilizes the membrane by restricting the lateral movement of phospholipids, thereby decreasing fluidity. At low temperatures, cholesterol acts to disrupt the regular, tight packing of the phospholipid fatty acid tails, preventing the membrane from crystallizing or solidifying, thus maintaining fluidity.

[1 mark] Award for option A.
- Reject B because cholesterol prevents (rather than encourages) tight packing at low temperatures.
- Reject C and D because cholesterol reduces membrane permeability to small polar/hydrophilic molecules by filling in spaces between phospholipids, and it does not form channels.

In a diploid cell where \(2n = 12\), the haploid number \(n = 6\). There are 6 homologous pairs of chromosomes.
1. During meiosis I, homologous chromosomes normally separate. If non-disjunction of one pair occurs during Anaphase I, one of the two daughter cells receives both homologues (\(6 + 1 = 7\) chromosomes, each with two chromatids), while the other daughter cell receives neither (\(6 - 1 = 5\) chromosomes, each with two chromatids).
2. Meiosis II proceeds normally, separating the sister chromatids in both cells.
3. The cell with 7 chromosomes divides into two gametes, each with 7 chromosomes.
4. The cell with 5 chromosomes divides into two gametes, each with 5 chromosomes.
5. Therefore, the resulting gametes will be two with \(n = 7\) and two with \(n = 5\).

[1 mark] Award for option A.
- Reject B because non-disjunction in Anaphase I affects all four resulting gametes, leaving none with the normal haploid number of 6.
- Reject C because it describes normal meiosis with no non-disjunction.
- Reject D because it does not represent the correct chromosomal separation steps of meiosis.

At time \(t_1\), adding pyruvate provides the necessary substrate for the link reaction and Krebs cycle, generating NADH and \(\text{FADH}_2\). These coenzymes feed electrons into the electron transport chain (ETC), resulting in the reduction of oxygen to water, which increases the rate of oxygen consumption. At time \(t_2\), oligomycin inhibits ATP synthase, stopping the flow of protons back into the matrix. This stalls the ETC due to the accumulation of an extremely high proton gradient, which subsequently decreases the rate of oxygen consumption.

Award [1] mark for selecting the correct option (B). Pyruvate increases respiration/oxygen consumption while ATP synthase inhibition blocks the ETC and decreases oxygen consumption.

A competitive inhibitor binds to the active site of the enzyme, directly competing with the substrate. Because it can be outcompeted at very high substrate concentrations, the maximum rate of reaction (\(V_{\max}\)) can still be reached, meaning \(V_{\max}\) is unchanged. However, more substrate is required to reach half of \(V_{\max}\), which increases the \(K_m\) value.

Award [1] mark for identifying both competitive inhibition and active site binding.

Using the Lincoln Index: \(N = \frac{n_1 \times n_2}{m}\), where \(n_1 = 80\), \(n_2 = 60\), and \(m = 15\). Population size \(N = \frac{80 \times 60}{15} = 320\). A key assumption is that the marks do not affect the survival rate or behaviour (such as predator vulnerability or movement patterns) of the beetles.

Award [1] mark for the correct calculation of 320 and identifying the correct assumption.

The total number of offspring is \(412 + 388 + 98 + 102 = 1000\). The recombinants are the non-parental classes, which are \(Ab\) and \(aB\) (\(98 + 102 = 200\)). The recombination frequency is \(\frac{200}{1000} \times 100\% = 20\%\). Since this frequency is significantly less than \(50\%\), it indicates that the two genes are linked on the same chromosome.

Award [1] mark for the correct calculation of 20% and identifying that the genes are linked on the same chromosome.

The osmolarity of the plant tissue corresponds to the concentration of the external solution where there is no net change in mass (0% mass change). This point lies between 0.2 M and 0.4 M, approximately at 0.35 M. When placed in a 0.6 M solution (which is hypertonic relative to the tissue), water moves out of the tissue by osmosis.

Award [1] mark for identifying the correct osmolarity estimate and the correct direction of water movement.

Because the populations are geographically overlapping or adjacent (in the same grassland area), this represents sympatric speciation rather than allopatric speciation. The reproductive barrier is a difference in flowering times, which is temporal isolation.

Award [1] mark for identifying sympatric speciation and temporal isolation.

(a) The highest rate of glucose production is at 40 °C, which represents the optimum temperature under these conditions.

(b) To calculate the percentage increase:
\(\text{Percentage Increase} = \frac{\text{Rate at } 40 ^\circ\text{C} - \text{Rate at } 20 ^\circ\text{C}}{\text{Rate at } 20 ^\circ\text{C}} \times 100\)
\(\text{Percentage Increase} = \frac{6.2 - 2.5}{2.5} \times 100 = \frac{3.7}{2.5} \times 100 = 148\%\).

(c) At 40 °C, the rate is at its peak because kinetic energy is high, facilitating substrate-active site collisions without destabilizing the protein structure. At 60 °C, the excessive thermal energy disrupts intramolecular bonds (hydrogen and ionic bonds) stabilizing the enzyme's tertiary structure. This leads to denaturation, causing the active site to lose its complementary shape to the substrate, preventing substrate binding and decreasing the reaction rate dramatically.

(d) An appropriate control involves using an identical setup but with inactive lactase (boiled/denatured) or substituting the enzyme solution with an equal volume of distilled water to confirm that glucose production is catalyzed by active enzyme molecules rather than occurring spontaneously.

Part (a) [1 mark]:
- Correctly states 40 °C (accept 40 with or without units).

Part (b) [1.25 marks]:
- Shows correct working using the values 6.2 and 2.5: \(((6.2 - 2.5) / 2.5) \times 100\) [0.5 marks]
- Correct calculation to give 148% (accept 148.0% or 150%) [0.75 marks]

Part (c) [2 marks max]:
- Explains high rate at 40 °C due to high kinetic energy / frequent successful collisions while active site is functional [1 mark].
- Explains low rate at 60 °C due to denaturation (breaking of hydrogen/ionic bonds) altering the active site shape [1 mark].
- Reject "enzyme is dead/killed".

Part (d) [2 marks max]:
- Identifies correct control medium: use boiled/denatured lactase OR replace lactase with distilled water/buffer [1 mark].
- Explains purpose: to demonstrate that glucose production is dependent on catalytic action of the functional enzyme [1 mark].

(a) The independent variable is the factor changed by the experimenter (sucrose concentration). The dependent variable is what is measured (mean percentage change in mass).

(b) The isotonic point is where there is no net movement of water, corresponding to 0% change in mass. Looking at the data, the mass change is +6.2% at 0.2 mol dm\(^{-3}\) and -1.5% at 0.4 mol dm\(^{-3}\). The point of 0% change lies between 0.34 and 0.38 mol dm\(^{-3}\) (linear interpolation gives approximately 0.36 mol dm\(^{-3}\)).

(c) A solution of 0.0 mol dm\(^{-3}\) sucrose (distilled water) is hypotonic to the cytoplasm of the potato cells (which contains dissolved solutes). Water moves from an area of higher water potential (the surrounding solution) to an area of lower water potential (inside the potato cells) across the semi-permeable cell membrane via osmosis, resulting in net water gain and an increase in mass.

(d) To ensure a fair test, variables such as the surface area and volume of the potato cylinders, the duration of time the cylinders remain in the solutions, the potato variety/source, and the temperature of the solutions must be controlled.

Part (a) [1.25 marks]:
- Correctly identifies independent variable as sucrose concentration [0.5 marks]
- Correctly identifies dependent variable as (percentage) change in mass [0.75 marks]

Part (b) [1 mark]:
- Provides an estimate between 0.34 and 0.38 mol dm\(^{-3}\) with correct units (accept mol/dm3 or M) [1 mark].

Part (c) [2 marks max]:
- Mentions that the 0.0 mol dm\(^{-3}\) solution has a higher water potential (or is hypotonic) compared to the potato cells [1 mark].
- Explains that water enters the potato cells down the water potential gradient by osmosis, increasing mass [1 mark].

Part (d) [2 marks max]:
- Award [1 mark] for each valid controlled variable listed (up to 2):
* Potato cylinder surface area / dimensions / volume
* Same potato source / variety / age
* Temperature of the solutions
* Duration of immersion
* Complete removal of excess surface liquid before weighing

(a) The role of potassium hydroxide (KOH) is to absorb the carbon dioxide (\(\text{CO}_2\)) produced by the cellular respiration of the germinating seeds. This ensures that any change in gas volume within the chamber is solely due to oxygen consumption.

(b) The rate is calculated as:
\(\text{Rate} = \frac{\text{Total distance moved}}{\text{Total time}} = \frac{47 \text{ mm}}{20 \text{ minutes}} = 2.35 \text{ mm min}^{-1}\).

(c) During aerobic respiration, germinating seeds absorb oxygen (\(\text{O}_2\)) gas from the air. Although they release \(\text{CO}_2\) gas in equal volume (assuming a respiratory quotient of 1.0), the \(\text{CO}_2\) is absorbed by the KOH. As a result, the total volume of gas in the sealed chamber decreases, creating a vacuum/lower air pressure inside. The higher external atmospheric pressure pushes the liquid droplet along the capillary tube toward the chamber.

(d) Increasing the temperature to 32 °C increases the kinetic energy of both enzymes (such as decarboxylases and dehydrogenases involved in the Link reaction and Krebs cycle) and substrates. This increases the frequency of successful collisions, speeding up cellular respiration. Consequently, more oxygen is consumed per unit time, resulting in a faster rate of droplet movement.

Part (a) [1.25 marks]:
- Correctly states that KOH absorbs carbon dioxide [1.25 marks].
- Accept chemical formula: CO2.

Part (b) [1 mark]:
- Correctly calculates \(47 / 20 = 2.35\) with units of mm min\(^{-1}\) [1 mark]. Accept 2.4 mm min\(^{-1}\).

Part (c) [2 marks max]:
- Explains that oxygen uptake decreases the gas volume/pressure inside the chamber [1 mark].
- States that carbon dioxide produced is removed by KOH, meaning there is no gas replacement, creating a pressure gradient that draws the fluid in [1 mark].

Part (d) [2 marks max]:
- Predicts that the rate of movement will increase [1 mark].
- Explains that higher temperature increases kinetic energy of enzymes/reactants, leading to more frequent collisions and a faster rate of aerobic respiration [1 mark].

(a) The relationship demonstrates a typical predator-prey cyclic interaction. There is a lag phase where changes in the predator (Species B) population follow changes in the prey (Species A) population. As Species A density peaks in Year 2, Species B rises to its peak in Year 3. This high predator pressure drives Species A density down to a minimum in Year 4, which subsequently causes Species B to crash in Year 5 due to lack of food.

(b) To calculate the percentage decrease:
\(\text{Decrease} = \frac{\text{Population in Year 2} - \text{Population in Year 4}}{\text{Population in Year 2}} \times 100\)
\(\text{Decrease} = \frac{85 - 15}{85} \times 100 = \frac{70}{85} \times 100 \approx 82.35\%\) (or 82.4%).

(c) Abiotic factors that influence ecosystems in controlled greenhouses include temperature, light intensity/duration, humidity, water supply, and carbon dioxide levels.

(d) A small mesocosm has structural and space limitations. The lack of complex spatial niches/refuges means the predator (Species B) can easily find and consume all prey (Species A), causing the prey to go extinct and subsequently causing the predator to starve. Additionally, a closed system can experience nutrient depletion or a collapse of the primary producer base if overgrazed by the herbivore.

Part (a) [2 marks max]:
- Identifies the relationship as predator-prey / cyclic oscillations [1 mark].
- Mentions the lag phase (predator population peaks/troughs after the prey population) [1 mark].

Part (b) [1.25 marks]:
- Shows correct working using 85 and 15: \(((85 - 15) / 85) \times 100\) [0.5 marks].
- Correct final value of 82.4% (accept 82.35% or 82%) [0.75 marks].

Part (c) [1 mark]:
- Lists two correct abiotic factors (e.g., temperature, humidity, light levels, water, CO2) [0.5 marks per factor].

Part (d) [2 marks max]:
- Explains that the limited space/lack of refuges allows the predator to overexploit/wipe out the prey [1 mark].
- Explains that resource depletion (e.g., loss of autotroph food supply for the herbivore, or build-up of metabolic wastes) can lead to system collapse [1 mark].

(a) Identify the initial increase (0 to 4 days) and the subsequent decrease (4 to 10 days) to describe the full trend.
(b) Since one species completely outcompetes and eliminates the other, this represents the competitive exclusion principle (Gause's principle).
(c) Common resources for aquatic heterotrophic ciliates include food/prey (bacteria) and dissolved oxygen or physical space.
(d) The carrying capacity is reached when the population size levels off (around 95-100 individuals per mL).

(a) [2 marks]
- 1 mark for stating that C. minor increases initially / peaks at Day 4 (32 individuals per mL).
- 1 mark for stating that it decreases after Day 4 / drops to zero by Day 10.

(b) [1 mark]
- 1 mark for competitive exclusion principle (accept Gause's principle).

(c) [2 marks]
- 1 mark per valid resource up to 2 (e.g., food/bacteria, space, dissolved oxygen). Reject light, carbon dioxide (non-photosynthetic organisms).

(d) [0.66 marks]
- 0.66 marks for stating 100 (accept range 95 to 100).

(a) Unlinked genes sort independently to yield a 1:1:1:1 ratio. Since parental combinations (Purple/cut and Green/potato) are highly overrepresented, the genes must be linked on the same chromosome.
(b) Recombinants are those offspring with non-parental phenotypes (Purple/potato = 12, Green/cut = 14). Total offspring = 306. Recombination frequency = \((26 / 306) \times 100 = 8.50\)% (accept 8.5%).
(c) Since the parents are AaBb and aabb, the recombinant gametes produced by the heterozygous parent are Ab and aB. Combined with the ab gamete from the test-cross parent, the recombinant genotypes are Aabb and aaBb.

(a) [2 marks]
- 1 mark for stating that the genes are linked / on the same chromosome.
- 1 mark for explaining that linkage prevents independent assortment / recombinants only arise through crossing over.

(b) [2 marks]
- 1 mark for correct working: sum of recombinants (26) divided by total offspring (306) multiplied by 100.
- 1 mark for correct final value of 8.5% (accept 8.49% to 8.53%).

(c) [1.66 marks]
- 0.83 marks for each correct genotype: Aabb and aaBb.

(a) Competitive inhibitors compete for the active site, meaning their effect can be minimized by high substrate concentrations (hence \(V_{max}\) is unchanged; Inhibitor X fits this). Non-competitive inhibitors bind to an allosteric site, decreasing active site function regardless of substrate level (hence reducing \(V_{max}\); Inhibitor Y fits this).
(b) By binding to the active site, competitive inhibitors raise the substrate concentration needed to reach half of \(V_{max}\) (increasing \(K_m\)), while \(V_{max}\) remains constant.
(c) With a non-competitive inhibitor, the active site structure is altered; increasing substrate concentration cannot displace the inhibitor, so the rate remains capped at a much lower limit.

(a) [2.66 marks]
- 0.5 marks for classifying Inhibitor X as competitive.
- 0.83 marks for justifying Inhibitor X with data (recovers original \(V_{max}\) of ~85 units).
- 0.5 marks for classifying Inhibitor Y as non-competitive.
- 0.83 marks for justifying Inhibitor Y with data (\(V_{max}\) is permanently reduced to 34 units).

(b) [2 marks]
- 1 mark for stating that \(V_{max}\) is unaffected / remains the same.
- 1 mark for stating that \(K_m\) is increased / affinity decreases.

(c) [1 mark]
- 1 mark for stating that it cannot overcome the inhibition / does not restore \(V_{max}\) / rate remains low.

(a) Yeast readily takes up and phosphorylates glucose, initiating glycolysis. Lactose requires lactose-specific transporters and beta-galactosidase (lactase), which are absent or not expressed in yeast.
(b) Anaerobic respiration in yeast (fermentation) converts glucose to ethanol and carbon dioxide.
(c) Even without external substrate, yeast cells can mobilize internal storage molecules (like glycogen) to generate ATP via respiration, yielding a small amount of carbon dioxide.
(d) During the energy payoff phase of glycolysis, glyceraldehyde-3-phosphate is oxidized. This requires \(NAD^+\) to accept electrons, converting it to NADH.

(a) [2 marks]
- 1 mark for stating that glucose is a monosaccharide directly utilized in glycolysis / yeast has the necessary enzymes.
- 1 mark for stating that yeast lacks the enzyme (lactase) to hydrolyze lactose.

(b) [1 mark]
- 1 mark for: Glucose \(\rightarrow\) ethanol + carbon dioxide. Accept: Glucose \(\rightarrow\) alcohol + carbon dioxide. Reject equations producing lactic acid.

(c) [1 mark]
- 1 mark for explaining that yeast uses internal/stored carbohydrates/reserves.

(d) [1.66 marks]
- 1 mark for stating that \(NAD^+\) is reduced to NADH / accepts hydrogen atoms or electrons.
- 0.66 marks for stating that this allows glycolysis to continue / allows oxidation of triose phosphate.

(a) Species richness is the total number of species present (3 in both). Species evenness is the relative abundance of each species (Community B is more even).
(b) Calculate the components for Community A:
- \(N = 80 + 10 + 10 = 100\)
- \(N(N-1) = 100 \times 99 = 9900\)
- \(\sum n(n-1) = 80(79) + 10(9) + 10(9) = 6320 + 90 + 90 = 6500\)
- \(D = 9900 / 6500 = 1.523\) (accept 1.52).
(c) In situ conservation maintains ecological integrity, keeps food webs intact, and avoids the selection pressures of artificial environments.

(a) [2 marks]
- 1 mark for stating that both have the same species richness (3 species).
- 1 mark for stating that Community B has higher species evenness (more balanced abundance).

(b) [2.66 marks]
- 1.66 marks for showing correct substitution into the formula (showing 9900 and 6500).
- 1 mark for correct calculation of \(D = 1.52\) (accept range 1.52 to 1.523).

(c) [1 mark]
- 1 mark for any valid advantage (e.g., preserves natural habitat, allows continued natural evolution, preserves ecological relationships, cheaper/less stressful for many species).

(a) By interpolating the data, the point of zero mass change lies between 0.2 M (positive change) and 0.4 M (negative change), close to 0.35 M.
(b) Since 0.8 M is highly concentrated, the water potential outside is lower than inside. Water moves out by osmosis, shrinking the cells and lowering mass.
(c) The investigator manipulates the concentration (independent) and measures the resulting mass change (dependent).

(a) [2 marks]
- 1 mark for estimating osmolarity in the range of 0.33 M to 0.37 M.
- 1 mark for explaining that this is where there is no net change in mass / isotonic point.

(b) [2 marks]
- 1 mark for stating that the solution is hypertonic to the cells / has a lower water potential than the tissue.
- 1 mark for stating that water leaves the cells by osmosis, resulting in mass loss.

(c) [1.66 marks]
- 0.83 marks for identifying the independent variable as sucrose concentration.
- 0.83 marks for identifying the dependent variable as percentage change in mass.

(a) Glycolysis:
- Occurs in the cytoplasm of the cell.
- Glucose is phosphorylated using two molecules of ATP to form hexose bisphosphate.
- This phosphorylation activates the glucose molecule, making it less stable.
- Lysis (splitting) of the hexose bisphosphate occurs, producing two molecules of triose phosphate.
- Triose phosphate is oxidized by the removal of hydrogen/electrons.
- NAD+ is reduced to NADH (or NADH + H+).
- Energy released during this oxidation is used to synthesize ATP via substrate-level phosphorylation.
- This results in a net yield of 2 ATP (4 produced, 2 consumed) and 2 molecules of pyruvate.

(b) Electron Transport Chain and ATP Synthesis:
- NADH and FADH2 (reduced coenzymes) deliver electrons and protons (hydrogen ions) to the electron transport chain (ETC) on the inner mitochondrial membrane (cristae).
- Electrons are passed along a series of electron carriers/proteins in the ETC.
- As electrons flow through the ETC, they release energy.
- This energy is used to pump protons (H+) from the mitochondrial matrix into the intermembrane space.
- A high concentration gradient of protons is established in the intermembrane space relative to the matrix.
- Protons diffuse down their concentration gradient back into the matrix through the enzyme ATP synthase (chemiosmosis).
- The flow of protons through ATP synthase causes it to rotate, providing the energy to phosphorylate ADP and inorganic phosphate (Pi) into ATP.
- Oxygen acts as the terminal electron acceptor, combining with electrons and protons to form water, maintaining the proton gradient.

(c) Anaerobic Respiration in Humans vs Yeast:
- In humans, pyruvate is converted into lactate (lactic acid); in yeast, pyruvate is converted into ethanol and carbon dioxide (CO2).
- No carbon dioxide is produced in human anaerobic respiration, whereas decarboxylation occurs in yeast to produce CO2.
- The conversion of pyruvate to lactate in humans is reversible (lactate can be converted back to pyruvate in the liver); the conversion in yeast is irreversible.
- Both processes occur in the cytoplasm and do not require oxygen.
- Both pathways regenerate oxidized NAD+ from NADH, allowing glycolysis to continue and produce a small yield of ATP (2 ATP per glucose molecule).

(Quality Mark):
- 1 mark is awarded if the response is well-structured, easy to follow, and uses clear, appropriate biological terminology across all three parts.

Part (a) [Max 5 marks]:
- cytoplasm: award 1 mark for stating glycolysis occurs in the cytoplasm.
- phosphorylation: award 1 mark for describing phosphorylation of glucose using 2 ATP.
- activation: award 1 mark for stating that phosphorylation activates glucose/makes it unstable.
- lysis: award 1 mark for describing the splitting of hexose bisphosphate into two triose phosphates.
- oxidation: award 1 mark for describing the oxidation of triose phosphate / reduction of NAD+ to NADH.
- ATP yield: award 1 mark for stating substrate-level phosphorylation produces a net of 2 ATP (and 2 pyruvate).

Part (b) [Max 6 marks]:
- electron source: award 1 mark for NADH/FADH2 delivering electrons to the inner membrane / cristae.
- electron flow: award 1 mark for electrons passing through a series of carriers in the ETC.
- proton pumping: award 1 mark for energy from electrons being used to pump protons into the intermembrane space.
- gradient: award 1 mark for the establishment of a proton / electrochemical gradient.
- chemiosmosis: award 1 mark for protons diffusing back into the matrix through ATP synthase.
- phosphorylation: award 1 mark for ATP synthase converting ADP and Pi to ATP.
- oxygen role: award 1 mark for oxygen acting as the terminal electron acceptor, forming water.

Part (c) [Max 4 marks]:
- human product: award 1 mark for stating human anaerobic respiration produces lactate.
- yeast product: award 1 mark for stating yeast anaerobic respiration produces ethanol and CO2.
- reversibility: award 1 mark for identifying that lactate production is reversible whereas ethanol production is irreversible.
- NAD+ regeneration: award 1 mark for explaining that both pathways regenerate NAD+ to allow glycolysis to continue.

Quality of Construction [1 mark]:
- Award 1 mark if the candidate writes in clear, continuous prose, structures the answers logically according to parts (a), (b), and (c), and uses correct scientific terminology throughout.