2022 Edexcel A-Level Biology B (9BI0) 模擬試題連答案詳解

(a) RuBisCO catalyzes the carboxylation reaction where carbon dioxide (\(CO_2\)) is combined with ribulose bisphosphate (RuBP), a 5-carbon compound. This produces an unstable 6-carbon intermediate which immediately splits into two molecules of glycerate 3-phosphate (GP).

(b) Substitute the given values into the equation:
\[v = \frac{120 \times 280}{280 + 350}\]
\[v = \frac{33600}{630}\]
\[v = 53.333... \mu\text{mol m}^{-2}\text{ s}^{-1}\]
Rounding to 3 significant figures gives \(53.3 \mu\text{mol m}^{-2}\text{ s}^{-1}\).

(c) Up to the optimum temperature, an increase in temperature increases the kinetic energy of both enzymes and substrates, leading to more frequent successful collisions and a higher rate of GP formation. However, at higher temperatures like \(40^\circ\text{C}\), RuBisCO\'s oxygenase activity increases faster than its carboxylase activity, leading to photorespiration. At very high temperatures, thermal energy disrupts the hydrogen and ionic bonds stabilizing the enzyme\'s tertiary structure, causing denaturation of the active site so that RuBP can no longer bind.

(a)
- 1 mark: Catalyzes the addition of carbon dioxide to RuBP / carboxylation of RuBP.
- 1 mark: Identifies RuBP as the 5C substrate and GP as the product.
- 1 mark: Explains that an unstable 6C intermediate is formed first before splitting into two 3C GP molecules.

(b)
- 1 mark: Correct substitution of values into the equation: \((120 \times 280) / (280 + 350)\).
- 1 mark: Correct calculation of 53.3 (allow 53.33).
- 1 mark: Correct units: \(\mu\text{mol m}^{-2}\text{ s}^{-1}\).

(c)
- 1 mark: Higher kinetic energy increases the frequency of effective collisions and enzyme-substrate complex formation up to the optimum temperature.
- 1 mark: At high temperatures (e.g. \(40^\circ\text{C}\)), RuBisCO exhibits increased affinity for oxygen relative to \(CO_2\), leading to photorespiration.
- 1 mark: Thermal energy disrupts hydrogen/ionic bonds in RuBisCO\'s tertiary structure.
- 1 mark: This denatures the enzyme, altering the shape of the active site so it is no longer complementary to RuBP.

(a) Plating raw milk directly would result in an excessive number of bacteria, leading to confluent growth (a lawn of bacteria) where individual colonies overlap and cannot be counted. Serial dilution reduces the concentration of bacteria to a range that produces separate, countable colonies.

(b) The colony count is 142. The dilution factor is \(10^{-5}\) and the volume plated is \(0.1\text{ cm}^3\).
First, calculate the concentration of bacteria in the \(10^{-5}\) dilution:
\[\text{Concentration in diluted sample} = \frac{142 \text{ colonies}}{0.1 \text{ cm}^3} = 1420 \text{ CFU cm}^{-3}\]
Now, multiply by the dilution factor to find the concentration in the original milk sample:
\[\text{Original concentration} = 1420 \times 10^5 = 1.42 \times 10^8 \text{ CFU cm}^{-3}\]

(c) A haemocytometer counts all cells present (both living and dead), resulting in a total cell count, whereas the viable plate count only counts living cells capable of dividing and forming colonies. The haemocytometer provides immediate results but requires specialized microscopes and grids, whereas plate counting requires an incubation period (usually 24-48 hours). The haemocytometer can suffer from inaccuracies if cell debris is mistaken for cells, whereas plate counts can underestimate cell numbers if multiple bacterial cells clump together to form a single colony.

(a)
- 1 mark: Direct plating would lead to too many colonies / confluent growth / overlapping colonies that are impossible to count.
- 1 mark: Serial dilution ensures colonies are isolated, allowing accurate counting of distinct colony-forming units.

(b)
- 1 mark: Calculation of diluted concentration (1420) or division by 0.1.
- 1 mark: Multiplying the count by the dilution factor (\(10^5\)).
- 1 mark: Correct final answer in standard form to 3 s.f. with units: \(1.42 \times 10^8\text{ CFU cm}^{-3}\).

(c) (Max 5 marks)
- 1 mark: Haemocytometer gives a total count (both living and dead cells) vs. plate count which gives a viable count (living cells only).
- 1 mark: Haemocytometer results are immediate vs. plate counts which require 24-48 hours incubation.
- 1 mark: Haemocytometer requires a microscope and specialized counting chamber/grid.
- 1 mark: Plate count can underestimate actual numbers if bacterial cells clump together (as one colony arises from a clump).
- 1 mark: Haemocytometer can suffer from counting non-cellular debris as cells, causing overestimation.

(a) Let the initial rate of diffusion be \(R_1\), with initial surface area \(A_1\) and initial thickness \(T_1\).
Let \(A_1 = 1.0\) and \(T_1 = 0.5 \mu\text{m}\).
\[R_1 \propto \frac{1.0}{0.5} = 2.0\]

Let the new rate of diffusion be \(R_2\).
The new surface area \(A_2\) is reduced by \(30\%\), so \(A_2 = 1.0 \times (1 - 0.3) = 0.7\).
The new thickness \(T_2 = 1.5 \mu\text{m}\).
\[R_2 \propto \frac{0.7}{1.5} \approx 0.4667\]

Calculate the percentage decrease:
\[\text{Percentage decrease} = \frac{R_1 - R_2}{R_1} \times 100\% = \frac{2.0 - 0.4667}{2.0} \times 100\% = \frac{1.5333}{2.0} \times 100\% = 76.667\%\]
Rounding to 3 significant figures gives \(76.7\%\) (or \(76.7\%\)).

(b) Mammalian alveoli are adapted by: having walls that are only one squamous epithelial cell thick to minimize diffusion distance; a vast network of capillaries providing a large surface area for contact; and a lining of moist surfactant which dissolves gases and prevents alveolar collapse.

(c) Ventilation constantly replenishes oxygen and removes carbon dioxide in the alveolar space, maintaining high oxygen concentration and low carbon dioxide concentration in the alveoli. Continuous blood flow brings deoxygenated blood (high carbon dioxide, low oxygen) to the alveoli and rapidly carries away newly oxygenated blood. Together, these processes maintain a steep concentration gradient across the alveolar membrane.

(a)
- 1 mark: Recognizes the new surface area is \(70\%\) (or 0.7) of the original surface area.
- 1 mark: Shows correct substitution for initial rate (e.g. \(1 / 0.5 = 2.0\)) and new rate (e.g. \(0.7 / 1.5 = 0.467\)).
- 1 mark: Correctly calculates the ratio of the new rate to the old rate (\(0.467 / 2.0 = 0.233\)).
- 1 mark: Correctly calculates percentage decrease: \(76.7\%\) (accept \(76.67\%\) or \(77\%\)).

(b)
- 1 mark: Thin alveolar wall / squamous epithelium only one cell thick to shorten diffusion distance.
- 1 mark: Extremely large surface area due to millions of highly folded alveoli.
- 1 mark: Moist surfactant layer to dissolve respiratory gases / prevent alveoli from collapsing.

(c)
- 1 mark: Ventilation continuously brings fresh air with high oxygen and removes air with high carbon dioxide from the alveoli.
- 1 mark: Continuous blood flow rapidly carries oxygen away from the lungs and delivers carbon-dioxide-rich blood.
- 1 mark: These processes ensure that the concentrations of gases on either side of the exchange surface do not reach equilibrium.

(a) Biodiversity refers to the variety of species in an ecosystem. Species richness is the number of different species present in a community. Species evenness is a measure of the relative abundance of the different species making up the richness. High biodiversity requires both high species richness and high species evenness.

(b) Let\'s calculate the terms for the formula:
First, calculate total \(N\):
\[N = 45 + 15 + 25 + 15 = 100\]
\[N(N-1) = 100 \times 99 = 9900\]

Now, calculate \(n(n-1)\) for each species:
- Species A: \(45 \times 44 = 1980\)
- Species B: \(15 \times 14 = 210\)
- Species C: \(25 \times 24 = 600\)
- Species D: \(15 \times 14 = 210\)

Sum of \(n(n-1)\):
\[\sum n(n-1) = 1980 + 210 + 600 + 210 = 3000\]

Now, substitute into the formula:
\[D = 1 - \frac{3000}{9900}\]
\[D = 1 - 0.30303...\]
\[D = 0.69696...\]
Rounding to 3 decimal places gives \(0.697\).

(c) A high value of \(D\) (close to 1) indicates high biodiversity and evenness. This means the ecosystem is highly stable because food webs are complex, with many interconnections. If one species declines due to an environmental change or disease, other species can fill the niche or serve as alternative food sources, making the community resilient.

(a)
- 1 mark: Defines species richness as the number of different species in a habitat.
- 1 mark: Defines species evenness as the relative abundance / proportion of each species.
- 1 mark: Notes that high biodiversity requires both high richness and high evenness (not dominated by a single species).

(b)
- 1 mark: Correctly calculates \(N = 100\) and \(N(N-1) = 9900\).
- 1 mark: Correctly calculates individual \(n(n-1)\) values (1980, 210, 600, 210).
- 1 mark: Correctly sums \(n(n-1)\) to 3000.
- 1 mark: Correct final calculation of \(D = 0.697\) (accept 0.70 if rounding error is explained, but 0.697 is accurate to 3 d.p.).

(c)
- 1 mark: High index indicates a highly diverse and stable ecosystem.
- 1 mark: Highly complex food webs / multiple alternative pathways for energy flow.
- 1 mark: Resilient to environmental changes/pests/diseases because if one species dies out, others can take its place.

(a) Competitive inhibitors have a similar shape to the substrate and bind directly to the active site, blocking substrate binding. This can be overcome by increasing substrate concentration, so the maximum rate of reaction (\(V_{max}\)) remains unchanged. Non-competitive inhibitors bind to an allosteric site (a site other than the active site), altering the tertiary structure of the enzyme and changing the active site shape. This prevents the substrate from binding/reacting, lowering the overall \(V_{max}\) regardless of substrate concentration.

(b) Calculate rates:
- Rate without inhibitor = \(\frac{4.8\text{ mg}}{5.0\text{ min}} = 0.96\text{ mg min}^{-1}\).
- Rate with inhibitor = \(\frac{1.8\text{ mg}}{5.0\text{ min}} = 0.36\text{ mg min}^{-1}\).
Calculate percentage reduction:
- Difference in rate = \(0.96 - 0.36 = 0.60\text{ mg min}^{-1}\).
- Percentage reduction = \(\frac{0.60}{0.96} \times 100\% = 62.5\%\).

(c) The inhibitory effect of a competitive inhibitor can be overcome by significantly increasing the concentration of the substrate. This increases the probability of a substrate molecule colliding with and binding to the active site rather than an inhibitor molecule, outcompeting the inhibitor and restoring the rate of reaction.

(a)
- 1 mark: Competitive inhibitor binds to the active site (similar shape to substrate) whereas non-competitive inhibitor binds to an allosteric site.
- 1 mark: Non-competitive binding alters the tertiary structure / shape of active site.
- 1 mark: Competitive inhibition does not alter the maximum rate of reaction (\(V_{max}\)) at high substrate concentrations.
- 1 mark: Non-competitive inhibition reduces the maximum rate of reaction (\(V_{max}\)) permanently.

(b)
- 1 mark: Correct rates calculated: \(0.96\text{ mg min}^{-1}\) (uninhibited) and \(0.36\text{ mg min}^{-1}\) (inhibited).
- 1 mark: Shows correct working for percentage reduction (e.g., \((0.96 - 0.36) / 0.96 \times 100\)).
- 1 mark: Correct percentage reduction: \(62.5\%\).

(c)
- 1 mark: Increase substrate concentration.
- 1 mark: Substrate molecules will outcompete/displace the competitive inhibitor.
- 1 mark: Greater likelihood of substrate-active site collisions/binding.

(a) An antigen-presenting cell (such as a macrophage) engulfs the pathogen via phagocytosis. It processes the pathogen\'s antigens and displays them on its cell surface membrane bound to MHC class II proteins. Helper T lymphocytes with complementary receptors bind to these presented antigens, activating the helper T cells to secrete cytokines.

(b) Clonal selection occurs when a specific B cell with membrane-bound antibodies complementary to a specific foreign antigen binds to that antigen. This selection is followed by clonal expansion, where the activated B cell divides rapidly by mitosis in response to cytokines released by helper T cells, producing a clone of identical B cells.

(c) During the primary response, there is a delay (lag phase) because it takes time to find the rare B cell with the complementary antibody and for clonal selection and expansion to occur. Memory B and T cells are produced and remain in circulation. Upon secondary exposure, these memory cells recognize the antigen immediately, rapidly dividing and differentiating into plasma cells. This results in the rapid, massive production of a high concentration of antibodies before symptoms develop.

(a)
- 1 mark: APC engulfs pathogen by phagocytosis, digests it, and displays antigens on outer membrane/MHC.
- 1 mark: Helper T cell with a complementary receptor binds to the presented antigen.
- 1 mark: This activates the helper T cell, triggering it to release cytokines/interleukins.

(b)
- 1 mark: Clonal selection: B cell binds to specific complementary antigen via its surface receptors.
- 1 mark: Cytokines from activated T helper cells stimulate this specific selected B cell.
- 1 mark: Clonal expansion: The selected B cell undergoes rapid mitosis to produce a large clone of genetically identical B cells.

(c)
- 1 mark: Primary response has a lag phase because there are few B cells with the correct receptor / clonal selection and expansion take time.
- 1 mark: Memory B and T cells are produced during the primary response and persist in the body.
- 1 mark: Upon secondary exposure, memory B cells recognize the antigen and rapidly divide/differentiate into plasma cells.
- 1 mark: Results in much faster antibody production and a significantly higher antibody concentration.

(a) Hydrogen ions (protons) are actively pumped out of companion cells into the surrounding cell wall using ATP. This establishes a high proton gradient outside the companion cells. Protons then diffuse back into the companion cells down their concentration gradient via a co-transporter protein, bringing sucrose molecules with them against their concentration gradient. Sucrose then diffuses into the sieve tube elements through plasmodesmata.

(b) The accumulation of sucrose in the sieve tube elements lowers the water potential. Consequently, water moves from the nearby xylem vessels into the sieve tubes by osmosis, creating a high hydrostatic pressure at the source. At the sink, sucrose is unloaded and used or stored, raising water potential so water exits by osmosis. This creates a low hydrostatic pressure at the sink, driving the bulk movement of sap from source to sink down the pressure gradient.

(c) Convert distance and time to the required units:
- Distance: \(45\text{ cm} = 0.45\text{ m}\).
- Time: \(30\text{ minutes} = 0.5\text{ hours}\).
Calculate velocity:
\[\text{Velocity} = \frac{\text{Distance}}{\text{Time}} = \frac{0.45\text{ m}}{0.5\text{ h}} = 0.9\text{ m h}^{-1}\]

(a)
- 1 mark: Protons (\(H^+\)) are actively pumped out of companion cells using ATP.
- 1 mark: Creates a high proton concentration gradient outside the companion cells.
- 1 mark: Protons co-transport sucrose back into companion cells through carrier proteins down proton gradient.
- 1 mark: Sucrose diffuses into sieve tube elements through plasmodesmata.

(b)
- 1 mark: Increased sucrose concentration lowers water potential, causing water to enter sieve tubes from xylem by osmosis.
- 1 mark: Entry of water raises hydrostatic pressure at the source (and unloading lowers pressure at the sink).
- 1 mark: High-to-low hydrostatic pressure gradient causes the mass flow of phloem sap.

(c)
- 1 mark: Converts distance from cm to m: \(45\text{ cm} = 0.45\text{ m}\).
- 1 mark: Converts time from minutes to hours: \(30\text{ minutes} = 0.5\text{ h}\).
- 1 mark: Correct velocity calculation with appropriate units: \(0.9\text{ m h}^{-1}\).

(a) Different photosynthetic pigments absorb different wavelengths of light across the electromagnetic spectrum. Having multiple pigments allows the plant to maximize light absorption for photosynthesis.

(b) Calculate \(R_f\) values using the formula:
\[R_f = \frac{\text{Distance traveled by pigment}}{\text{Distance traveled by solvent front}}\]
- Chlorophyll a \(R_f = \frac{3.5\text{ cm}}{8.4\text{ cm}} = 0.4166... \approx 0.42\).
- Carotene \(R_f = \frac{8.0\text{ cm}}{8.4\text{ cm}} = 0.9523... \approx 0.95\).

(c) Separation depends on the solubility of the pigments in the mobile phase (solvent) and their affinity (adsorption) to the stationary phase (chromatography plate). More soluble pigments (like carotene) travel faster and further with the solvent front, whereas pigments that adsorb more strongly to the stationary phase (like chlorophylls) travel slower.

(d) Pigments can be extracted in a solvent, and the extract placed in a spectrophotometer. Light of varying wavelengths is passed through the extract to measure light absorption. Plotting an absorption spectrum will show distinct peaks of absorption; Chlorophyll a and Chlorophyll b have different peak absorption wavelengths in the red and blue regions of the spectrum (e.g. Chlorophyll a peaks around \(430\text{ nm}\) and \(660\text{ nm}\), while Chlorophyll b peaks around \(450\text{ nm}\) and \(640\text{ nm}\)).

(a)
- 1 mark: Different pigments absorb different wavelengths of light.
- 1 mark: This widens the action spectrum / maximizes energy captured for light-dependent reactions.

(b)
- 1 mark: Stating or showing use of the correct formula for \(R_f\).
- 1 mark: Chlorophyll a \(R_f = 0.42\) (must be rounded to 2 d.p.).
- 1 mark: Carotene \(R_f = 0.95\) (must be rounded to 2 d.p.).

(c)
- 1 mark: Pigments with higher solubility in the mobile phase travel faster / further.
- 1 mark: Pigments with higher adsorption / affinity for the stationary phase travel slower / less far.

(d)
- 1 mark: Use of a spectrophotometer to measure absorption of light at different wavelengths.
- 1 mark: Generating/comparing the absorption spectrum for both pigments.
- 1 mark: Identifying that Chlorophyll a and Chlorophyll b peak at different, specific wavelengths in the blue and red regions.

### Part (a)
1. **Determine the time spent in the exponential growth phase (\(t\)):**
\[t = 8.0\text{ hours} - 1.5\text{ hours (lag phase)} = 6.5\text{ hours}\]

2. **Calculate the log values:**
- \(N_0 = 1.2 \times 10^3\text{ cells cm}^{-3}\)
- \(\log_{10}(N_0) = \log_{10}(1.2 \times 10^3) = 3.0792\)
- \(N_t = 3.4 \times 10^7\text{ cells cm}^{-3}\)
- \(\log_{10}(N_t) = \log_{10}(3.4 \times 10^7) = 7.5315\)

3. **Substitute the values into the formula:**
\[k = \frac{7.5315 - 3.0792}{0.301 \times 6.5}\]
\[k = \frac{4.4523}{1.9565} \approx 2.2756\text{ hours}^{-1}\]

4. **Round to 3 significant figures and state units:**
- \(k = 2.28\)
- Units: \(\text{hour}^{-1}\) (or \(\text{h}^{-1}\) or \(\text{generations hour}^{-1}\))

### Part (b)(i)
- Label 5 test tubes from \(10^{-1}\) to \(10^{-5}\) and add exactly \(9.0\text{ cm}^3\) of sterile water (or sterile diluent/nutrient broth) into each tube.
- Pipette \(1.0\text{ cm}^3\) of the original culture into the first tube (\(10^{-1}\)) and mix thoroughly.
- Pipette \(1.0\text{ cm}^3\) of the mixture from the first tube into the second tube (\(10^{-2}\)), mix thoroughly, and repeat this serial transfer sequentially up to the fifth tube (\(10^{-5}\)), ensuring a fresh sterile pipette tip is used for each transfer to prevent carryover of cells.

### Part (b)(ii)
- The undiluted culture contains a very high concentration of bacteria (\(3.4 \times 10^7\text{ cells cm}^{-3}\)). Plating this directly would result in colonies overlapping to form a continuous sheet of bacterial growth (confluent growth or a 'lawn' of bacteria).
- This makes it impossible to distinguish and count individual, discrete colonies.
- Plating multiple dilutions ensures that at least one dilution will produce a countable number of distinct colonies (ideally between 30 and 300) from which the original concentration can be calculated back accurately.

### Part (a) [4 marks]
- **MP1:** Correct calculation of exponential phase duration (\(t = 6.5\text{ hours}\)).
- **MP2:** Correct calculation of the difference in logs (\(\log_{10} N_t - \log_{10} N_0 = 4.45\)).
- **MP3:** Correct final value calculated to 3 significant figures: \(2.28\) (Accept range \(2.27\) to \(2.28\)).
- **MP4:** Correct unit stated: \(\text{hour}^{-1}\), \(\text{h}^{-1}\), or \(\text{generations hour}^{-1}\) (or per hour).

### Part (b)(i) [3 marks]
- **MP1:** Add \(9.0\text{ cm}^3\) of sterile diluent/water/saline to each of 5 tubes. (Accept alternative correct proportions, e.g., \(0.9\text{ cm}^3\) diluent and \(0.1\text{ cm}^3\) culture).
- **MP2:** Transfer \(1.0\text{ cm}^3\) of culture to first tube, mix, and transfer \(1.0\text{ cm}^3\) from this tube to the next, repeating sequentially up to the fifth tube.
- **MP3:** Use a fresh/sterile pipette tip (or syringe) for each transfer stage.

### Part (b)(ii) [3 marks]
- **MP1:** High cell density in undiluted culture would lead to overlapping colonies / confluent growth / a lawn of bacterial growth.
- **MP2:** Therefore, individual colonies cannot be distinguished or counted.
- **MP3:** Diluting ensures that a plate is obtained with a countable/measurable range of distinct colonies (e.g., 30 to 300) to allow accurate back-calculation of original numbers.

(a) DCPIP acts as an electron acceptor that intercepts electrons from the electron transport chain (replacing NADP). When reduced, it changes color from blue to colorless, leading to a decrease in absorbance at 600 nm. (b) Change in absorbance = 0.82 - 0.18 = 0.64. Time taken = 4 minutes and 30 seconds = 4.5 minutes. Rate of reduction = 0.64 / 4.5 = 0.1422... which rounds to 0.14 absorbance units per minute. (c) The inhibitor blocks electron transfer from photosystem II to plastoquinone, which stops the flow of electrons along the electron transport chain. Consequently, DCPIP cannot be reduced, causing the rate of the Hill reaction to drop to zero. Furthermore, the lack of electron flow prevents protons from being pumped into the thylakoid lumen, so no proton gradient is established. Without a proton motive force, there is no flow of protons through ATP synthase (chemiosmosis), completely halting photophosphorylation and ATP synthesis.

[Part a] 1 mark: DCPIP acts as an electron acceptor/intercepts electrons from the transport chain (or replaces NADP). 1 mark: reduced DCPIP changes from blue to colorless. 1 mark: this color change causes a measurable decrease in absorbance at 600 nm. [Part b] 1 mark: Correct calculation of absorbance change (0.64) and conversion of time to minutes (4.5). 1 mark: Correct calculation of rate to 2 decimal places (0.14). [Part c] 1 mark: Inhibitor stops electron transfer, so DCPIP is not reduced (Hill reaction ceases/absorbance does not decrease). 1 mark: Interruption of the electron transport chain prevents the pumping of protons into the thylakoid lumen. 1 mark: This prevents the establishment of a proton gradient / proton motive force. 1 mark: No flow of protons through ATP synthase (chemiosmosis). 1 mark: Consequently, ATP synthesis/photophosphorylation ceases.

(a) Species richness is the number of different species in a community or habitat. Species evenness is a measure of the relative abundance of each of these species in the community. (b) Total population N = 45 + 15 + 10 = 70. For Species X: (n/N)^2 = (45/70)^2 = 0.413265. For Species Y: (n/N)^2 = (15/70)^2 = 0.045918. For Species Z: (n/N)^2 = (10/70)^2 = 0.020408. Sum of (n/N)^2 = 0.413265 + 0.045918 + 0.020408 = 0.479591. D = 1 - 0.479591 = 0.520409, which rounds to 0.520. (c) Field A has a higher Simpson's Index of Diversity (0.520) than Field B (0.342), indicating that the community in Field A is more biodiverse and stable. A higher diversity index means the ecosystem is less susceptible to environmental changes because there are alternative food webs and ecological pathways. Intensive grazing in Field B has reduced biodiversity, likely because dominant or grazing-tolerant plant species outcompete others, or because the physical pressure of intensive grazing destroys sensitive species, lowering both species richness and evenness.

[Part a] 1 mark: Correct definition of species richness (number of different species). 1 mark: Correct definition of species evenness (relative abundance of each species). [Part b] 1 mark: Calculates total N = 70. 1 mark: Correctly calculates individual (n/N)^2 values or their sum (0.479591). 1 mark: Correct final calculation of D to 3 decimal places (0.520). [Part c] 1 mark: States Field A has higher diversity/stability than Field B. 1 mark: Explains that higher diversity makes the ecosystem more stable / less vulnerable to environmental change. 1 mark: Explains that higher diversity provides alternative food webs / ecological niches. 1 mark: Explains that intensive grazing reduces species richness/evenness. 1 mark: Suggests a reason for reduction in Field B (e.g., grazing-tolerant species dominant, physical damage, lack of recovery time).

(a) According to Fick's Law, rate is proportional to (Surface Area x Concentration Difference) / Thickness. Let the original rate be 1. If Surface Area is multiplied by 3, the rate increases by 3 times. If Thickness is halved (divided by 0.5), the rate increases by 1/0.5 = 2 times. Total factor of increase = 3 x 2 = 6. (b) Alveolar ventilation continuously replaces stale air with fresh, oxygen-rich air, maintaining high alveolar oxygen concentration. Meanwhile, continuous blood flow in the surrounding capillary network rapidly removes oxygenated blood and replaces it with deoxygenated blood, keeping blood oxygen levels low. The thin squamous epithelial cell layers minimize diffusion distance but do not maintain the gradient itself, while ventilation and circulation ensure the concentration difference remains high. (c) In insects, oxygen is delivered directly through an extensive trachea and tracheole system, where air-filled tubes branch directly to respiring cells, bypassing the circulatory system entirely. In mammals, oxygen must diffuse from the alveoli into the blood, where it binds to haemoglobin in red blood cells and is transported via a high-pressure cardiovascular system to respiring tissues.

[Part a] 1 mark: Identifies that halving thickness doubles the rate, or shows correct algebraic manipulation (e.g., 3 / 0.5). 1 mark: Correct factor of 6. [Part b] 1 mark: Ventilation/breathing in brings in fresh air high in oxygen. 1 mark: Ventilation/breathing out removes air high in carbon dioxide. 1 mark: Blood flow/circulation constantly removes oxygenated blood from the lungs. 1 mark: Blood flow constantly brings deoxygenated blood to the lungs. [Part c] 1 mark: Insects deliver oxygen directly via tracheoles / gas-filled tubes to cells. 1 mark: Insects do not use a circulatory system / blood for oxygen transport. 1 mark: Mammals transport oxygen via blood / red blood cells / haemoglobin. 1 mark: Mammals require a transport system / heart pump because of greater transport distances / lower surface area to volume ratio.

(a) A total cell count determines the total number of cells in a sample, including both living (viable) and dead (non-viable) cells, typically using a haemocytometer under a microscope. A viable cell count only measures living cells that are capable of dividing and forming colonies when plated onto a solid nutrient medium. (b) Dilution factor = 10^-5. Volume plated = 0.1 cm^3. Viable count in diluted sample = 54 / 0.1 = 540 CFU per cm^3. Concentration in original culture = 540 / 10^-5 = 5.4 x 10^7 CFU per cm^3. (c) During the lag phase, bacteria do not divide; instead, they adapt to the new medium by active gene transcription, synthesizing respiratory and metabolic enzymes, replicating DNA, and growing in size. During the stationary phase, the growth rate equals the death rate (no net change in population size) because essential nutrients become depleted, space becomes limited, and toxic waste products accumulate.

[Part a] 1 mark: Total cell count measures both living and dead cells. 1 mark: Viable cell count measures only living/colony-forming cells. 1 mark: Identifies an appropriate method for each (e.g., total count uses a haemocytometer/microscope, viable count uses dilution plating). [Part b] 1 mark: Correctly accounts for the plated volume (multiplies by 10 to get 540 per cm^3). 1 mark: Correctly applies the dilution factor (multiplies by 10^5). 1 mark: Expresses the correct final answer in standard form (5.4 x 10^7 CFU cm^-3 or 5.4 x 10^7). [Part c] 1 mark: Lag phase: cells do not divide but synthesize enzymes / replicate DNA / grow. 1 mark: Lag phase: cells adjust to the new environmental conditions. 1 mark: Stationary phase: birth rate/division rate equals death rate. 1 mark: Stationary phase: caused by depletion of nutrients / accumulation of toxic wastes.

(a) Clonal selection occurs when a specific B cell with a complementary membrane-bound antibody/receptor binds to a specific foreign antigen on the virus. The B cell presents this antigen on MHC Class II molecules to activated T helper cells, which release cytokines. Clonal expansion is the subsequent rapid mitotic division of this specific activated B cell, producing a large clone of genetically identical plasma cells and memory B cells. (b) T helper cells release cytokines (e.g., interleukins) that coordinate the overall immune response by stimulating B cells to divide and differentiate into plasma cells, and activating T killer cells. T killer (cytotoxic) cells directly destroy infected host cells by binding to viral antigens presented on MHC Class I molecules and releasing perforins/granzymes to induce lysis or apoptosis. (c) Monoclonal antibodies are produced from a single clone of identical hybridoma cells, meaning they all have the exact same primary structure (amino acid sequence). This primary structure determines a unique tertiary structure with a specific shape of the variable region (antigen-binding site), which is complementary to only one specific epitope on the target viral antigen.

[Part a] 1 mark: Clonal selection involves binding of specific viral antigen to a complementary B cell receptor. 1 mark: B cell presents antigen (on MHC Class II) to T helper cells. 1 mark: Cytokines from T helper cells stimulate B cells. 1 mark: Clonal expansion involves rapid mitosis of the selected B cell to produce plasma and memory cells. [Part b] 1 mark: T helper cells secrete cytokines/interleukins to coordinate/activate other immune cells (B cells and T killer cells). 1 mark: T killer cells target and destroy infected host cells. 1 mark: T killer cells release perforins/granzymes to induce lysis/apoptosis of infected cells (or recognize antigen on MHC I). [Part c] 1 mark: Monoclonal antibodies are derived from a single clone of hybridoma cells (so are identical). 1 mark: They have a specific primary structure/amino acid sequence that determines a unique tertiary structure. 1 mark: This creates a uniquely shaped variable region/antigen-binding site complementary to only one specific antigen/epitope.

(a) The Michaelis constant (Km) is defined as the substrate concentration at which the reaction rate is half of the Vmax. Since 12.5 is exactly half of Vmax (25.0), the Km is 2.0 x 10^-3 mol dm^-3. A lower Km indicates high affinity of the enzyme for its substrate, as less substrate is needed to half-saturate the active sites. (b) Competitive inhibitors have a structure similar to the substrate and bind to the active site via temporary intermolecular forces, blocking the substrate from entering. Non-competitive inhibitors bind to an allosteric site (a separate site), forming intermolecular bonds that alter the tertiary structure of the enzyme. This changes the precise conformation of the active site so that the substrate can no longer bind or the enzyme cannot catalyze the reaction. (c) A non-competitive inhibitor decreases the Vmax because it effectively reduces the concentration of active enzyme molecules, and this cannot be overcome by adding more substrate. The Km remains unchanged because the remaining uninhibited enzyme molecules have the same affinity for the substrate as before.

[Part a] 1 mark: States Km = 2.0 x 10^-3 mol dm^-3 (with units). 1 mark: Defines Km as the substrate concentration at half Vmax. 1 mark: Explains that lower Km means higher substrate affinity / higher Km means lower substrate affinity. [Part b] 1 mark: Competitive inhibitors are similar in shape to the substrate and bind to the active site. 1 mark: Competitive inhibitors block the active site, preventing substrate binding. 1 mark: Non-competitive inhibitors bind to an allosteric site. 1 mark: Non-competitive binding changes the tertiary structure/conformation of the active site, preventing binding/catalysis. [Part c] 1 mark: Vmax decreases. 1 mark: Km remains unchanged. 1 mark: Explains that increasing substrate concentration cannot overcome non-competitive inhibition (lowering Vmax), but does not alter the affinity of unaffected enzymes (Km unchanged).

(a) Companion cells use ATP to actively pump hydrogen ions (H+) out of their cytoplasm into the apoplast (cell wall) via proton pumps. This generates a high concentration gradient of H+ outside the cell. H+ ions then diffuse back down their electrochemical gradient into the companion cell through co-transporter proteins, which simultaneously transport sucrose molecules against their concentration gradient. Once inside, sucrose diffuses into the sieve tube elements via plasmodesmata. (b) Distance = 18.0 cm. Time = 45 minutes = 45 / 60 = 0.75 hours. Velocity = Distance / Time = 18.0 cm / 0.75 hours = 24.0 cm hour^-1. (c) At the source, the active loading of sucrose into sieve tube elements lowers their water potential. Water then moves into the sieve tubes from the adjacent xylem by osmosis, creating a high hydrostatic pressure. At the sink, sucrose is unloaded, which increases the water potential inside the sieve tube. Water leaves the sieve tube by osmosis into surrounding tissues, creating a low hydrostatic pressure. This difference in pressure generates a hydrostatic gradient that drives mass flow from source to sink.

[Part a] 1 mark: H+ ions actively pumped out of companion cells into apoplast/cell wall using ATP. 1 mark: Creates a high H+ concentration gradient. 1 mark: H+ diffuses back in through co-transporter proteins carrying sucrose against its gradient. 1 mark: Sucrose diffuses from companion cell into sieve tube element via plasmodesmata. [Part b] 1 mark: Converts 45 minutes to 0.75 hours or shows correct division (18.0 / 45 x 60). 1 mark: Correct velocity (24.0) with correct units (cm hour^-1). [Part c] 1 mark: Loading of sucrose lowers water potential in sieve tubes at source. 1 mark: Water enters by osmosis from xylem, generating high hydrostatic pressure. 1 mark: Unloading of sucrose at sink raises water potential, so water leaves by osmosis. 1 mark: Creates a hydrostatic pressure gradient driving mass flow from high to low pressure.

(a) Penicillin is bactericidal because it actively kills bacterial cells. It inhibits transpeptidase enzymes, preventing peptidoglycan cross-linking in the cell wall, which weakens the wall and causes osmotic lysis. Tetracycline is bacteriostatic because it only inhibits bacterial growth and reproduction rather than killing them directly. It binds to the 30S subunit of bacterial ribosomes, blocking translation and protein synthesis. (b) Total zone radius (R) = 24.0 / 2 = 12.0 mm. Total area = pi x R^2 = 3.142 x 12.0^2 = 3.142 x 144 = 452.448 mm^2. Disc radius (r) = 6.0 / 2 = 3.0 mm. Disc area = pi x r^2 = 3.142 x 3.0^2 = 3.142 x 9 = 28.278 mm^2. Net zone of inhibition area = Total area - Disc area = 452.448 - 28.278 = 424.17 mm^2, which rounds to 424 mm^2 to 3 significant figures. (c) Penicillin specifically targets peptidoglycan cell walls, which are present in bacterial cells but entirely absent in viruses. Furthermore, viruses do not have metabolic processes, ribosomes, or their own protein-synthetic machinery, meaning they do not possess any of the biochemical targets that antibiotics act upon.

[Part a] 1 mark: Bactericidal means kills bacteria; bacteriostatic means inhibits growth/reproduction. 1 mark: Penicillin inhibits transpeptidase / peptidoglycan cross-linking. 1 mark: Weakens the cell wall, leading to cell lysis. 1 mark: Tetracycline binds to ribosomal 30S subunit / inhibits translation / protein synthesis. [Part b] 1 mark: Correctly calculates radii (12.0 mm and 3.0 mm) or shows correct formula application. 1 mark: Calculates total area (452.448 mm^2) and disc area (28.278 mm^2). 1 mark: Correct net area of 424 mm^2 (to 3 significant figures). [Part c] 1 mark: Penicillin targets the bacterial cell wall (peptidoglycan), which viruses do not have. 1 mark: Viruses lack cell structure / metabolism / organelles / ribosomes of their own. 1 mark: Therefore, viruses do not possess the biological targets of antibiotics.

(a) Step 1: Convert capillary measurements to cm to match the required units. Diameter = \(0.8\text{ mm} = 0.08\text{ cm}\), so radius \(r = 0.04\text{ cm}\). Distance moved \(h = 45\text{ mm} = 4.5\text{ cm}\). Step 2: Calculate the volume of water absorbed using the cylinder volume formula: \(V = \pi r^2 h = \pi \times 0.04^2 \times 4.5 = 3.14159 \times 0.0016 \times 4.5 \approx 0.02262\text{ cm}^3\). Step 3: Convert the time into hours. \(15\text{ minutes} = 0.25\text{ hours}\). Step 4: Calculate the volume uptake rate per hour: \(0.02262\text{ cm}^3 \div 0.25\text{ hours} = 0.09048\text{ cm}^3\text{ hour}^{-1}\). Step 5: Divide this hourly rate by the total leaf surface area: \(0.09048\text{ cm}^3\text{ hour}^{-1} \div 32.0\text{ cm}^2 = 0.002827\text{ cm}^3\text{ cm}^{-2}\text{ hour}^{-1}\). Step 6: Convert to standard form to 3 significant figures: \(2.83 \times 10^{-3}\text{ cm}^3\text{ cm}^{-2}\text{ hour}^{-1}\). (b) Xylem vessels are long, continuous hollow tubes with no living contents (no cytoplasm or organelles), which minimizes resistance to the upward flow of water. Their end walls are completely broken down, allowing water to flow in an uninterrupted column. The cell walls are heavily thickened with spiral or annular bands of lignin, which prevents the vessels from collapsing inward under the tension (negative pressure) generated by transpiration pulling force. Furthermore, pits (unlignified regions in the walls) allow lateral movement of water between adjacent vessels to bypass any blockages. (c) First, some of the water taken up is used directly in photosynthesis rather than being transpired. Second, some water remains in the plant cells to maintain cell turgor and volume, or is produced/consumed in metabolic respiration, so not all water uptake translates directly into water loss via stomata.

(a) MP1: Correctly converts capillary dimensions to cm: radius \(r = 0.04\text{ cm}\) AND distance \(h = 4.5\text{ cm}\) (or calculates volume in \(mm^3\) as \(22.62\text{ mm}^3\) and correctly converts to \(0.02262\text{ cm}^3\)). MP2: Correctly calculates volume of water uptake as \(0.0226\text{ cm}^3\) (accept \(0.0226\) to \(0.0230\)). MP3: Correctly adjusts the rate for time (multiplies by 4 to get rate per hour) and divides by leaf area (\(32.0\text{ cm}^2\)). MP4: Gives correct final answer in standard form and to 3 s.f.: \(2.83 \times 10^{-3}\) (accept \(2.82 \times 10^{-3}\) to \(2.83 \times 10^{-3}\)). (b) Any four from: MP1: Dead cells/hollow tubes with no cytoplasm/organelles to allow unobstructed/low resistance water flow. MP2: Lack of end walls/perforated end plates to allow continuous columns of water. MP3: Lignification of cell walls provides strength to withstand high tension/negative pressure (preventing vessel collapse). MP4: Hydrophilic nature of lignin/cellulose walls facilitates adhesion of water molecules. MP5: Pits in walls allow lateral transport of water (to bypass air bubbles/blockages). (c) Any two from: MP1: Potometer measures water uptake rather than water loss/transpiration directly. MP2: Some water is used in metabolic processes/photosynthesis. MP3: Some water is used to maintain cell turgidity.

(a) Tube C is a control to show that light alone does not reduce or decolourise DCPIP without chloroplasts. Tube D is a control to show that the reduction of DCPIP is an enzyme-controlled process (or depends on active, un-denatured thylakoid membrane proteins) rather than a non-biological reaction.
(b) The buffer must be ice-cold to slow down enzyme activity and prevent autolysis/degradation of chloroplasts during isolation. It must be isotonic to maintain the same water potential as the chloroplast stroma, preventing the net movement of water by osmosis so the chloroplasts do not swell and burst (lyse) or shrink.
(c) Change in absorbance = 0.82 - 0.16 = 0.66 over 10 minutes. Rate = 0.66 / 10 = 0.066 absorbance units per minute.
(d) DCPIP acts as an artificial electron acceptor. During the light-dependent stage, water is photolysed, releasing electrons. These electrons are accepted by DCPIP, reducing the blue oxidized dye to colourless reduced DCPIP, resulting in a decrease in absorbance at 600 nm.

(a) Tube C: Control to prove light alone does not reduce DCPIP (1 mark); Tube D: Control to prove the reaction requires active/intact/undenatured enzymes or proteins in chloroplasts (1 mark); by demonstrating boiling denatures these components preventing reduction (1 mark).
(b) Ice-cold: reduces enzyme activity / autolysis / prevents degradation of thylakoid membranes (1 mark); Isotonic: maintains same water potential / no net water movement by osmosis (1 mark); preventing chloroplasts from swelling/bursting or shrinking (1 mark).
(c) Correct calculation of difference: 0.82 - 0.16 = 0.66 (1 mark); Correct rate: 0.066 absorbance units per min (accept min^-1) (1 mark).
(d) DCPIP is a blue dye that acts as an artificial electron acceptor (1 mark); it is reduced by electrons released during the light-dependent stage / photolysis of water (1 mark); reduced DCPIP is colourless, so absorbance at 600 nm decreases (0.9 marks).

(a) Grind leaves with a mortar and pestle using an organic solvent like propanone/acetone. Filter the extract to obtain a concentrated pigment solution. Draw a pencil line (the origin) about 1-2 cm from the bottom of the TLC plate. Use a capillary tube to apply a small, concentrated spot of pigment on the pencil line, letting it dry between applications to keep the spot small and focused.
(b) Rf = distance moved by pigment / distance moved by solvent front = 2.1 / 8.2 = 0.256... which rounds to 0.26.
(c) To compare quantitatively, the student could measure the surface area of the resulting pigment spots or use a densitometer to measure the intensity/optical density of the spots. Alternatively, scrape the spots off the plate, dissolve the pigments back into a solvent, and measure their absorbance using a spectrophotometer. Controlled variables include using the same mass of leaf tissue for extraction, the same volume of extraction solvent, and loading the same volume of pigment extract onto the origin line of each plate.

(a) Grind leaves with propanone/acetone (1 mark); draw origin line in pencil (not ink) above solvent level (1 mark); use capillary tube to apply spot of pigment extract on origin line (1 mark); allow to dry and repeat to build up a small, concentrated spot (1 mark).
(b) Rf = 2.1 / 8.2 (1 mark); = 0.26 (must be 2 d.p.) (1 mark).
(c) Methods of quantitative comparison: measure area of the pigment spots (1 mark) OR scrape spot off plate, dissolve in solvent and measure absorbance with a colorimeter (1 mark). Control variables (any two for 1.9 marks): same mass of leaf tissue extracted (0.95 marks), same volume of extraction solvent (0.95 marks), same volume of pigment spot loaded onto chromatogram (0.95 marks).

(a) Set up a coordinate grid over the sampling area using two tape measures at right angles. Use a random number generator to obtain coordinates for quadrat placement to avoid bias. Place a quadrat (e.g. 0.5m x 0.5m) at each coordinate. Count the number of individuals of each species or estimate percentage cover. Repeat this process at least 10 times per meadow to obtain a representative mean.
(b) N = 240, so N(N-1) = 240 x 239 = 57,360. Sum of n(n-1) = 14,200. Simpson's Index (D) = 1 - (14,200 / 57,360) = 1 - 0.24756 = 0.75244... D = 0.752.
(c) The ungrazed meadow has a higher Simpson's Index of Diversity (0.812) than the grazed meadow (0.752). This indicates higher species richness and species evenness. A higher index of diversity suggests a more stable, complex ecosystem with more ecological niches and food webs, making it less vulnerable to environmental changes. In the grazed meadow, grazing pressure likely selectively removes palatable plant species, reducing overall diversity.

(a) Use grid/tape measures at right angles to define the area (1 mark); use random number generator to select coordinates to avoid bias (1 mark); place quadrats at coordinates and count individuals of each species / estimate percentage cover (1 mark); repeat at least 10 times to obtain a representative mean (1 mark).
(b) Correct calculation of N(N-1) = 240 x 239 = 57360 (1 mark); correct division: 14200 / 57360 = 0.2476 (1 mark); correct final subtraction and rounding to 3 d.p.: 0.752 (1 mark).
(c) Ungrazed meadow has higher biodiversity/index of diversity than grazed meadow (1 mark); higher index of diversity means a more complex food web / more ecological niches (1 mark); ungrazed meadow is more stable / less susceptible to environmental changes (0.95 marks); grazing reduces diversity by selectively removing palatable species OR grazing maintains diversity of some species but overall reduces it here (0.95 marks).

(a) Potassium hydroxide (KOH) absorbs carbon dioxide (\(CO_2\)) produced by the woodlice during respiration. This ensures that any change in gas volume/pressure in the tube is due solely to the consumption of oxygen (\(O_2\)).
(b) Radius of capillary tube (r) = 1.2 mm / 2 = 0.6 mm = 0.06 cm. Distance moved (h) = 45 mm = 4.5 cm. Volume of oxygen consumed (V) = \(\pi \times 0.06^2 \times 4.5 = 3.14159 \times 0.0036 \times 4.5 = 0.05089\ cm^3\). Mass of woodlice = 4.2 g. Time = 15 minutes = 0.25 hours. Rate of oxygen consumption = 0.05089 / (4.2 * 0.25) = 0.05089 / 1.05 = 0.04847\ cm^3\ g^{-1}\ h^{-1}\. This rounds to 0.048\ cm^3\ g^{-1}\ h^{-1}\.
(c) Controlled variables: temperature (using a water bath) and equilibration time. A suitable control experiment would involve setting up an identical respirometer tube but replacing the living woodlice with an equal mass/volume of non-living, inert material (such as glass beads or boiled/dead woodlice). This accounts for any movement of the colored liquid due to expansion/contraction of gas caused by ambient temperature or pressure fluctuations.

(a) Absorbs carbon dioxide (\(CO_2\)) produced by respiration (1 mark); ensures change in volume/pressure inside the tube is solely due to oxygen uptake (1 mark).
(b) Correct conversion of units (radius = 0.06 cm, length = 4.5 cm) (1 mark); correct calculation of gas volume: \(\pi \times 0.06^2 \times 4.5 = 0.0509\ cm^3\) (accept 0.051 or \(50.9\ mm^3\)) (1 mark); correct adjustment for time (multiplied by 4 to get per hour) (1 mark); final correct rate: \(0.048\ cm^3\ g^{-1}\ h^{-1}\) (accept \(0.048 - 0.049\)) (1 mark).
(c) Control variables (any two): Temperature (using a water bath) (0.95 marks); equilibration time before closing tap (0.95 marks). Suitable control experiment: Replace woodlice with glass beads / inert material of equal volume/mass (1 mark); to show movement of liquid is due to respiration and not fluctuations in temperature/pressure (1 mark).

(a) Wipe down the workspace with disinfectant before and after. Work near a lit Bunsen burner to create an upward convection current of warm air that carries away airborne contaminants. Flame the neck of the bottle containing the *E. coli* culture before and after opening. Only lift the lid of the Petri dish slightly (at a 45-degree angle) when inoculating to prevent dust/spores from falling in. Use sterile equipment (e.g. sterile spreader or sterile pipette).
(b) Total diameter = 18 mm, so total radius R = 9 mm. Disc diameter = 6 mm, so disc radius r = 3 mm. Area of zone + disc = \(\pi R^2 = 3.142 \times 9^2 = 3.142 \times 81 = 254.502\ mm^2\). Area of disc = \(\pi r^2 = 3.142 \times 3^2 = 3.142 \times 9 = 28.278\ mm^2\). Area of zone of inhibition = 254.502 - 28.278 = 226.224\ mm^2\, which rounds to 226.2\ mm^2\.
(c) Incubating at 37 °C (human body temperature) would encourage the growth of human pathogens (microorganisms that are harmful to humans). Incubating at a lower temperature (25 °C) reduces this risk while still allowing the non-pathogenic or environmental bacteria to grow sufficiently for measurement.

(a) Clean work surfaces with disinfectant and sterilize loops/pipettes or use sterile single-use equipment (1 mark); work near a Bunsen burner to create an upward convection current of air (1 mark); flame the neck of the bacterial culture bottle to prevent contamination (1 mark); open Petri dish lid at an angle / for minimum time possible (1 mark).
(b) Calculation of radii: total radius = 9 mm, disc radius = 3 mm (1 mark); calculation of total area (\(254.5\ mm^2\)) and disc area (\(28.3\ mm^2\)) (1 mark); subtraction to give correct final area: \(226.2\ mm^2\) (accept \(226\ mm^2\)) (1 mark).
(c) 37 °C is human body temperature (1 mark); incubating at this temperature could promote the growth of human pathogens / harmful bacteria (1 mark); incubating at 25 °C allows *E. coli* growth while minimizing the safety hazard of culturing pathogens (1.9 marks).

(a) The macrophage recognizes foreign antigens on the yeast cell. The macrophage cell membrane folds around the yeast cell, engulfing it to form a phagocytic vacuole (phagosome). Lysosomes fuse with the phagosome, releasing hydrolytic/digestive enzymes (such as lysozymes) which digest and destroy the yeast cell.
(b) Prepare slides of the macrophage and stained yeast mixture incubated at each temperature. View the slides under high power using a light microscope. Count a fixed total number of macrophages (e.g., 100 macrophages) and count how many stained yeast cells have been engulfed inside them. Record the start and end time of incubation (30 minutes) and calculate the rate as the mean number of yeast cells engulfed per macrophage per hour.
(c) At 10 °C, value = 1.2. At 37 °C, value = 8.1. Percentage increase = ((8.1 - 1.2) / 1.2) * 100 = (6.9 / 1.2) * 100 = 575%.
(d) At 60 °C, high thermal energy causes hydrogen and ionic bonds within the proteins of the cell membrane and enzymes of the macrophage to break. This denatures the membrane receptor proteins (preventing binding to yeast antigens) and denatures lysosomal enzymes, halting the phagocytic process.

(a) Phagocyte/macrophage membrane engulfs the yeast cell to form a phagosome / phagocytic vacuole (1 mark); lysosomes fuse with the phagosome (1 mark); releasing hydrolytic/digestive enzymes (lysozymes) that hydrolyse/destroy the yeast cell (1 mark).
(b) Take samples from each temperature incubation and prepare microscope slides (1 mark); observe under high power magnification (1 mark); count a set number of macrophages and count the number of engulfed red-stained yeast cells inside them (1 mark); divide the average number of engulfed yeast cells by the time of incubation to get the rate (1 mark).
(c) Correct calculation of difference: 8.1 - 1.2 = 6.9 (1 mark); correct percentage increase calculation: (6.9 / 1.2) * 100 = 575% (1 mark).
(d) Proteins (receptors / lysosomal enzymes / cytoskeletal proteins) are denatured due to loss of tertiary structure/hydrogen bonding at 60 °C (1 mark); preventing receptor binding or preventing the mechanical cell engulfment/fusion of lysosomes (0.9 marks).

(a) The iodine solution turns from blue-black to orange-brown (the color of iodine itself) when starch is fully broken down. The time taken to reach this endpoint is measured. The rate can be calculated as the reciprocal of time taken (1 / time taken).
(b) Use a colorimeter to measure the absorbance or transmission of light. Prepare a calibration curve using known concentrations of starch mixed with iodine. Take samples from the reaction mixture at regular intervals, add to iodine, and read the absorbance at a fixed wavelength (e.g., 620 nm) to determine the exact concentration of starch.
(c) The value -0.045 g dm^-3 s^-1 represents the initial rate of reaction. The rate is highest at the start because there is a high concentration of substrate (starch) molecules, making collisions between substrate and enzyme active sites highly frequent, forming maximum enzyme-substrate complexes per unit time.
(d) The color change is subjective, making it difficult to judge the exact moment the blue-black color completely disappears. Additionally, the sampling interval (e.g., 30 seconds) limits the resolution of the time measurement.

(a) Iodine solution remains yellow/brown/orange / fails to turn blue-black (1 mark); record the time taken to reach this point (1 mark); rate is calculated as reciprocal of time (1/t) (1 mark).
(b) Use a colorimeter (1 mark); calibrate using known starch concentrations with iodine to make a calibration curve (1 mark); measure absorbance/transmission of reaction samples over time to find exact starch concentrations (1 mark).
(c) Represents the initial rate of reaction (1 mark); at the start, substrate concentration is at its maximum/is not limiting (1 mark); maximum frequency of successful collisions between starch and amylase active sites / highest rate of enzyme-substrate complex formation (1 mark).
(d) Subjective determination of color change endpoint (1 mark); large time intervals between sampling (e.g., 30s) reduces accuracy of the exact endpoint time (0.9 marks).

(a) Three precautions include: 1. Cut the shoot underwater to prevent air from entering and locking the xylem. 2. Assemble the entire potometer apparatus underwater to prevent air bubbles from forming in the capillary tube. 3. Apply petroleum jelly (Vaseline) to all joints to ensure the system is completely airtight.
(b) Volume of water uptake = cross-sectional area * distance = 0.8 mm^2 * 60 mm = 48 mm^3. Time = 10 minutes. Rate of water uptake = 48 mm^3 / 10 minutes = 4.8 mm^3 min^-1.
(c) Increased wind speed blows away the water vapor that accumulates around the stomata on the leaf surface. This reduces the humidity of the boundary layer, maintaining a steep water vapor potential gradient between the air spaces inside the leaf and the external air, which increases the rate of diffusion of water vapor out of the stomata.
(d) Some of the water taken up by the shoot is used as a reactant in photosynthesis. Additionally, some water is used to maintain cell turgor pressure (turgidity) within the plant cells rather than being transpired into the atmosphere.

(a) Cut shoot underwater to prevent air locks in the xylem (1 mark); assemble apparatus underwater to ensure no air bubbles are trapped (1 mark); seal joints with waterproof jelly/Vaseline to make it airtight (1 mark).
(b) Calculate total volume of water: 0.8 mm^2 x 60 mm = 48 mm^3 (1 mark); divide by time: 48 / 10 (1 mark); correct final answer: 4.8 mm^3 min^-1 (1 mark).
(c) Wind removes the boundary layer of saturated air/water vapor from the leaf surface (1 mark); maintaining a steeper water vapor potential gradient between inside the leaf and the atmosphere (1 mark); increasing rate of diffusion of water vapor out through stomata (0.9 marks).
(d) Not all water taken up is transpired; some is used in photosynthesis (1 mark); some water is used to maintain turgidity of plant cells (1 mark).

(a) Blend spinach leaves in cold, isotonic, pH-buffered isolation medium. Filter the homogenate through muslin to remove large cell debris. Centrifuge the filtrate at a low speed to sediment nuclei and cell walls, then pour the supernatant into a new tube and centrifuge at a high speed to obtain a pellet of intact chloroplasts. Resuspend the pellet in a cold, pH-buffered buffer. (b)(i) Relative light intensity = \(1 / 20^2 = 0.0025\) \(\text{cm}^{-2}\). Rate of reaction = \(1000 / 92 = 10.87\) \(\text{s}^{-1}\). (ii) As light intensity increases, the rate of the Hill reaction increases because more light energy is absorbed by chlorophyll in photosystem II. This increases the rate of photolysis of water and excitation of electrons, which are then accepted by DCPIP at a faster rate, causing quicker decolourisation. (c) Variables to control: Temperature (controlled by placing a clear water shield between the lamp and the test tube to absorb heat, or using a thermostatically controlled water bath) and the concentration/volume of DCPIP added (controlled by using a graduated pipette to add the exact same volume of a stock DCPIP solution to each reaction tube).

Part (a): 4 marks total. 1 mark: Blend/grind leaves in cold, isotonic, and pH-buffered medium (accept reasons: cold to slow enzyme activity, isotonic to prevent osmotic lysis, buffer to maintain enzyme structure). 1 mark: Filter through muslin/cheesecloth. 1 mark: Centrifuge filtrate to obtain supernatant and centrifuge again at higher speed to pellet chloroplasts. 1 mark: Resuspend chloroplast pellet in ice-cold buffer. Part (b)(i): 2 marks total. 1 mark for correct relative light intensity: 0.0025 (or \(2.5 \times 10^{-3}\)) \(\text{cm}^{-2}\). 1 mark for correct rate: 10.87 (or 10.9) \(\text{s}^{-1}\). Part (b)(ii): 3 marks total. 1 mark: Higher light intensity increases light energy absorption by photosynthetic pigments. 1 mark: Increases rate of photolysis / excitation of electrons in the light-dependent stage. 1 mark: More electrons are transferred to DCPIP per unit time, resulting in faster decolourisation. Part (c): 2 marks total. 1 mark: Identify two valid control variables (e.g., temperature, DCPIP volume/concentration, chloroplast concentration). 1 mark: Explain a practical control method (e.g., use a water bath / glass heat shield to absorb heat from the light source).

(a) Disinfect all work surfaces before and after the practical. Work near a lit Bunsen burner to create an upward convective sterile air current that prevents airborne contaminants from settling. Flame the necks of culture bottles before and after transferring bacteria. Keep petri dish lids slightly open at an angle and for the minimum time possible when seeding agar. Autoclave all equipment and media before use. (b) Points in agreement: Extract C produces a significantly larger zone of inhibition than A and B, indicating higher antimicrobial efficacy. There is no overlapping of standard deviations between C (18.2 +/- 0.5 mm) and B (15.8 +/- 1.2 mm), indicating the difference is likely significant. Point C uses less extract than D (75% vs 100%) but has almost identical efficacy (18.2 vs 18.5 mm) with overlapping standard deviations, so C is cost-effective. Points in disagreement/limitation: Penicillin is still significantly more effective (22.1 mm) than Extract C (18.2 mm) with no overlap in SD. Standard deviation only indicates spread around the mean; a statistical test (e.g., t-test) is needed to confirm significance. In vitro agar plate results do not necessarily reflect efficacy in vivo (human body). Extract might be toxic to human cells. (c) Pipette 5.0 \(\text{cm}^3\) of the 100% stock solution into a test tube containing 5.0 \(\text{cm}^3\) of sterile distilled water to produce 10.0 \(\text{cm}^3\) of 50% solution. Mix thoroughly. Take 5.0 \(\text{cm}^3\) of this 50% solution and transfer to another tube containing 5.0 \(\text{cm}^3\) of water to produce 25% solution. Repeat the process taking 5.0 \(\text{cm}^3\) of the 25% solution and transferring it to 5.0 \(\text{cm}^3\) of water to produce 12.5% solution.

Part (a): 4 marks total. 1 mark: Disinfect bench or wash hands before and after work. 1 mark: Work near Bunsen burner to create sterile updraft. 1 mark: Flame neck of culture bottle. 1 mark: Open petri dish lid minimally / at an angle. Part (b): 4 marks total. 1 mark: Extract C is significantly more effective than A and B as there is no overlap in standard deviations. 1 mark: Extract C has almost the same efficacy as D (overlapping SDs) so is more cost-effective. 1 mark: Penicillin is still significantly more effective than Extract C (22.1 vs 18.2 mm, no SD overlap). 1 mark: Cannot conclude absolute efficacy without statistical tests (e.g., t-test) or in vivo trials / test for toxicity. Part (c): 3 marks total. 1 mark: Use equal volumes of stock and sterile water to halve the concentration (e.g. 5 cm3 stock + 5 cm3 water for 50%). 1 mark: Describe transferring the same volume from the previous tube to the next tube of water (serial dilution). 1 mark: Ensure thorough mixing at each stage.

(a) Woodlice are placed in the experimental tube, and a carbon dioxide absorbent (such as potassium hydroxide or soda lime) is placed at the bottom, isolated from the organisms. As woodlice respire, they consume oxygen and release carbon dioxide. The carbon dioxide is absorbed, creating a drop in pressure inside the tube, which causes the colored liquid in the capillary tube to move towards the chamber. The distance moved by the liquid is measured over a set period of time using a ruler and stopwatch. (b)(i) Rate of \(O_2\) consumption = (volume of capillary tube moved) / time. Volume = cross-sectional area * distance. At 15 degrees C: Volume = \(0.80\ \text{mm}^2 \times 12\ \text{mm} = 9.6\ \text{mm}^3\). Rate = \(9.6\ \text{mm}^3 / 5\ \text{min} = 1.92\ \text{mm}^3\ \text{min}^{-1}\). At 25 degrees C: Volume = \(0.80\ \text{mm}^2 \times 28\ \text{mm} = 22.4\ \text{mm}^3\). Rate = \(22.4\ \text{mm}^3 / 5\ \text{min} = 4.48\ \text{mm}^3\ \text{min}^{-1}\). (ii) \(Q_{10} = \text{Rate at } 25^\circ\text{C} / \text{Rate at } 15^\circ\text{C} = 4.48 / 1.92 = 2.33\). (c) Fluctuations in temperature affect the kinetic energy of gas molecules inside the closed tube. An increase in temperature causes the gas inside the tube to expand, which would push the colored liquid away from the chamber, opposing the effect of oxygen consumption and leading to an underestimation of the respiration rate.

Part (a): 4 marks total. 1 mark: Use of potassium hydroxide/soda lime to absorb carbon dioxide. 1 mark: Oxygen consumption by woodlice decreases volume/pressure of gas in chamber. 1 mark: Measure distance moved by colored liquid in a set time. 1 mark: Keep mass of organisms constant / use a syringe to reset liquid between runs. Part (b)(i): 3 marks total. 1 mark: Calculation of volume at 15 C (9.6 mm3) and 25 C (22.4 mm3). 1 mark: Correct rate at 15 C (1.92 mm3 min-1). 1 mark: Correct rate at 25 C (4.48 mm3 min-1). Part (b)(ii): 2 marks total. 1 mark: Correct formula or substitution (Rate at 25 C / Rate at 15 C). 1 mark: Correct Q10 calculation (2.33) (accept 2.3). Part (c): 2 marks total. 1 mark: Temperature changes gas pressure/volume (Charles's law / expansion of air). 1 mark: Increase in temperature causes gas expansion which pushes liquid back, resulting in inaccurate/underestimated rates.