2024 AQA IAS-Level Biology (9610) 模擬試題連答案詳解

For part (a), based on descriptions, A is a chloroplast, B is a mitochondrion, and C is a ribosome. For part (b), convert 45 mm to micrometres: \(45 \times 1000 = 45,000\ \mu\text{m}\). Magnification is calculated using the formula: \(\text{Magnification} = \frac{\text{Image Size}}{\text{Actual Size}} = \frac{45,000\ \mu\text{m}}{3.0\ \mu\text{m}} = 15,000\). For part (c), TEM has higher resolution and transmits electrons through the specimen, showing the internal ultrastructure, whereas SEM scans the surface to produce a 3-D image.

(a) A: Chloroplast [1 mark], B: Mitochondrion [1 mark], C: Ribosome [1 mark]. (b) Convert 45 mm to \(45,000\ \mu\text{m}\) [1.7 marks]. Calculate magnification: \(45,000 / 3.0 = 15,000\) [2 marks]. (c) TEM has higher resolution than SEM [1 mark]; uses a transmitted electron beam to pass through a thin specimen [1 mark]; which allows internal details/ultrastructure to be visualised [1 mark]; SEM only provides information about the surface/3D structure [1 mark].

For (a), the independent variable is the concentration of hydrogen peroxide, and the dependent variable is the volume of oxygen produced per unit time. For (b), as the reaction proceeds, the substrate concentration decreases because it is converted into products, which slows the rate. The initial rate represents the true rate when the substrate is not limiting. For (c), 30 seconds is equal to 0.5 minutes. Rate = \(\frac{4.5\text{ cm}^3}{0.5\text{ min}} = 9.0\text{ cm}^3\text{ min}^{-1}\). For (d), heat breaks hydrogen and ionic bonds, altering the enzyme's tertiary structure. This changes the active site's shape so it is no longer complementary, preventing substrate binding and enzyme-substrate complexes.

(a) Independent variable: hydrogen peroxide/substrate concentration [1 mark]; Dependent variable: volume of oxygen collected / rate of oxygen production [1 mark]. (b) Substrate concentration decreases over time as it is converted to product [1 mark]; substrate concentration becomes a limiting factor [1 mark]; initial rate ensures substrate concentration is at the specified start value [0.7 marks]. (c) Convert 30 seconds to 0.5 minutes [1 mark]. Rate = \(4.5\text{ cm}^3 / 0.5\text{ min} = 9.0\text{ cm}^3\text{ min}^{-1}\) [1 mark]. (d) Hydrogen bonds and ionic bonds break [1 mark]; tertiary structure alters [1 mark]; active site changes shape / no longer complementary [1 mark]; no enzyme-substrate complexes form [1 mark].

For part (a), first perform a Benedict's test to confirm the absence of reducing sugars. Then, boil a fresh sample with dilute HCl to hydrolyse the glycosidic bond into reducing sugars. Neutralise with sodium hydrogencarbonate. Finally, perform the Benedict's test again; a change from blue to green/orange/brick-red confirms a non-reducing sugar. For part (b), the link between glucose and fructose in sucrose is a glycosidic bond, and it is broken by a hydrolysis reaction. For part (c), dilution factor = \(2.0 / 0.4 = 5\). Volume of stock = \(20 / 5 = 4\text{ cm}^3\). Volume of water = \(20 - 4 = 16\text{ cm}^3\).

(a) Heat with Benedict's reagent to show no change / negative result [1 mark]; heat a fresh sample with dilute hydrochloric acid [1 mark]; neutralise with sodium hydrogencarbonate [1 mark]; heat with Benedict's reagent again [1 mark]; observe brick-red precipitate [0.7 marks]. (b) Glycosidic bond [1 mark]; hydrolysis [1 mark]. (c) Dilution factor is 5 [1 mark]; volume of stock = \(4\text{ cm}^3\) [1.5 marks]; volume of distilled water = \(16\text{ cm}^3\) [1.5 marks].

For part (a), excess water left on the surface would contribute to the measured mass, leading to errors in the percentage mass change. Drying removes surface liquid. For part (b), because the initial masses of the potato cylinders are not identical, calculating percentage changes allows a valid comparison. For part (c), since there is no net movement of water, the water potential inside the cells is equal to the water potential of the 0.3 \(\text{mol dm}^{-3}\) sucrose solution. For part (d), since they are in equilibrium, the water potential of the potato cells is equal to the solution, which is -780 kPa.

(a) To remove excess water/liquid on the surface [1 mark]; which would otherwise increase the measured mass / create inaccuracy [1 mark]. (b) Initial mass of the cylinders is different [1 mark]; percentage change normalises the values [1 mark]; allowing direct/valid comparison [0.7 marks]. (c) Water potential of the potato cells is equal to the water potential of the 0.3 \(\text{mol dm}^{-3}\) solution [1 mark]; no net movement of water by osmosis [1 mark]; osmosis is at equilibrium [1 mark]; cells are neither turgid nor plasmolysed [1 mark]. (d) Water potential is \(-780\text{ kPa}\) [2 marks] (1 mark for correct value, 1 mark for unit kPa).

For part (a), X refers to the gill filaments and Y refers to the gill lamellae. For part (b), in a counter-current system, blood and water flow in opposite directions. This prevents the reach of equilibrium, meaning a concentration gradient of oxygen is maintained across the entire length of the lamellae, enabling continuous oxygen diffusion. For part (c), the change is \(78 - 42 = 36\). The percentage increase is calculated as: \(\frac{36}{42} \times 100 = 85.71\%\).

(a) X: Gill filaments [1 mark], Y: Gill lamellae [1 mark]. (b) Blood and water flow in opposite directions [1 mark]; water with high oxygen always meets blood with lower oxygen [1 mark]; maintaining a concentration gradient [1 mark]; across the entire length of the gas exchange surface [1 mark]; parallel flow would reach equilibrium so diffusion would stop [1 mark]. (c) Increase in rate is 36 [1 mark]; percentage calculation: \((36/42) \times 100\) [1.7 marks]; correct final answer of 85.7% (accept 86%) [1 mark].

For part (a), a DNA nucleotide consists of deoxyribose sugar, a phosphate group, and a nitrogenous base. For part (b), Cytosine (34%) pairs with Guanine, so G is also 34%. Together C + G = 68%. This leaves 32% for Adenine and Thymine (A + T). Since A pairs with T, Thymine is \(32 / 2 = 16\%\). For part (c), features include: 1. A sugar-phosphate backbone with strong covalent phosphodiester bonds, which protect the internal base sequence. 2. A double helix shape, which physically shields the organic bases from biochemical attacks.

(a) Deoxyribose [1 mark]; phosphate group [1 mark]; nitrogenous base [1 mark]. (b) Cytosine = Guanine = 34% [1 mark]; C + G = 68% [1 mark]; A + T = 32% [1 mark]; T = 16% [0.7 marks]. (c) Feature 1: Sugar-phosphate backbone with strong covalent phosphodiester bonds [1 mark]; which protects the internal sequence of bases from chemical degradation [1 mark]. Feature 2: Double helix structure [1 mark]; which physically shields the bases / many hydrogen bonds provide collective stability [1 mark].

For part (a), find total N = \(18 + 12 + 35 + 5 = 70\). Calculate each \((n/N)^2\): Buttercup: \((18/70)^2 \approx 0.0661\); Dandelion: \((12/70)^2 \approx 0.0294\); Clover: \((35/70)^2 = 0.2500\); Daisy: \((5/70)^2 \approx 0.0051\). Sum of squares = \(0.0661 + 0.0294 + 0.2500 + 0.0051 = 0.3506\). Therefore, \(D = 1 - 0.3506 = 0.6494 \approx 0.65\). For part (b), Meadow A (D = 0.65) has higher biodiversity than Meadow B (D = 0.45). A high D value means the ecosystem has high species richness and evenness, making it more stable because if one species is affected by a threat, others remain to support the community. For part (c), use tape measures to create grid coordinates, then select positions randomly using a random number generator.

(a) Calculate N = 70 [1 mark]; calculate separate squared terms [2 marks]; sum of squared terms = 0.35 [0.7 marks]; subtract from 1 to find D = 0.65 (accept 0.649) [1 mark]. (b) Meadow A has a higher biodiversity [1 mark]; high D means high species richness and evenness [1 mark]; this increases ecosystem stability because alternative species are present if one population falls [1 mark]. (c) Establish a coordinate grid system using tape measures [1 mark]; use random number generator to select coordinates [1 mark]; place the quadrat at selected intersections without bias [1 mark].

(a)
Percentage increase = \(\frac{\text{Binding of mAb-Y} - \text{Binding of mAb-X}}{\text{Binding of mAb-X}} \times 100\)
\(\text{Percentage increase} = \frac{65 - 35}{35} \times 100 = \frac{30}{35} \times 100 = 85.71\%\)
(Accept \(85.7\%\) or \(86\%\))

(b)
At low antigen concentrations (e.g. \(0.2\text{ nmol dm}^{-3}\)), mAb-Y has a significantly higher binding percentage (\(65\%\)) compared to mAb-X (\(35\%\)). This means mAb-Y has a higher affinity for the antigen. Thus, mAb-Y will yield a stronger/more detectable signal at low viral loads, preventing false-negative results.

(c)
B-lymphocytes (specifically plasma cells) are isolated from the spleen of an immunized animal (e.g., a mouse) because they produce the specific antibody matching the antigen. However, B-cells have a limited lifespan and cannot divide in culture. Therefore, they are fused with tumor cells (myeloma cells) using a helper chemical like PEG to create hybridoma cells. These hybridoma cells possess both the ability to produce the specific antibody (from the B-lymphocyte) and the ability to divide indefinitely in culture.

(d)
Active immunity involves the individual's own immune system producing antibodies and memory cells in response to an antigen (either via infection or vaccination), providing long-term protection. Passive immunity involves the introduction of external antibodies from an outside source (e.g., across the placenta, via breastmilk, or by injection), which provides immediate but temporary protection as no memory cells are produced and the external antibodies are eventually broken down.

(a)
- Mark 1: Correct working shown: \(\frac{65-35}{35} \times 100\) (or equivalent).
- Mark 2: Correct answer of \(85.7\%\) or \(86\%\).

(b)
- Mark 1: Identifies that mAb-Y has a higher percentage of binding / higher affinity at low concentrations.
- Mark 2: Explains that this leads to a more sensitive detection / stronger signal / lower limit of detection.
- Mark 3: Connects this to a reduced rate of false negatives (early-stage diagnosis).

(c)
- Mark 1: B-lymphocytes/plasma cells are harvested from an animal (e.g. mouse) injected with the antigen to provide the specific antibody gene/production capability.
- Mark 2: B-lymphocytes are fused with myeloma (cancer/tumor) cells.
- Mark 3: To form hybridoma cells which can divide continuously/indefinitely.
- Mark 0.7: Monoclonal antibodies are harvested from the cloned hybridoma cells.

(d)
- Mark 1: Active immunity produces memory cells, whereas passive immunity does not.
- Mark 2: Active immunity involves the body producing its own antibodies, whereas passive immunity involves receiving antibodies from an external source.

(a)
Change in LDL clearance rate = \(9.6 - 2.5 = 7.1\%\text{ per hour}\)
Change in functional LDL receptors = \(100 - 20 = 80\%\)
Rate of increase = \(\frac{7.1}{80} = 0.08875\text{ \% clearance per hour per \% receptor}\) (accept \(0.089\))

(b)
High levels of blood cholesterol (specifically LDLs) lead to accumulation of LDLs in the endothelial walls of arteries. The endothelium becomes damaged (due to factors like high blood pressure or toxins). White blood cells (macrophages) ingest the cholesterol and form 'foam cells', which accumulate together with lipids, fibers, and dead cells to form an atheroma (fatty plaque). This plaque narrows the arterial lumen, restricting blood flow. If the atheroma ruptures, it triggers blood clot formation (thrombosis). If this thrombosis occurs in the coronary arteries, it blocks the blood flow, depriving cardiac muscle of oxygen, preventing aerobic respiration, and leading to myocardial cell death (myocardial infarction).

(c)
One lifestyle modification is regular aerobic exercise. Regular exercise helps lower blood pressure, which reduces the mechanical stress on arterial endothelial linings, thus decreasing the likelihood of endothelial damage and subsequent plaque formation. Additionally, it can increase high-density lipoprotein (HDL) levels, which helps transport excess cholesterol back to the liver for excretion. Alternatively, stopping smoking reduces carbon monoxide (which binds to hemoglobin reducing oxygen transport) and nicotine (which constricts blood vessels and increases platelet stickiness, elevating clot risk).

(a)
- Mark 1: Shows correct working: \(\frac{9.6 - 2.5}{100 - 20}\) or \(\frac{7.1}{80}\).
- Mark 2: Correct answer of \(0.08875\) or \(0.089\) with appropriate units (\(\%\text{ clearance per hour per \% receptor}\)).

(b)
- Mark 1: Damage occurs to the endothelium of an artery.
- Mark 2: High concentration of LDL cholesterol accumulates beneath the endothelium/in the wall.
- Mark 3: Macrophages engulf lipids to form foam cells, creating an atheroma/plaque.
- Mark 4: Plaque narrows the lumen / restricts blood flow, which may lead to thrombus (clot) formation.
- Mark 5: Blockage of coronary artery deprives heart muscle of oxygen, leading to myocardial infarction.

(c)
- Mark 1: Identifies lifestyle modification (e.g., regular exercise / stopping smoking / stress reduction).
- Mark 2: Links modification to a physical mechanism (e.g., exercise lowers blood pressure, reducing endothelial damage; or stopping smoking reduces nicotine/platelet aggregation).
- Mark 0.7: Explains how this directly reduces atheroma formation or thrombosis risk.

(a)
Time interval: From \(0.25\text{ s}\) to \(0.55\text{ s}\).
Reason: The semi-lunar valve opens when the pressure inside the left ventricle exceeds the pressure inside the aorta, forcing the valve open to allow blood to exit.

(b)
Length of one cardiac cycle = \(0.80\text{ s}\).
Heart rate = \(\frac{60}{0.80} = 75\text{ beats per minute (bpm)}\).

(c)
Stroke volume = \(130\text{ cm}^3 - 55\text{ cm}^3 = 75\text{ cm}^3\).
Cardiac output = \(\text{Stroke Volume} \times \text{Heart Rate}\)
\(\text{Cardiac output} = 75\text{ cm}^3 \times 75\text{ min}^{-1} = 5625\text{ cm}^3\text{ min}^{-1}\)
In \(\text{dm}^3\text{ min}^{-1}\), divide by \(1000\):
\(\text{Cardiac output} = 5.625\text{ dm}^3\text{ min}^{-1}\).

(d)
- Arteries have a thicker tunica media containing more elastic fibers and smooth muscle than veins. This structure allows arteries to stretch and recoil to maintain high blood pressure and resist bursting.
- Veins have a wider lumen than arteries, which minimizes resistance to blood flow under low pressure.
- Veins possess pocket valves along their length to prevent the backflow of blood, which is not necessary in arteries due to high-pressure flow.

(a)
- Mark 1: Correct time interval of \(0.25\text{ s}\) to \(0.55\text{ s}\).
- Mark 2: Reason states that ventricular pressure must exceed aortic pressure for the semi-lunar valve to open.

(b)
- Mark 1: Identifies formula: \(\frac{60}{\text{cycle length}}\).
- Mark 2: Correctly calculates \(75\text{ bpm}\).

(c)
- Mark 1: Correct stroke volume calculation: \(75\text{ cm}^3\).
- Mark 2: Multiplies stroke volume by heart rate (\(75 \times 75\)).
- Mark 3: Correct conversion to \(5.625\text{ dm}^3\text{ min}^{-1}\) (allow ecf from part b).

(d)
- Mark 1: Contrast in muscle/elastic layer thickness (arteries thicker than veins to withstand/maintain high pressure).
- Mark 2: Contrast in lumen size (veins wider than arteries to reduce resistance under low pressure).
- Mark 3: Contrast in valves (veins have valves, arteries do not to prevent blood backflow).
- Mark 0.7: Clear, structured comparison linking the anatomical structures directly to physiological differences.

(a)
Difference = \(2.8 - 1.1 = 1.7\text{ g dm}^{-2}\text{ h}^{-1}\)
Percentage difference relative to wild-type = \(\frac{1.7}{1.1} \times 100 = 154.55\%\) (accept \(154.5\%\) or \(155\%\)).

(b)
High relative humidity means there is a high concentration of water vapor in the atmosphere surrounding the leaf. This increases the water potential of the external air, which narrows (makes less steep) the water potential gradient between the moist air spaces inside the leaf (sub-stomatal cavities) and the external environment. Therefore, the rate of diffusion of water vapor out of the stomata via osmosis/diffusion decreases.

(c)
Sunken stomata are located inside pits or depressions in the leaf epidermis. This structure traps moist air/water vapor close to the stomatal pore, preventing it from being blown away by wind currents. This microenvironment increases the local humidity around the stomata, which reduces the water potential gradient between the inside and outside of the leaf, thereby decreasing the rate of transpiration.

(d)
Water evaporates from the cell walls of mesophyll cells into air spaces and diffuses out of the stomata (transpiration). This lowers the water potential of mesophyll cells, drawing water from adjacent cells and ultimately the xylem. Water molecules form hydrogen bonds with one another (cohesion), creating a continuous column of water extending from the roots up to the leaves. As water is pulled upwards, tension is created within the xylem. Additionally, water molecules adhere to the hydrophilic cellulose walls of the xylem vessels (adhesion), preventing the column of water from breaking under gravity.

(a)
- Mark 1: Correct substitution shown: \(\frac{2.8 - 1.1}{1.1} \times 100\).
- Mark 2: Correct final value of \(154.5\%\), \(154.55\%\) or \(155\%\).

(b)
- Mark 1: Higher relative humidity increases the water potential of the external atmosphere.
- Mark 2: This reduces the steepness of the water potential gradient between the sub-stomatal cavity and the atmosphere.
- Mark 3: Resulting in a reduced rate of diffusion of water vapor out of the stomata.

(c)
- Mark 1: Sunken stomata trap a layer of moist air/water vapor in the pit.
- Mark 2: This moist air layer reduces the water potential gradient between the internal leaf space and the immediate external air.
- Mark 0.7: Minimizes diffusion/transpiration rate under dry/windy conditions.

(d)
- Mark 1: Water transpires/evaporates from leaves, creating tension (negative pressure) that pulls the water column.
- Mark 2: Cohesion due to hydrogen bonding between water molecules maintains a continuous, unbroken column.
- Mark 3: Adhesion between water molecules and the xylem vessel walls helps support the water column against gravity.

(a)
Number of cells in mitosis (Control Group) = \(50\text{ (Prophase)} + 40\text{ (Metaphase)} + 30\text{ (Anaphase)} + 60\text{ (Telophase)} = 180\).
Total cells = \(1000\).
Mitotic Index = \(\frac{180}{1000} = 0.18\) (or \(18\%\)).

(b)
Compound X most likely inhibits Metaphase (or the metaphase-to-anaphase transition). In the control, only \(4\%\) (\(40 / 1000\)) of cells are in metaphase. In the Compound X treated group, \(61\%\) (\(580 / 950\)) of the cells are arrested in metaphase, whereas the proportion of cells in subsequent stages (anaphase and telophase) has dropped dramatically compared to the control.

(c)
Spindle fibers are responsible for attaching to centromeres of sister chromatids during metaphase and contracting to pull the separated sister chromatids to opposite poles of the cell during anaphase. If the shortening of spindle fibers is prevented, sister chromatids cannot be separated or moved apart. The cell will remain arrested in metaphase, preventing the completion of mitosis and leading to cell death (apoptosis).

(d)
- Similiarity: Both processes result in two genetically identical daughter cells and involve replication of DNA before division begins.
- Differences: Mitosis involves the condensation of chromosomes, breakdown of a nuclear envelope, and formation of a spindle apparatus, none of which occur during binary fission. In addition, binary fission involves the replication and separation of a single circular DNA molecule and plasmids, whereas mitosis involves linear eukaryotic chromosomes.

(a)
- Mark 1: Correct identification of mitotic cells (\(50+40+30+60 = 180\)) and division by total cells (\(1000\)).
- Mark 2: Correct mitotic index of \(0.18\) or \(18\%\).

(b)
- Mark 1: Identifies Metaphase (or the transition to Anaphase) as the inhibited stage.
- Mark 2: Cites data showing a very large accumulation of cells in metaphase in the Compound X group (e.g., \(580/950\) or \(61\%\) vs \(40/1000\) or \(4\%\)).
- Mark 3: Mentions the corresponding decrease of cells in subsequent stages (Anaphase/Telophase) in the treatment group.

(c)
- Mark 1: Identifies that sister chromatids will not be pulled to opposite poles of the cell.
- Mark 2: Explains that anaphase cannot occur / sister chromatids remain at the equator.
- Mark 0.7: Concludes that mitosis/cell division is halted / cell death occurs.

(d)
- Mark 1: Both replicate their DNA prior to division / produce two genetically identical daughter cells.
- Mark 2: Mitosis involves spindle fibers, chromosome condensation, and nuclear envelope breakdown; binary fission does not.
- Mark 3: Mitosis divides linear chromosomes; binary fission divides a single circular DNA loop (and plasmids).

(a)
CD4+ count at Year 5 = \(350\text{ cells mm}^{-3}\)
CD4+ count at Year 10 = \(80\text{ cells mm}^{-3}\)
Decline in CD4+ count = \(350 - 80 = 270\text{ cells mm}^{-3}\)
Time interval = \(10 - 5 = 5\text{ years}\)
Rate of decline = \(\frac{270}{5} = 54\text{ cells mm}^{-3}\text{ year}^{-1}\).

(b)
CD4+ T-lymphocytes (specifically T-helper cells) play a crucial role in coordinating the immune response. When activated by antigen-presenting cells, they release cytokines that stimulate B-lymphocytes to divide (clonal selection) and differentiate into antibody-producing plasma cells. They also activate cytotoxic T-cells to destroy infected cells. When CD4+ cell count drops critically low, B-cell activation and antibody production are severely impaired, and the cellular immune response is compromised. Without this functional adaptive immune response, the body cannot defend itself against normally harmless, opportunistic pathogens.

(c)
HIV is a retrovirus containing single-stranded RNA. Once inside the host CD4+ T-cell, reverse transcriptase uses the viral RNA template to synthesize a complementary single strand of DNA. The enzyme then synthesizes a second complementary DNA strand, producing a double-stranded viral DNA molecule. This viral DNA can then enter the nucleus and integrate into the host cell's genome, allowing the virus to be transcribed and translated by the host cell's machinery.

(d)
Antibiotics target specific cellular structures or metabolic processes unique to prokaryotic cells (bacteria), such as cell wall synthesis (peptidoglycan) or bacterial ribosomes. Viruses are non-living, non-cellular particles without a cell wall or independent metabolic pathways; they rely entirely on host cell cellular machinery for replication, so antibiotics have no effect on them.

(a)
- Mark 1: Shows correct working: \(\frac{350 - 80}{5}\) or \(\frac{270}{5}\).
- Mark 2: Correct rate of \(54\) with correct units (\(\text{cells mm}^{-3}\text{ year}^{-1}\)).

(b)
- Mark 1: CD4+ cells are helper T-cells which secrete cytokines to coordinate immune responses.
- Mark 2: Lack of helper T-cells means B-cells are not stimulated/activated to clonal selection.
- Mark 3: Consequently, antibody production by plasma cells is severely reduced.
- Mark 4: Loss of helper T-cells also impairs cytotoxic T-cell function, leaving the host vulnerable to pathogens (opportunistic infections).

(c)
- Mark 1: Reverse transcriptase uses viral single-stranded RNA as a template.
- Mark 2: To synthesize a complementary strand of viral DNA.
- Mark 3: It then synthesizes the second DNA strand to make it double-stranded, which integrates into host genome.

(d)
- Mark 1: Antibiotics target bacterial-specific cell walls / structures / metabolic processes.
- Mark 0.7: Viruses lack these targets/structures / exist inside host cells / use host cell machinery.

(a)
At \(10.0\ \mu\text{g cm}^{-3}\):
Net water secretion rate = \(3.10\text{ cm}^3\text{ g}^{-1}\text{ h}^{-1}\)
Net \(\text{Cl}^-\) secretion rate = \(12.4\ \mu\text{mol g}^{-1}\text{ h}^{-1}\)
Ratio = \(\frac{3.10}{12.4} = 0.25\) (or \(1 : 4\) or \(0.25\text{ cm}^3\ \mu\text{mol}^{-1}\)).

(b)
When *Vibrio cholerae* colonizes the small intestine, it secretes cholera toxin. The toxin binds to receptors on the epithelial cells of the intestinal wall, leading to the continuous activation of a membrane protein (adenylyl cyclase), which increases cyclic AMP (cAMP) levels. This causes the active transport of chloride ions (\(\text{Cl}^-\)) out of the epithelial cells into the lumen of the intestine. The high concentration of solute (\(\text{Cl}^-\)) in the lumen lowers its water potential. Consequently, a water potential gradient is established, forcing water to move out of the epithelial cells and blood capillaries into the intestinal lumen by osmosis, resulting in severe watery diarrhea and life-threatening dehydration.

(c)
Using pure water alone is ineffective because the osmotic gradient is pulling water into the lumen. ORT contains glucose and sodium ions to exploit the sodium-glucose co-transport proteins present in the membranes of epithelial cells in the small intestine. As sodium ions and glucose are co-transported together into the epithelial cells, the concentration of solutes inside these cells increases. This lowers the water potential of the epithelial cells, creating a water potential gradient that draws water out of the lumen back into the cells and blood vessels by osmosis, thereby rehydrating the patient.

(a)
- Mark 1: Correctly substitutes values from Table 4: \(\frac{3.10}{12.4}\).
- Mark 2: Correctly calculated ratio of \(0.25\) or \(1 : 4\).

(b)
- Mark 1: Cholera toxin binds to receptors on intestinal epithelial cells, raising cAMP levels.
- Mark 2: This stimulates the active transport of chloride ions (\(\text{Cl}^-\)) into the lumen.
- Mark 3: The accumulation of chloride ions in the lumen lowers the water potential of the lumen.
- Mark 4: A water potential gradient is established (from epithelial cells to lumen).
- Mark 5: Water moves from the blood/cells into the lumen by osmosis, resulting in diarrhea/dehydration.

(c)
- Mark 1: Pure water cannot be easily absorbed due to the unfavorable osmotic gradient.
- Mark 2: ORT contains glucose and sodium to target sodium-glucose co-transporters in epithelial cell membranes.
- Mark 3: Active absorption of sodium and glucose lowers the water potential inside the epithelial cells.
- Mark 0.7: Allowing water to move out of the lumen into blood by osmosis (rehydration).

a) Parent 1 (pink, homozygous lobed): \(C^R C^W LL\). Parent 2 (pink, smooth): \(C^R C^W ll\).

b) Gametes:
- Parent 1: \(C^R L\) and \(C^W L\)
- Parent 2: \(C^R l\) and \(C^W l\)

Crossing these gametes in a 2x2 Punnett square yields:
- \(C^R L \times C^R l \rightarrow C^R C^R Ll\) (Red, lobed)
- \(C^R L \times C^W l \rightarrow C^R C^W Ll\) (Pink, lobed)
- \(C^W L \times C^R l \rightarrow C^R C^W Ll\) (Pink, lobed)
- \(C^W L \times C^W l \rightarrow C^W C^W Ll\) (White, lobed)

Ratio: 1 Red, lobed : 2 Pink, lobed : 1 White, lobed.

c) Evaluation:
- The student is incorrect. The probability of getting a pink flower with smooth leaves is 0.
- Reason: All gametes from Parent 1 carry the dominant allele \(L\) because Parent 1 is homozygous dominant for leaf shape (\(LL\)).
- Therefore, all \(F_1\) offspring must inherit one \(L\) allele (and thus have genotype \(Ll\)).
- Because \(L\) is dominant to \(l\), 100% of the offspring will have lobed leaves; none can have smooth leaves (\(ll\)).

a) [2 marks total]
- 1 mark for Parent 1 genotype: \(C^R C^W LL\) (accept alternative letters if symbols remain consistent)
- 1 mark for Parent 2 genotype: \(C^R C^W ll\)

b) [5 marks total]
- 1 mark for identifying correct gametes from Parent 1: \(C^R L\) and \(C^W L\)
- 1 mark for identifying correct gametes from Parent 2: \(C^R l\) and \(C^W l\)
- 1 mark for correct Punnett square or crossing lines showing all 4 offspring genotypes (\(C^R C^R Ll\), \(C^R C^W Ll\), \(C^W C^W Ll\))
- 1 mark for matching genotypes to correct phenotypes (Red lobed, Pink lobed, White lobed)
- 1 mark for correct ratio: 1 Red lobed : 2 Pink lobed : 1 White lobed (or equivalent probabilities: 0.25 : 0.50 : 0.25)

c) [3.7 marks total]
- 1 mark for stating that the student's claim is incorrect / the probability is 0.
- 1 mark for explaining that Parent 1 can only pass on the dominant lobed allele (\(L\)) / is homozygous dominant.
- 1 mark for explaining that all offspring will be heterozygous (\(Ll\)) for leaf shape.
- 0.7 marks for concluding that because lobed is dominant, no offspring can display the smooth phenotype (which requires homozygous recessive genotype \(ll\)).

a) Conversion of ammonium to nitrate is called nitrification.
- It occurs in two stages, both of which are oxidation reactions carried out by aerobic nitrifying bacteria.
- Stage 1: Ammonium ions (\(NH_4^+\)) are converted into nitrite ions (\(NO_2^-\)) by nitrifying bacteria (such as Nitrosomonas).
- Stage 2: Nitrite ions (\(NO_2^-\)) are converted into nitrate ions (\(NO_3^-\)) by another group of nitrifying bacteria (such as Nitrobacter).

b) Soil waterlogging deprives the soil of oxygen, creating anaerobic conditions.
- Anaerobic conditions favor denitrifying bacteria (e.g., Pseudomonas).
- These bacteria carry out denitrification, using nitrate ions as electron acceptors during anaerobic respiration.
- This converts nitrate ions (\(NO_3^-\)) into nitrogen gas (\(N_2\)), which escapes into the atmosphere, causing a decrease in soil nitrate concentration.

c) Nitrification inhibitors slow down the conversion of ammonium to nitrates.
- Ammonium ions (\(NH_4^+\)) carry a positive charge and bind electrostatically to negatively charged clay particles in the soil, resisting leaching.
- Nitrate ions (\(NO_3^-\)) carry a negative charge and do not bind to soil particles, making them highly soluble and easily washed out of the soil (leached) during heavy rain.
- By maintaining nitrogen as ammonium, inhibitors prevent leaching, ensuring nitrogen remains in the soil for plants to absorb.

a) [4 marks total]
- 1 mark for naming the process as nitrification.
- 1 mark for stating that it is an aerobic/oxidation reaction.
- 1 mark for describing the conversion of ammonium to nitrite (\(NH_4^+\rightarrow NO_2^-\)) by nitrifying bacteria.
- 1 mark for describing the conversion of nitrite to nitrate (\(NO_2^-\rightarrow NO_3^-\)) by nitrifying bacteria.

b) [3.7 marks total]
- 1 mark for explaining that waterlogging leads to anaerobic/low oxygen conditions.
- 1 mark for naming the process of denitrification.
- 1 mark for identifying that denitrifying bacteria are responsible.
- 0.7 marks for explaining that nitrates are converted into nitrogen gas (\(N_2\)), thereby reducing nitrate concentration in the soil.

c) [3 marks total]
- 1 mark for stating that ammonium ions (\(NH_4^+\)) are positively charged and bind to soil/clay particles.
- 1 mark for stating that nitrate ions (\(NO_3^-\)) are negatively charged and are easily leached/washed out of the soil.
- 1 mark for concluding that inhibitors reduce nitrogen leaching/loss, making nitrogen available to plants for longer, which promotes growth/protein synthesis.

a) i) Frequency of homozygous recessive genotype (\(q^2\)) = \(18 / 200 = 0.09\).
- Frequency of recessive allele \(b\) (\(q\)) = \(\sqrt{0.09} = 0.30\).
- Since \(p + q = 1\), the frequency of the dominant allele \(B\) (\(p\)) = \(1 - 0.30 = 0.70\).

ii) Frequency of heterozygotes = \(2pq = 2 \times 0.70 \times 0.30 = 0.42\).
- Expected number of heterozygous mice in a sample of 200 = \(0.42 \times 200 = 84\).

b) Statistical Interpretation:
- The calculated \(\chi^2\) value (5.42) is greater than the critical value (3.84) at \(p = 0.05\).
- Therefore, the probability (\(p\)) of the observed differences between expected and observed frequencies occurring by chance is less than 0.05 (less than 5%).
- We reject the null hypothesis.
- This indicates that there is a statistically significant difference in allele/phenotype frequencies between the two generations.
- Biologically, this means the population is no longer in Hardy-Weinberg equilibrium.
- Fragmentation of the forest into smaller, isolated habitats has likely led to genetic drift (random fluctuations in small populations) or changed selection pressures (natural selection), leading to evolutionary change in the fragment.

a) i) [2 marks total]
- 1 mark for correct calculation of recessive allele frequency: \(q = 0.30\) (or showing working \(\sqrt{0.09}\))
- 1 mark for correct calculation of dominant allele frequency: \(p = 0.70\)

ii) [3 marks total]
- 1 mark for formula or use of heterozygote frequency expression: \(2pq\)
- 1 mark for correct calculation of heterozygote frequency: \(2 \times 0.70 \times 0.30 = 0.42\)
- 1 mark for correct final expected number: 84

b) [5.7 marks total]
- 1 mark for stating that the calculated \(\chi^2\) value (5.42) is greater than the critical value (3.84).
- 1 mark for stating that the difference is statistically significant.
- 1 mark for defining the probability value: the probability of this difference occurring by chance is less than 5% (or \(p < 0.05\)).
- 1 mark for stating that the null hypothesis is rejected.
- 1 mark for explaining that the population is no longer in Hardy-Weinberg equilibrium / allele frequencies have changed.
- 0.7 marks for linking this change to genetic drift (due to smaller population in the fragment) or natural selection/selective advantage.

a) Carbon dioxide (\(CO_2\)) from the air is fixed by combining with the 5-carbon compound ribulose bisphosphate (RuBP).
- This reaction is catalyzed by the enzyme rubisco.
- It forms an unstable 6-carbon intermediate compound which immediately splits into two molecules of the 3-carbon compound glycerate 3-phosphate (GP).

b) Effect of turning light off:
- The light-dependent reactions cease, stopping the production of ATP and reduced NADP (NADPH).
- ATP and reduced NADP are required to reduce GP into triose phosphate (TP).
- Because GP cannot be reduced to TP, GP accumulates, and its concentration initially increases.
- The conversion of RuBP to GP continues (as long as \(CO_2\) is present), but RuBP cannot be regenerated from TP because that process also requires ATP. Thus, the concentration of RuBP rapidly falls.

c) Effect of reducing \(CO_2\) concentration:
- The concentration of RuBP increases.
- This is because there is less \(CO_2\) available to combine with RuBP, so less RuBP is converted into GP.
- However, under constant light, the conversion of existing GP to TP and the regeneration of TP into RuBP continue normally. Thus, RuBP is regenerated but not used up, leading to its accumulation.

a) [3 marks total]
- 1 mark for carbon dioxide combining with ribulose bisphosphate (RuBP).
- 1 mark for role of rubisco enzyme in catalyzing this reaction.
- 1 mark for stating that the resulting unstable 6-carbon compound splits into two molecules of glycerate 3-phosphate (GP).

b) [4.7 marks total]
- 1 mark for stating that light-dependent reactions stop, so no ATP or reduced NADP are produced.
- 1 mark for explaining that GP cannot be reduced to triose phosphate (TP) (causing GP concentration to rise/accumulate).
- 1 mark for explaining that RuBP continues to react with carbon dioxide to form GP.
- 1 mark for explaining that RuBP cannot be regenerated from TP because regeneration requires ATP.
- 0.7 marks for concluding that RuBP concentration falls.

c) [3 marks total]
- 1 mark for stating that RuBP concentration increases.
- 1 mark for explaining that less carbon dioxide is available to react with/fix RuBP.
- 1 mark for explaining that regeneration of RuBP from TP still continues because ATP and reduced NADP from the light-dependent reactions are still available.

a) The type of competition is interspecific competition.
- Both species occupy similar ecological niches and compete for the same limiting food source (bacteria).
- *P. aurelia* is better adapted/more efficient at obtaining the food source.
- This leads to competitive exclusion, where *P. aurelia* survives and *P. caudatum* is driven to local extinction.

b) i) Null Hypothesis:
- There is no significant difference in the mean maximum population density of *Paramecium aurelia* grown at 25 °C compared to 15 °C (any observed difference is due to chance alone).

ii) Interpretation of Student's t-test:
- Compare values: The calculated \(t\)-value of 3.12 is greater than the critical value of 2.82 at the \(p = 0.01\) significance level.
- Significance: The difference between the mean population densities is highly statistically significant.
- Probability definition: The probability of this difference occurring by chance is less than 1% (\(p < 0.01\)).
- Null Hypothesis: Reject the null hypothesis.
- Conclusion: Temperature has a highly significant effect on the growth rate/population density of *P. aurelia*, with 25 °C yielding a significantly higher mean density than 15 °C.

a) [3 marks total]
- 1 mark for naming interspecific competition.
- 1 mark for stating that both species compete for the same limiting resource / food supply.
- 1 mark for explaining competitive exclusion: *P. aurelia* is a stronger competitor/better adapted, causing the population of *P. caudatum* to crash.

b) i) [1.7 marks total]
- 1.7 marks for stating there is no significant difference in the mean maximum population density of *P. aurelia* between 25 °C and 15 °C (accept correct phrasing containing 'no significant difference').

ii) [6 marks total]
- 1 mark for stating the calculated \(t\)-value (3.12) is greater than the critical value (2.82).
- 1 mark for stating that the difference is statistically significant.
- 1 mark for defining probability: the probability that this difference occurred by chance is less than 1% / less than 0.01.
- 1 mark for stating that the null hypothesis is rejected.
- 1 mark for stating that the alternative hypothesis is supported / temperature has a significant effect.
- 1 mark for stating that the population density at 25 °C is significantly higher than at 15 °C.

a) Speciation via geographic isolation:
- The motorway acts as a physical barrier, preventing gene flow (interbreeding) between the two separated snail populations.
- The two environments on either side of the motorway may have different abiotic conditions (e.g., microclimate, moisture) or biotic conditions (e.g., different predators, food availability).
- Consequently, the two populations experience different selection pressures.
- Random mutations occur independently in the gene pools of each population, introducing different new alleles.
- Natural selection acts on these mutations: alleles that offer a selective advantage in each specific environment are favored and increase in frequency.
- Over many generations, genetic differences accumulate (genetic divergence).
- Eventually, the populations become reproductively isolated; even if reunited, they can no longer interbreed to produce fertile offspring, constituting separate species.

b) Natural selection role:
- In any population, there is variation in phenotypes due to different alleles (arising from mutations).
- Snails produce more offspring than the environment can support, leading to a struggle for survival (competition).
- Certain alleles give rise to phenotypes that are better adapted to the local environment (e.g., shell color camouflage protecting against bird predators).
- Individuals possessing these advantageous alleles are more likely to survive selection pressures and successfully reproduce (differential reproductive success).
- These individuals pass on their advantageous alleles to their offspring.
- Over generations, the frequency of these advantageous alleles increases in the population's gene pool, while disadvantageous alleles decrease.

a) [6 marks total]
- 1 mark for stating that geographic isolation prevents gene flow / interbreeding between populations.
- 1 mark for stating that different environments have different selection pressures (abiotic/biotic).
- 1 mark for stating that random mutations occur independently in both populations.
- 1 mark for explaining that natural selection operates on these different alleles, changing the allele frequencies in each gene pool.
- 1 mark for stating that genetic divergence occurs over many generations.
- 1 mark for stating that reproductive isolation occurs, meaning they cannot produce fertile offspring.

b) [4.7 marks total]
- 1 mark for stating that variation exists within the population due to mutations.
- 1 mark for identifying that selection pressures exist / there is a struggle for survival.
- 1 mark for stating that individuals with advantageous alleles are more likely to survive and reproduce.
- 1 mark for explaining that these individuals pass on their advantageous alleles to their offspring.
- 0.7 marks for concluding that the frequency of the advantageous allele increases in the gene pool over time.

a) Production of lactate:
- During glycolysis, glucose is converted into pyruvate, producing a net yield of 2 ATP and converting NAD to reduced NAD.
- Under anaerobic conditions, oxygen is not available to act as the final electron acceptor in the electron transport chain, stopping oxidative phosphorylation, the Krebs cycle, and the Link reaction.
- To allow glycolysis to continue, pyruvate is reduced to lactate.
- This reaction is catalyzed by the enzyme lactate dehydrogenase.
- During this reduction, pyruvate accepts hydrogen atoms from reduced NAD, which oxidizes reduced NAD back to NAD.
- The regeneration of oxidized NAD is essential because it is a required coenzyme for the triose phosphate oxidation step of glycolysis. This allows glycolysis to continue, providing a small continuous yield of ATP (2 ATP per glucose) in the absence of oxygen.

b) Metabolism of lactate and oxygen debt:
- Lactate diffuses out of muscle cells into the bloodstream and is transported to the liver.
- In hepatocytes (liver cells), lactate is oxidized back into pyruvate.
- This pyruvate has two main metabolic fates:
1. It can enter the mitochondria to be fully oxidized via the Link reaction, Krebs cycle, and oxidative phosphorylation to produce carbon dioxide, water, and ATP.
2. It can be converted back into glucose or glycogen via a process called gluconeogenesis.
- Gluconeogenesis is an anabolic, energy-requiring pathway that consumes ATP.
- The ATP required for gluconeogenesis and general metabolic recovery must be generated by aerobic respiration, which requires oxygen. This elevated oxygen consumption post-exercise to metabolize lactate is known as paying off the 'oxygen debt'.

c) Net yield:
- Anaerobic respiration yields a net of 2 ATP molecules per glucose molecule.
- Aerobic respiration yields significantly more, approximately 30 to 38 ATP molecules per glucose molecule.

a) [5 marks total]
- 1 mark for stating that glycolysis produces pyruvate, ATP, and reduced NAD.
- 1 mark for describing the reduction of pyruvate to lactate by accepting hydrogen from reduced NAD.
- 1 mark for naming the enzyme lactate dehydrogenase.
- 1 mark for explaining that this oxidizes / regenerates NAD.
- 1 mark for explaining that regenerated NAD allows glycolysis to continue (by acting as a hydrogen acceptor in triose phosphate oxidation), maintaining a small ATP output.

b) [4.7 marks total]
- 1 mark for stating that lactate is transported via blood to the liver where it is oxidized back to pyruvate.
- 1 mark for stating that pyruvate can be used in aerobic respiration or converted to glucose/glycogen (gluconeogenesis).
- 1 mark for explaining that gluconeogenesis / glycogen synthesis is an active process requiring ATP.
- 1 mark for explaining that this ATP must be supplied by aerobic respiration, which requires oxygen.
- 0.7 marks for concluding that this is why extra oxygen (oxygen debt) must be inhaled after exercise.

c) [1 mark total]
- 1 mark for stating 2 ATP for anaerobic respiration, compared to 30-38 ATP (accept any number in this range) for aerobic respiration.