2025 IB DP Biology Practice Paper with Answers

The Lincoln index formula used for capture-mark-recapture assumes that the marking technique does not affect the survival rate or behavior of the marked individuals. If the paint made the woodlice more vulnerable to predators or altered their movement, the proportion of marked individuals recaptured would not represent the true population ratio.

Award [1] for the correct option (C). Reject all other options as they are either incorrect assumptions or violate the validity of the Lincoln index.

When two species compete for the same resource, natural selection favors individuals of each species that avoid direct competition. This leads to evolutionary changes in physical traits (like beak size) in the zone of overlap, a phenomenon known as character displacement due to competition.

Award [1] for the correct option (A). Options B, C, and D are incorrect ecological descriptions of this scenario.

The pink-flowered plant has the genotype C^R C^W and the white-flowered plant has the genotype C^W C^W. Crossing these two plants (C^R C^W x C^W C^W) yields offspring genotypes C^R C^W (pink) and C^W C^W (white) in a 1:1 ratio. Therefore, the expected phenotypic ratio is 1 pink : 1 white.

Award [1] for the correct option (C). All other options represent incorrect phenotypic ratios based on codominant genetic crosses.

Injecting pre-formed antibodies from another organism provides passive immunity. Because the patient's own B lymphocytes are not activated by an antigen to divide and produce memory cells, this treatment does not result in long-term immunological memory.

Award [1] for the correct option (D). Options A, B, and C are incorrect because antivenom provides passive, not active, immunity, which does not generate memory cells.

Competitive inhibitors bind to the active site of the enzyme and compete directly with the substrate. Increasing the substrate concentration increases the probability that a substrate molecule, rather than an inhibitor molecule, will bind to the active site, thereby overcoming the inhibition and restoring Vmax.

Award [1] for the correct option (B). Non-competitive and irreversible inhibitors cannot be overcome by increasing substrate concentration, making options A, C, and D incorrect.

High blood glucose levels are detected by the beta cells of the pancreas, which respond by secreting insulin. Insulin stimulates body cells to take up glucose and directs the liver and muscle cells to convert glucose into glycogen (glycogenesis), lowering blood glucose levels.

Award [1] for the correct option (D). Alpha cells produce glucagon when glucose is low, and insulin promotes glycogen synthesis rather than breakdown, ruling out A, B, and C.

DNA polymerase I possesses a 5' to 3' exonuclease activity that allows it to remove RNA primers of Okazaki fragments and replace them with DNA nucleotides. DNA polymerase III is responsible for the main elongation, DNA ligase joins the nicks, and helicase unwinds the double helix.

Award [1] for the correct option (A). Options B, C, and D identify other replication enzymes with distinct functions.

Shrinking or crenation of animal cells occurs when they are placed in a hypertonic solution (a solution with a higher osmolarity than the cytoplasm). Water leaves the cells by osmosis, moving down its concentration gradient (from high water potential to low water potential).

Award [1] for the correct option (B). Option A is incorrect because water would enter in a hypotonic solution; C is incorrect because isotonic solutions cause no net movement; D is incorrect because water transport in this scenario is passive osmosis, not active transport.

Using the Lincoln index formula: \(N = \frac{n_1 \times n_2}{m_2}\), where \(n_1 = 80\) (initially marked), \(n_2 = 100\) (second sample size), and \(m_2 = 16\) (marked recaptures). \(N = \frac{80 \times 100}{16} = \frac{8000}{16} = 500\). A critical assumption of the capture-mark-recapture method is that marked individuals mix randomly and redistribute evenly back into the population before the second sample is taken.

Award 1 mark for the correct calculation (500) and the correct identifying assumption. Reject all other combinations.

The woman must be heterozygous for blood group A (\(I^A i\)) because her father had blood group O (\(ii\)) and could only pass on the \(i\) allele. The man has blood group AB, so his genotype is \(I^A I^B\). Crossing these two genotypes (\(I^A i \times I^A I^B\)) yields the following possible offspring genotypes: \(I^A I^A\) (Group A), \(I^A I^B\) (Group AB), \(I^A i\) (Group A), and \(I^B i\) (Group B). Out of 4 possible combinations, 2 result in blood group A. The probability is therefore \(2/4 = 0.50\).

Award 1 mark for identifying the correct probability of 0.50 (B). Other responses are incorrect.

The primary immune response occurs upon first exposure to an antigen, characterized by a lag phase while naive B-cells are selected, cloned, and differentiated into plasma cells. The secondary immune response occurs upon re-exposure, where memory B-cells rapidly respond, producing a much larger quantity of antibodies in a shorter period of time.

Award 1 mark for the correct option (C). Other options describe incorrect immune system features or swap the characteristics of the two responses.

A competitive inhibitor binds to the active site of the enzyme, competing directly with the substrate. Because it competes for the same site, increasing the substrate concentration increases the likelihood that a substrate molecule, rather than the inhibitor, will bind to the active site, thereby overcoming the inhibition.

Award 1 mark for the correct option (C). Non-competitive inhibitors bind to an allosteric site and their inhibition cannot be overcome by increasing substrate concentration.

Following a meal rich in carbohydrates, blood glucose levels rise. In response, beta cells in the pancreatic islets secrete insulin. Insulin stimulates liver and muscle cells to take up glucose from the blood and store it as glycogen, returning blood glucose levels to the normal range.

Award 1 mark for the correct cell type and hormone action (B). Glucagon is secreted by alpha cells when blood glucose is low, not high.

Energy at the producer level (phytoplankton) = \(12,000 \text{ kJ m}^{-2} \text{ yr}^{-1}\). Energy transferred to the primary consumers (zooplankton) = \(12,000 \times 0.10 = 1,200 \text{ kJ m}^{-2} \text{ yr}^{-1}\). Energy transferred to secondary consumers (small fish) = \(1,200 \times 0.12 = 144 \text{ kJ m}^{-2} \text{ yr}^{-1}\).

Award 1 mark for the correct calculation (A). Reject all other numeric values.

The man's genotype is \(X^B Y\) (normal vision) and the woman's genotype is \(X^B X^b\) (carrier). The possible genotypes for their offspring are: \(X^B X^B\) (normal female), \(X^B X^b\) (carrier female), \(X^B Y\) (normal male), and \(X^b Y\) (color blind male). Each of these four outcomes has an equal probability of \(25\%\) (or \(0.25\)). Therefore, the probability that their first child is a color-blind son (\(X^b Y\)) is \(25\%\).

Award 1 mark for the correct percentage of 25% (B). Option C (50%) represents the probability of a son being color blind *given* that the child is male, but the question asks for the probability of the child being both a son and color-blind.

Monoclonal antibodies are produced by fusing antigen-activated B-lymphocytes (which produce the desired antibody but have a limited lifespan) with myeloma (cancerous B-cell tumor) cells, which can divide endlessly in culture. The resulting hybridoma cell retains both properties.

Award 1 mark for identifying the correct cell fusion partner types (C). Other choices present incorrect immune cell combinations.

The Lincoln index formula is \( N = \frac{M \times n}{m} \), where \( N \) is the estimated population size, \( M \) is the number of individuals marked on the first day, \( n \) is the total number of individuals recaptured, and \( m \) is the number of marked individuals recaptured. Because 10% of the original marks washed off, the actual number of successfully marked snails remaining in the population is \( 120 \times 0.90 = 108 \). Placing these corrected values into the formula gives \( N = \frac{108 \times 150}{30} = 540 \). If the loss of marks had been ignored, the estimate would have been 600, which is incorrect.

[1 mark] for option B. Award 1 mark for calculating the corrected marked individuals (108) and correctly applying the Lincoln index formula to get 540. Deduct 0 marks for wrong answers.

Ecological efficiency is calculated as the ratio of net productivity at a higher trophic level to the net productivity at the preceding trophic level, expressed as a percentage. To find the transfer efficiency from primary consumers to secondary consumers: \( \text{Efficiency} = \frac{\text{Net productivity of secondary consumers}}{\text{Net productivity of primary consumers}} \times 100 = \frac{172.8}{2160} \times 100 = 8.0\% \).

[1 mark] for option B. Award 1 mark for correct calculation: (172.8 / 2160) * 100 = 8%.

The parental phenotypes are gray body/normal wings and black body/vestigial wings, which represent the original linkage groups. The recombinant phenotypes are gray body/vestigial wings (105) and black body/normal wings (95). Total offspring = \( 415 + 385 + 105 + 95 = 1000 \). The total number of recombinants is \( 105 + 95 = 200 \). The recombination frequency is calculated as: \( \frac{\text{Recombinants}}{\text{Total offspring}} \times 100 = \frac{200}{1000} \times 100 = 20\% \).

[1 mark] for option B. Award 1 mark for identifying the recombinant classes, summing them to find 200, and dividing by the total of 1000 offspring to obtain 20%.

The woman must be heterozygous \( I^A i \) because her father was blood group O (genotype \( ii \)). The man must be heterozygous \( I^B i \) because his mother was blood group O (genotype \( ii \)). A cross between \( I^A i \) and \( I^B i \) produces offspring genotypes: \( I^A I^B \) (group AB, probability 1/4), \( I^A i \) (group A, probability 1/4), \( I^B i \) (group B, probability 1/4), and \( ii \) (group O, probability 1/4). The probability of the first child being AB (1/4) AND the second child being O (1/4) is: \( \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} = 0.0625 \) or 6.25%.

[1 mark] for option A. Award 1 mark for correctly determining the parental genotypes, finding individual offspring probabilities (25% each), and applying the product rule for independent events (0.25 * 0.25 = 0.0625).

On Day 28, the exposure to Antigen X triggers a secondary immune response because memory cells specific to Antigen X already exist from the Day 0 exposure. This results in a rapid and high-volume production of anti-X antibodies. In contrast, the exposure to Antigen Y on Day 28 represents a primary immune response, which has a lag phase and produces a much lower concentration of anti-Y antibodies. Therefore, anti-X antibody levels will be significantly higher than anti-Y antibody levels on Day 35.

[1 mark] for option B. Award 1 mark for recognizing that a secondary immune response (to X) is faster and of greater magnitude than a primary immune response (to Y).

An individual with blood group B negative (B-) has B antigens on their red blood cells and produces anti-A antibodies. Being Rh-negative, they also have the potential to produce anti-Rh antibodies if exposed to Rh-positive blood. Therefore, they cannot receive blood containing the A antigen (groups A and AB) or the Rh antigen (Rh-positive groups). The only safe donor options are B- and O- (universal donor), as O- lacks A, B, and Rh antigens.

[1 mark] for option C. Award 1 mark for correctly identifying that O negative is compatible because it lacks A, B, and Rh antigens, whereas the other options present incompatible antigens.

A non-competitive inhibitor binds to an allosteric site (rather than the active site) of the enzyme. This binding reduces the overall rate of catalysis (decreasing \( V_{\max} \)), and this effect cannot be overcome by adding more substrate. However, because the inhibitor does not compete with the substrate for the active site, the affinity of the enzyme for the substrate (represented by \( K_m \)) remains unchanged.

[1 mark] for option B. Award 1 mark for identifying that a decrease in V_max with an unchanged K_m is characteristic of non-competitive inhibition.

Dehydration increases the solute concentration (osmolarity) of the blood. Osmoreceptors in the hypothalamus detect this increase and stimulate the posterior pituitary gland to secrete more antidiuretic hormone (ADH) into the blood. ADH travels to the kidneys, where it binds to receptors on the collecting duct cells, promoting the insertion of aquaporins into the membrane. This increases the permeability of the collecting ducts to water, allowing more water to be reabsorbed back into the body.

[1 mark] for option B. Award 1 mark for correctly identifying the physiological sequence of events involved in osmoregulation during dehydration.

Density-dependent factors are those whose effects on the size or growth of the population vary with the population density. Examples include resource depletion, predation, and disease transmission. In this scenario, a lethal virus will transmit more rapidly as the population density increases, slowing down growth. The other options (volcanic eruptions, cold winters, and rainfall changes) are physical, abiotic factors that act independently of population density (density-independent).

[1 mark] B is the correct answer because disease transmission rate is density-dependent. Award 0 marks for any other option.

The parental phenotypes are tall red and short white, which are represented by the majority of the offspring (412 and 408). The recombinant phenotypes are the non-parental combinations: tall white (89) and short red (91). To calculate the recombination frequency, sum the recombinants and divide by the total number of offspring: \((89 + 91) / (412 + 89 + 91 + 408) \times 100 = 180 / 1000 \times 100 = 18\%\).

[1 mark] B is the correct answer. Recombinants make up 180 out of 1000 offspring. Award 0 marks for any other option.

The secondary immune response is mediated by immunological memory. During the primary response, some activated B-lymphocytes differentiate into memory B-cells. Upon secondary exposure, these pre-existing memory cells rapidly proliferate and differentiate into antibody-secreting plasma cells, bypassing the slower steps of naive cell activation.

[1 mark] B is the correct answer because memory B-cells are the basis of immunological memory. Award 0 marks for any other option.

Competitive inhibitors bind to the active site and block substrate binding, but this can be overcome by high substrate concentrations (\(V_{max}\) is unchanged). Non-competitive inhibitors bind to an allosteric site and decrease the overall catalytic rate regardless of substrate concentration (\(V_{max}\) decreases).

[1 mark] A is the correct answer as Inhibitor A is competitive and Inhibitor B is non-competitive. Award 0 marks for any other option.

Following a meal, blood glucose increases. In response, beta cells in the islets of Langerhans of the pancreas secrete insulin. Insulin acts on hepatocytes and body cells to promote glucose uptake and convert glucose to glycogen (glycogenesis), thereby lowering blood glucose.

[1 mark] D is the correct answer because beta cells produce insulin which promotes glucose uptake and glycogenesis. Award 0 marks for any other option.

The competitive exclusion principle states that two species competing for the exact same resources cannot stably coexist. The species that has even a slight advantage will dominate and drive the other to local extinction.

[1 mark] A is the correct answer. This is a classic demonstration of competitive exclusion. Award 0 marks for any other option.

Vaccines expose the body to safe forms of antigens from a pathogen. This exposure stimulates a primary immune response, producing memory B and T cells. If the actual pathogen is encountered later, these memory cells enable a rapid, powerful secondary response before disease develops.

[1 mark] B is the correct answer because active immunity requires the production of memory cells by the individual's own immune system. Award 0 marks for any other option.

Let \(X^N\) be the dominant normal allele and \(X^n\) be the recessive color-blind allele. The mother is carrier \(X^N X^n\) and the father has normal vision \(X^N Y\). The possible offspring genotypes are \(X^N X^N\) (normal female), \(X^N X^n\) (carrier female), \(X^N Y\) (normal male), and \(X^n Y\) (color-blind male). Each outcome has a probability of 0.25 (25%). The probability of having a child who is both male and color-blind is therefore 25%.

[1 mark] B is correct. Out of all possible offspring, only 1 in 4 (25%) is a color-blind male. Award 0 marks for any other option.

Initially, 80 beetles were marked. Since 20% of the marks rubbed off, only \(80 \times 0.8 = 64\) beetles in the population remained marked. Using the Lincoln Index formula, \(P = \frac{M \times n}{r}\), where \(M\) is the corrected number of marked individuals (64), \(n\) is the second sample size (60), and \(r\) is the number of recaptured marked individuals (15). Therefore, \(P = \frac{64 \times 60}{15} = 256\).

Award 1 mark for the correct option (B). Correct calculation steps: 1. Adjust marked population: \(80 \times 0.8 = 64\). 2. Apply Lincoln index formula: \(\frac{64 \times 60}{15} = 256\).

The parental phenotypes are the most frequent classes: Round purple (220) and Wrinkled white (230). The recombinant phenotypes are Round white (80) and Wrinkled purple (70). The total number of offspring is \(220 + 80 + 70 + 230 = 600\). The recombination frequency is calculated as: \(\text{Recombination frequency} = \frac{\text{Number of recombinants}}{\text{Total offspring}} \times 100 = \frac{80 + 70}{600} \times 100 = 25.0\\%\).

Award 1 mark for the correct option (B). Award 0 marks for incorrect options, with common errors leading to 12.5% (A) or 50% (C).

Plasma B cells produce specific antibodies but have a limited lifespan and do not divide in culture. Myeloma (tumor) cells divide rapidly and indefinitely but do not produce the desired antibodies. Fusing them produces a hybridoma cell line that has both properties: longevity/immortality and the production of a single specific antibody type.

Award 1 mark for the correct option (C).

In competitive inhibition, the inhibitor competes with the substrate for the active site. At very high substrate concentrations, the substrate outcompetes the inhibitor, allowing the enzyme to reach its normal maximum rate (\(V_{max}\) is unchanged). However, more substrate is needed to reach half-\(V_{max}\), meaning the \(K_m\) is increased.

Award 1 mark for the correct option (B). No partial marks.

Dehydration increases blood solute concentration, which is detected by osmoreceptors in the hypothalamus. The hypothalamus signals the posterior pituitary gland to release antidiuretic hormone (ADH). ADH increases the water permeability of the collecting ducts in the kidney, allowing more water to be reabsorbed into the blood, resulting in a small volume of highly concentrated urine.

Award 1 mark for the correct option (A).

The fundamental niche of a species is the complete range of conditions and resources it can potentially utilize. The realized niche is the actual range of resources used when competitors are present. Because Species B expanded its feeding behavior to the upper branches after Species A was removed, it shows that the upper branches are part of Species B's fundamental niche, and its realized niche was previously restricted by competition with Species A.

Award 1 mark for the correct option (D).

The woman's father was color-blind (\(X^bY\)), so she must have inherited the recessive allele and is a carrier (\(X^BX^b\)). The man has normal vision (\(X^BY\)). The possible offspring genotypes are: \(X^BX^B\) (normal female), \(X^BX^b\) (carrier female), \(X^BY\) (normal male), and \(X^bY\) (color-blind male). There is a 1 in 4 (25%) chance that any child they have will be a color-blind male.

Award 1 mark for the correct option (B).

The primary immune response occurs when an antigen is encountered for the first time; it is relatively slow and produces fewer antibodies. The secondary immune response is much faster, more robust, and produces a far greater concentration of antibodies because immunological memory (memory B and T cells) is already established, allowing rapid differentiation into antibody-producing plasma cells upon re-exposure.

Award 1 mark for the correct option (C).

Part A: In isolation, Species B grows exponentially and stabilizes at a carrying capacity of 120 individuals per gram. In co-culture, Species B's population size initially increases slightly but then declines steadily until it reaches 0 (extinction) after 150 days due to competitive exclusion by Species A. Part B: The competitive exclusion principle (Gause's Law) states that two species competing for the exact same limiting resources cannot stably coexist. The species with even a slight advantage (Species A) will dominate, leading to the local extinction of the weaker competitor (Species B) unless an evolutionary shift (niche differentiation) occurs. Part C: Abiotic factors include: 1. Temperature (influences metabolic rate, developmental speed, and reproductive rate), 2. Humidity/moisture content of the flour (critical for preventing desiccation of eggs and larvae).

Part A (Max 3 marks): Award 1 mark for noting Species B reaches a stable plateau/carrying capacity in isolation. Award 1 mark for noting the decline to zero/extinction in co-culture. Award 1 mark for comparative data (e.g., 120 individuals/g in isolation versus 0/g after 150 days in co-culture). Part B (Max 3 marks): Award 1 mark for naming the competitive exclusion principle or Gause's law. Award 1 mark for stating they share the same ecological niche or compete for the exact same resource. Award 1 mark for explaining that the more fit species outcompetes and drives the other to extinction. Part C (Max 2.75 marks): Award 1.375 marks for each correct abiotic factor listed (e.g., temperature, space/volume of container, moisture/humidity, oxygen levels) up to a maximum of 2.75 marks.

Part A: Let \(X^H\) be the normal allele and \(X^h\) be the hemophilia allele. II-1 is an affected male, so his genotype is \(X^hY\). II-2 is an unaffected female whose mother (I-2) is a carrier (\(X^HX^h\)) and father (I-1) is normal (\(X^HY\)). Therefore, II-2 has a 50% chance of being \(X^HX^H\) and a 50% chance of being a carrier (\(X^HX^h\)). Part B: II-2 has a 0.5 probability of being a carrier (\(X^HX^h\)). If she is a carrier, she has a 0.5 chance of passing the recessive allele \(X^h\) to her son III-2. The overall probability that III-2 inherits the hemophilia allele is \(0.5 \times 0.5 = 0.25\) (or 25%). Part C: Males are hemizygous (they only possess one X chromosome), meaning they will express the recessive trait if they inherit a single copy of the allele from their mother. Females possess two X chromosomes, so they must inherit two copies of the recessive allele (one from each parent) to express the phenotype, making X-linked recessive conditions far more common in males.

Part A (Max 3 marks): Award 1 mark for defining correct allele symbols (e.g., \(X^H\) and \(X^h\)). Award 1 mark for correct genotype of II-1 (\(X^hY\)). Award 1 mark for identifying that II-2 could be either \(X^HX^H\) or \(X^HX^h\). Part B (Max 3 marks): Award 1 mark for identifying that II-2 has a 50% chance of being a carrier. Award 1 mark for showing that a carrier mother has a 50% chance of passing the recessive allele to a male offspring. Award 1 mark for correct final calculated probability of 0.25 / 25% / 1 in 4. Part C (Max 2.75 marks): Award 1.375 marks for explaining that males are hemizygous and require only one allele to express the phenotype. Award 1.375 marks for explaining that females need two copies to express the phenotype and can remain unaffected as carriers.

Part A: The primary immune response has a long lag phase (7 days before detection) compared to the secondary response (under 24 hours). The peak antibody concentration is significantly lower in the primary response (10 units) compared to the secondary response (150 units). Part B: Upon the first exposure to Antigen X, clonal selection and expansion occur, producing plasma cells and long-lived memory B cells. During the second exposure, these pre-existing memory B cells recognize the antigen immediately, rapidly proliferating and differentiating into antibody-secreting plasma cells, producing a faster and much larger quantity of antibodies. Part C: Active immunity involves the production of antibodies by the organism's own immune system following exposure to a pathogen or vaccine, leading to long-term memory. Passive immunity is the acquisition of pre-formed antibodies from another source (e.g., maternal antibodies or artificial antibody injection), providing immediate but temporary protection without forming memory cells.

Part A (Max 3 marks): Award 1 mark for contrasting lag time (7 days versus less than 1 day). Award 1 mark for contrasting peak antibody concentration (10 units versus 150 units). Award 1 mark for overall comparative statement showing the secondary response is faster and larger. Part B (Max 3 marks): Award 1 mark for noting the creation and persistence of memory B/T cells from the primary response. Award 1 mark for explaining that memory cells are activated directly and undergo rapid clonal expansion. Award 1 mark for stating that memory B cells differentiate into antibody-producing plasma cells in large numbers. Part C (Max 2.75 marks): Award 1.375 marks for active immunity definition (antibodies produced by own body, creates immunological memory). Award 1.375 marks for passive immunity definition (antibodies received from external source, short-term, no memory cells made).

Part A: Competitive inhibitors bind directly to the active site, competing with the substrate. At extremely high substrate concentrations, the substrate outcompetes the inhibitor, occupying all active sites and achieving the same maximum rate (\(V_{\text{max}}\)). However, a higher concentration of substrate is required to reach half \(V_{\text{max}}\), which increases the apparent \(K_m\). Part B: A non-competitive inhibitor binds to an allosteric site, structurally altering the enzyme's active site so it cannot catalyze the reaction effectively. This decreases \(V_{\text{max}}\) because it reduces the concentration of active, functional enzymes. It does not alter the substrate binding affinity of unmodified enzymes, so the \(K_m\) remains unchanged. Part C: Using the Michaelis-Menten equation with the inhibitor present: \(\text{Rate}_{\text{inhibited}} = \frac{80 \times 5}{15 + 5} = \frac{400}{20} = 20\) micromoles per minute. The uninhibited rate at this substrate concentration is 40 micromoles per minute. The factor by which the rate decreases is \(\frac{40}{20} = 2\) (meaning the rate is halved).

Part A (Max 3 marks): Award 1 mark for stating competitive inhibitors bind to the active site. Award 1 mark for explaining that high substrate concentration overcomes/outcompetes the inhibitor, keeping \(V_{\text{max}}\) the same. Award 1 mark for explaining that more substrate is needed to reach half \(V_{\text{max}}\), hence increasing \(K_m\). Part B (Max 3 marks): Award 1 mark for stating non-competitive inhibitors bind to an allosteric site. Award 1 mark for stating \(V_{\text{max}}\) decreases due to structural inactivation of the enzyme. Award 1 mark for stating \(K_m\) remains unchanged as substrate binding affinity is unaffected. Part C (Max 2.75 marks): Award 1 mark for setting up the equation correctly with values: \(\frac{80 \times 5}{15 + 5}\). Award 1 mark for calculating the inhibited rate as 20 micromoles per minute. Award 0.75 marks for correctly determining that the rate decreases by a factor of 2 (or is halved).

(a) Dry biomass of *P. lanceolata* in monoculture increases continuously with increasing soil nitrogen concentration. The increase is steepest between 10 and 50 mg kg\(^{-1}\) (from 1.2 g to 3.4 g) and begins to level off/plateau between 100 and 200 mg kg\(^{-1}\) (from 5.8 g to 6.2 g).

(b) At low nitrogen (10-50 mg kg\(^{-1}\)), *T. repens* shows higher biomass and is less affected by competition (biomass barely changes from monoculture to diculture), indicating it is more competitive at low nitrogen. At high nitrogen (100-200 mg kg\(^{-1}\)), *P. lanceolata* is more competitive because its biomass in diculture continues to rise, whereas *T. repens* biomass decreases significantly (from 2.9 g at 50 mg kg\(^{-1}\) to 1.5 g at 200 mg kg\(^{-1}\)) due to interspecific competition.

(c) The fundamental niche is the potential niche space an organism can occupy in the absence of competition. This is represented by *T. repens* maintaining stable biomass (2.5g to 3.3g) across all nitrogen levels in monoculture. The realized niche is the actual niche occupied in the presence of competition. In diculture (presence of *P. lanceolata*), *T. repens* biomass drops significantly at higher nitrogen levels (to 1.5g), showing its realized niche is restricted to lower nitrogen environments where it can successfully compete.

[Max 2 marks for part (a)]
- Biomass increases as soil nitrogen increases (1 mark)
- Increase is rapid between 10 and 50/100 mg kg\(^{-1}\) OR levels off/plateaus between 100 and 200 mg kg\(^{-1}\) (1 mark)

[Max 2 marks for part (b)]
- At low nitrogen (10-50 mg kg\(^{-1}\)), *T. repens* is more competitive/less affected by competition than *P. lanceolata* (1 mark)
- At high nitrogen (100-200 mg kg\(^{-1}\)), *P. lanceolata* is more competitive/strongly suppresses *T. repens* growth (1 mark)

[Max 3 marks for part (c)]
- Fundamental niche defined as potential niche/range in the absence of competition (1 mark)
- Monoculture data shows *T. repens* can grow across all nitrogen concentrations (up to 3.3 g at 200 mg kg\(^{-1}\)) (1 mark)
- Realized niche defined as actual niche in the presence of competitors (1 mark)
- Diculture data shows *T. repens* growth is severely restricted/reduced at high nitrogen due to competition with *P. lanceolata* (1 mark)

(a) Keystone species.

(b) Top-down control occurs when predators limit the population density of lower trophic levels (e.g., sea otters controlling sea urchins, which prevents overgrazing of kelp forests). Bottom-up control occurs when the availability of resources (like nutrients or primary producers) limits the energy flow and population sizes of higher trophic levels (e.g., nutrients limiting phytoplankton blooms, which in turn limits zooplankton and fish abundance).

(c) An invasive herbivore competes with native herbivores for food resources (producers). This competitive exclusion can decrease the population of native herbivores. Consequently, there is less food energy available for primary predators, which lowers their carrying capacity. Alternatively, if primary predators can switch to consuming the invasive herbivore, the abundance of prey may increase, leading to an increase in the primary predators' carrying capacity.

[1 mark for part (a)]
- Keystone species (1 mark)

[Max 3 marks for part (b)]
- Top-down: community structure is regulated by predation/higher trophic levels (1 mark)
- Example: sea otters/wolves/sea stars (1 mark)
- Bottom-up: community structure is regulated by resource availability/primary productivity (1 mark)
- Example: nutrients/nitrogen in soil/phytoplankton (1 mark)

[Max 3 marks for part (c)]
- Invasive herbivores reduce food availability for native herbivores (competition) (1 mark)
- Reduced native herbivore biomass reduces food supply for primary predators, lowering their carrying capacity (1 mark)
- Alternatively, if predators successfully prey on the invader, the overall prey abundance increases, which can raise the predator carrying capacity (1 mark)

(a) Male genotype: \(W^S W^B X^{a^s} Y\) (mosaic, short-antennaed male).
Female genotype: \(W^S W^S X^{A^L} X^{a^s}\) (spotted, heterozygous long-antennaed female).

(b)
Gametes produced by the male parent:
\(W^S X^{a^s}\), \(W^S Y\), \(W^B X^{a^s}\), \(W^B Y\)

Gametes produced by the female parent:
\(W^S X^{A^L}\), \(W^S X^{a^s}\)

Punnett Grid:
- \(W^S X^{A^L}\) x \(W^S X^{a^s}\) \(\rightarrow\) \(W^S W^S X^{A^L} X^{a^s}\) (Spotted, long-antennaed female)
- \(W^S X^{A^L}\) x \(W^S Y\) \(\rightarrow\) \(W^S W^S X^{A^L} Y\) (Spotted, long-antennaed male)
- \(W^S X^{A^L}\) x \(W^B X^{a^s}\) \(\rightarrow\) \(W^S W^B X^{A^L} X^{a^s}\) (Mosaic, long-antennaed female)
- \(W^S X^{A^L}\) x \(W^B Y\) \(\rightarrow\) \(W^S W^B X^{A^L} Y\) (Mosaic, long-antennaed male)
- \(W^S X^{a^s}\) x \(W^S X^{a^s}\) \(\rightarrow\) \(W^S W^S X^{a^s} X^{a^s}\) (Spotted, short-antennaed female)
- \(W^S X^{a^s}\) x \(W^S Y\) \(\rightarrow\) \(W^S W^S X^{a^s} Y\) (Spotted, short-antennaed male)
- \(W^S X^{a^s}\) x \(W^B X^{a^s}\) \(\rightarrow\) \(W^S W^B X^{a^s} X^{a^s}\) (Mosaic, short-antennaed female)
- \(W^S X^{a^s}\) x \(W^B Y\) \(\rightarrow\) \(W^S W^B X^{a^s} Y\) (Mosaic, short-antennaed male)

(c) The spotted, long-antennaed male offspring has the genotype \(W^S W^S X^{A^L} Y\). Out of 8 possible offspring combinations, only 1 matches this genotype/phenotype. Thus, the probability is 1/8 (or 0.125 or 12.5%).

[Max 2 marks for part (a)]
- Male: \(W^S W^B X^{a^s} Y\) (1 mark)
- Female: \(W^S W^S X^{A^L} X^{a^s}\) (1 mark)

[Max 4 marks for part (b)]
- Correct gametes identified for both parents (1 mark)
- Punnett grid constructed correctly with 8 offspring genotypes (1 mark)
- Corresponding phenotypes mapped correctly to genotypes (1 mark)
- Distinguishes sex-linked traits correctly for male (XY) and female (XX) offspring (1 mark)

[1 mark for part (c)]
- 1/8 OR 0.125 OR 12.5% (1 mark)

(a) Day 14 (with 12 arbitrary units).

(b) Both responses involve the production of specific antibodies in response to antigen injection on Day 28. However, the response to Antigen A is much more rapid and of a much higher magnitude (peaks at 120 units on Day 35) because it is a secondary response. In contrast, the response to Antigen B is slower and of a much lower magnitude (peaks at 15 units on Day 42) because it is a primary response. Additionally, Antigen A antibody levels begin to slowly decline by Day 42, whereas Antigen B antibody levels are still rising until Day 42.

(c) The primary injection of Antigen A on Day 0 led to clonal selection and expansion of specific B lymphocytes, resulting in the formation of memory B cells. When Antigen A is injected again on Day 28, these memory B cells immediately recognize the antigen. They rapidly divide and differentiate into antibody-secreting plasma cells, bypassing the slower steps of initial antigen presentation and helper T-cell activation. This leads to a faster and much larger production of antibodies.

[1 mark for part (a)]
- Day 14 (1 mark)

[Max 3 marks for part (b)]
- Similarity: both produce specific antibodies following injection on Day 28 (1 mark)
- Difference (magnitude): Anti-A concentration is much higher than Anti-B (peaks at 120 vs 15 units) (1 mark)
- Difference (speed/timing): Anti-A peaks faster (7 days post-injection / Day 35) than Anti-B (14 days post-injection / Day 42) (1 mark)

[Max 3 marks for part (c)]
- Primary exposure (Day 0) produced memory B cells (1 mark)
- Memory cells persist in blood/lymph nodes and rapidly detect Antigen A on Day 28 (1 mark)
- Memory B cells differentiate rapidly into plasma cells (without needing helper T cell activation steps again), producing large amounts of antibodies (1 mark)

(a) When blood vessels are damaged, platelets release clotting factors. These clotting factors, along with factors released from the damaged tissue, initiate a cascade of reactions. This cascade converts the inactive plasma protein prothrombin into the active enzyme thrombin. Thrombin then acts as a catalyst to convert the soluble plasma protein fibrinogen into insoluble, fibrous fibrin. Fibrin forms a mesh of threads that trap blood cells and platelets, forming a semi-solid clot that seals the wound.

(b) Coronary thrombosis is caused by the rupture of an atheroma (cholesterol/plaque buildup) in the wall of a coronary artery. This rupture exposes collagen fibers, triggering platelet aggregation and releasing clotting factors that form a blood clot (thrombus) inside the coronary lumen. This clot blocks the flow of oxygenated blood to the heart muscle (myocardium) downstream, depriving it of oxygen and nutrients. This leads to myocardial infarction (heart attack) where cardiac muscle cells cannot respire aerobically, undergo cell death, and fail to contract.

[Max 4 marks for part (a)]
- Damaged tissue/platelets release clotting factors (1 mark)
- Clotting factors trigger a cascade of reactions (1 mark)
- Prothrombin is converted to thrombin (active enzyme) (1 mark)
- Thrombin catalyzes conversion of soluble fibrinogen to insoluble fibrin (1 mark)
- Fibrin forms a mesh that traps blood cells/platelets to form a clot (1 mark)

[Max 3 marks for part (b)]
- Rupture of atheroma/plaque in coronary arteries triggers platelet clotting cascade (1 mark)
- Clot/thrombus blocks coronary artery, preventing blood/oxygen delivery to heart muscle (1 mark)
- Heart muscle cells die/cannot perform aerobic respiration, leading to myocardial infarction/heart attack (1 mark)

(a) 40°C.

(b) At 40°C, the enzyme has its highest activity (6.8 mg/min) because molecules have high kinetic energy, causing frequent successful collisions between enzyme active sites and substrates. However, at 60°C, the rate drops drastically to 0.2 mg/min because the high thermal energy disrupts the hydrogen and ionic bonds stabilizing the enzyme's tertiary structure. This causes the enzyme to denature, changing the shape of its active site so that the substrate (starch) can no longer bind.

(c) Copper sulfate acts as a non-competitive inhibitor. This is because at all temperatures, including the optimum (40°C), the rate of reaction is significantly reduced (from 6.8 to 1.6 mg/min). Non-competitive inhibitors bind to an allosteric site (rather than the active site) and alter the enzyme's conformation, lowering the maximum rate of reaction (\(V_{max}\)) regardless of the substrate concentration. (Even if substrate concentration is high, the enzyme activity remains low, which is characteristic of non-competitive inhibition).

[1 mark for part (a)]
- 40°C (1 mark)

[Max 3 marks for part (b)]
- At 40°C, the kinetic energy is high, resulting in frequent collisions between enzyme and substrate (1 mark)
- At 60°C, high temperature disrupts bonds (hydrogen/ionic/hydrophobic) in the protein structure (1 mark)
- The enzyme denatures, altering the shape of the active site (1 mark)
- Substrate can no longer bind to the active site, resulting in a dramatic decrease in the rate of reaction (1 mark)

[Max 3 marks for part (c)]
- Non-competitive inhibitor (1 mark)
- Inhibitor significantly reduces the rate of reaction across all temperatures / cannot be overcome (1 mark)
- Non-competitive inhibitors bind to an allosteric site, altering the active site's shape so the reaction cannot proceed (1 mark)

(a) A decrease in blood glucose concentration is detected by alpha (\(\alpha\)) cells in the islets of Langerhans in the pancreas. In response, alpha cells secrete the hormone glucagon into the bloodstream. Glucagon travels to the liver, where it binds to specific receptors on hepatocytes. This triggers glycogenolysis (the breakdown of stored glycogen into glucose) and gluconeogenesis (the synthesis of glucose from non-carbohydrate sources). The glucose is then released into the blood, raising blood glucose levels back to the normal set point.

(b) Type I diabetes is caused by an autoimmune destruction of beta (\(\beta\)) cells in the pancreas, leading to an inability to produce insulin. It is treated primarily with regular insulin injections. Type II diabetes is caused by insulin resistance, where body cells (target cells) fail to respond to insulin, often associated with obesity, diet, and genetics. It is treated with lifestyle changes (diet, exercise) and oral medications that improve insulin sensitivity, rather than relying solely on insulin injections.

[Max 4 marks for part (a)]
- Low blood glucose is detected by alpha (\(\alpha\)) cells in the pancreas (1 mark)
- Alpha cells secrete glucagon into the blood (1 mark)
- Glucagon binds to receptors on liver cells/hepatocytes (1 mark)
- Promotes glycogenolysis/glycogen breakdown into glucose (1 mark)
- Promotes gluconeogenesis (1 mark)
- Glucose is released into blood to raise concentration to normal levels (1 mark)

[Max 3 marks for part (b)]
- Type I caused by autoimmune destruction of beta (\(\beta\)) cells/lack of insulin production, while Type II is caused by insulin resistance of body cells (1 mark)
- Type I treatment involves insulin injections, while Type II treatment involves lifestyle changes/diet/exercise/medication to increase insulin sensitivity (1 mark)
- Type I is typically early-onset/genetic, whereas Type II is typically late-onset/associated with obesity (1 mark)

(a) Barriers to pathogen entry:
- The skin acts as a primary physical barrier to pathogens. The outer layer (epidermis) is composed of tough, keratinized, closely packed dead cells that prevent entry.
- Sebum/oil secreted by sebaceous glands in the skin lowers the pH (making it acidic), which chemically inhibits the growth of many fungi and bacteria.
- Mucous membranes line body cavities exposed to the exterior (e.g., respiratory, digestive, and reproductive tracts). They act as a physical barrier by secreting sticky mucus that traps pathogens.
- Cilia in the trachea and nasal passages sweep the trapped mucus upward towards the throat to be swallowed and destroyed.
- Tears and saliva contain lysozyme, an enzyme that chemically digests the cell walls of certain bacteria.
- Acid in the stomach (hydrochloric acid, low pH) kills ingested pathogens.

(b) Clonal selection and antibody production:
- A pathogen enters the body, and its antigens are recognized as foreign by immune cells.
- Macrophages/dendritic cells engulf the pathogen through phagocytosis and present the pathogen's antigens on their cell surface membrane (acting as antigen-presenting cells or APCs).
- Helper T-cells with a complementary/matching receptor bind to the presented antigen on the APC.
- This binding activates the helper T-cell, which then selectively binds to and activates specific B-cells that have matching receptors for the same antigen (clonal selection).
- Once activated, these B-cells undergo rapid cell division by mitosis (clonal expansion).
- The cloned B-cells differentiate into two populations: plasma cells and memory B-cells.
- Plasma cells produce and secrete large quantities of specific antibodies into the blood or lymph, which target and neutralize the specific antigen.
- Memory B-cells persist in the body for long-term immunity, allowing a rapid response if the same pathogen is encountered again.

(c) Active vs Passive Immunity:
- Active immunity arises from the production of antibodies by the individual's own body/immune system, whereas passive immunity involves receiving antibodies produced by another organism.
- Active immunity results in the production of memory cells, providing long-term protection, whereas passive immunity does not produce memory cells, providing only immediate, short-term protection.
- Active immunity can be acquired naturally by contracting an infection/disease, or artificially through vaccination (injecting weakened or dead pathogens/antigens).
- Passive immunity can be acquired naturally via the transfer of antibodies across the placenta or through breast milk (colostrum) to the infant, or artificially via injection of antibodies/antiserum (e.g., antivenom for snake bites).

Part (a): [4 marks maximum]
- Award 1 mark for skin acting as a physical barrier / tough keratinized cells.
- Award 1 mark for sebaceous glands secreting sebum / low pH of skin.
- Award 1 mark for mucous membranes trapping pathogens in sticky mucus.
- Award 1 mark for cilia sweeping mucus away.
- Award 1 mark for lysozyme in tears/saliva digesting bacterial cell walls.
- Award 1 mark for stomach acid killing ingested pathogens.

Part (b): [6 marks maximum]
- Award 1 mark for macrophages/APCs engulfing pathogens and presenting antigens.
- Award 1 mark for helper T-cells binding to antigens on the APC.
- Award 1 mark for activated helper T-cells activating specific B-cells (clonal selection).
- Award 1 mark for activated B-cells dividing rapidly by mitosis (clonal expansion).
- Award 1 mark for differentiation of B-cells into plasma cells and memory cells.
- Award 1 mark for plasma cells secreting large quantities of specific antibodies.
- Award 1 mark for memory B-cells persisting to provide long-term immunity/rapid secondary response.

Part (c): [5 marks maximum]
- Award 1 mark for stating active immunity involves antibodies made by the individual's own body, while passive involves receiving antibodies from an external source.
- Award 1 mark for stating active immunity produces memory cells, while passive does not.
- Award 1 mark for stating active immunity is long-lasting, while passive is short-term/temporary.
- Award 1 mark for describing natural and artificial acquisition of active immunity (infection vs. vaccination).
- Award 1 mark for describing natural and artificial acquisition of passive immunity (colostrum/placenta vs. antibody injection/antivenom).

Quality of Communication: [1 mark]
- Award 1 mark if the candidate's answers to all parts are well-structured, logical, and consistently use accurate scientific terminology throughout.

(a) Negative feedback in homeostasis:
- Homeostasis is the maintenance of a stable internal environment within narrow limits.
- Negative feedback is a mechanism where a change in a physiological variable triggers a response that counteracts or reverses the direction of that change, returning the variable to its set point.
- In thermoregulation, body temperature is monitored by thermoreceptors in the skin and the hypothalamus in the brain (the control center).
- If body temperature rises above the set point, the hypothalamus triggers cooling mechanisms: sweating (heat loss via evaporation) and vasodilation of skin arterioles (bringing blood closer to the skin surface to radiate heat).
- If body temperature falls below the set point, the hypothalamus triggers heating mechanisms: shivering (rapid muscle contraction generating metabolic heat) and vasoconstriction of skin arterioles (reducing blood flow to skin to conserve heat).
- Once the temperature returns to the normal range, the corrective actions are switched off, preventing over-correction.

(b) Blood glucose regulation by the pancreas and liver:
- When blood glucose is high (e.g., after a meal), this change is detected by the beta (\(\beta\)) cells in the islets of Langerhans in the pancreas.
- Beta cells secrete insulin into the bloodstream.
- Insulin targets the liver and muscle cells, stimulating them to increase their absorption of glucose from the blood.
- In the liver, insulin stimulates glycogenesis, which is the conversion of glucose into insoluble glycogen for storage.
- When blood glucose is low (e.g., during exercise or fasting), this is detected by the alpha (\(\alpha\)) cells in the islets of Langerhans in the pancreas.
- Alpha cells secrete glucagon into the bloodstream.
- Glucagon targets liver cells and stimulates glycogenolysis (the breakdown of stored glycogen back into glucose) and gluconeogenesis (the production of glucose from non-carbohydrate sources).
- This newly released glucose passes out of liver cells into the blood, raising blood glucose levels back to the normal range.

(c) Type I vs Type II diabetes:
- Cause of Type I: An autoimmune condition where the body's immune system attacks and destroys the insulin-producing beta cells in the pancreas, leading to an inability to produce insulin.
- Cause of Type II: Target cells (e.g., liver, muscle) develop insulin resistance, meaning they no longer respond effectively to the insulin that is produced. This is strongly associated with lifestyle factors like obesity and physical inactivity.
- Age of onset: Type I usually develops during childhood or adolescence (early onset), whereas Type II typically develops in adulthood (late onset), though it is increasingly seen in younger people due to rising obesity rates.
- Treatment of Type I: Requires daily monitoring of blood glucose levels and regular injections or infusions of synthetic insulin.
- Treatment of Type II: Managed primarily through lifestyle adjustments (healthier diet, weight loss, increased physical activity) and oral medications, though some severe cases may eventually require insulin.

Part (a): [5 marks maximum]
- Award 1 mark for defining homeostasis as maintaining a stable internal environment within narrow limits.
- Award 1 mark for explaining negative feedback (response counteracts/reverses the initial change).
- Award 1 mark for identifying the hypothalamus as the sensor/control center for thermoregulation.
- Award 1 mark for describing mechanisms to lower temperature (sweating and/or vasodilation of skin arterioles).
- Award 1 mark for describing mechanisms to raise temperature (shivering and/or vasoconstriction of skin arterioles).
- Award 1 mark for stating that feedback loops turn off once the set point is reached.

Part (b): [6 marks maximum]
- Award 1 mark for stating high blood glucose is detected by beta (\(\beta\)) cells in the pancreas.
- Award 1 mark for stating beta cells secrete insulin.
- Award 1 mark for stating insulin causes liver/muscle cells to absorb glucose from the blood.
- Award 1 mark for stating insulin promotes glycogenesis (conversion of glucose to glycogen in the liver).
- Award 1 mark for stating low blood glucose is detected by alpha (\(\alpha\)) cells in the pancreas.
- Award 1 mark for stating alpha cells secrete glucagon.
- Award 1 mark for stating glucagon stimulates the liver to break down glycogen into glucose (glycogenolysis) / gluconeogenesis.

Part (c): [4 marks maximum]
- Award 1 mark for distinguishing the causes: Type I is autoimmune destruction of beta cells (no insulin produced) vs. Type II is insulin resistance of target cells.
- Award 1 mark for distinguishing onset: Type I is early onset/childhood vs. Type II is late onset/adulthood/associated with obesity.
- Award 1 mark for distinguishing primary treatments: Type I is treated with insulin injections vs. Type II is treated with diet/exercise/lifestyle adjustments/oral medications.
- Award 1 mark for distinguishing risk factors: Type I is mainly genetic/unrelated to lifestyle vs. Type II is heavily linked to sedentary lifestyle/diet/obesity.

Quality of Communication: [1 mark]
- Award 1 mark if the candidate's answers to all parts are well-structured, logical, and consistently use accurate scientific terminology throughout.