2022 OCR AS Level Biology A - H020 模擬試題連答案詳解

Starch reacts with iodine in potassium iodide solution to give a blue-black colour, so the iodine test is positive. Sucrose is a non-reducing sugar, so performing the Benedict's test directly on the sample yields a negative (blue) result. Hydrolysis with hydrochloric acid breaks the glycosidic bonds in sucrose to produce glucose and fructose (both reducing sugars). After neutralisation with sodium hydrogencarbonate, these reducing sugars will react with Benedict's reagent to produce a brick-red precipitate.

1 mark for selecting the correct row (A) showing a positive iodine test, a negative raw Benedict's test, and a positive Benedict's test after acid hydrolysis.

First, calculate the proportion of cells in anaphase: \(12 / 400 = 0.03\). To find the duration, multiply this proportion by the total length of the cell cycle: \(0.03 \times 20\) hours = 0.6 hours. Convert hours to minutes: \(0.6 \times 60 = 36\) minutes.

1 mark for selecting C, showing correct calculation of 36 minutes based on the proportion of cells in anaphase.

As temperature increases, phospholipids gain kinetic energy and move faster, which increases the fluidity and permeability of the lipid bilayer. Additionally, at high temperatures like 70 °C, membrane proteins (such as transport proteins) denature, creating large gaps in the membrane and allowing pigment molecules to leak out more easily.

1 mark for selecting C, which correctly identifies both the increased kinetic energy/fluidity of phospholipids and the denaturation of membrane proteins.

In actively respiring tissues, carbon dioxide is produced and diffuses into red blood cells. Carbonic anhydrase catalyses its reaction with water to form carbonic acid (\(H_2CO_3\)), which dissociates into hydrogen ions (\(H^+\)) and hydrogencarbonate ions (\(HCO_3^-\)). The excess \(H^+\) ions bind to oxyhaemoglobin to form haemoglobinic acid, which forces haemoglobin to release its bound oxygen. This is the Bohr effect, shifting the curve to the right.

1 mark for selecting B, which correctly outlines the role of carbonic anhydrase and hydrogen ions in reducing haemoglobin's oxygen affinity.

Protons (hydrogen ions) are actively pumped out of the companion cell into the apoplast (cell wall space) using ATP. This establishes a proton concentration gradient. The protons then diffuse back down their concentration gradient into the companion cell via a co-transporter protein, which simultaneously transports sucrose against its concentration gradient into the cell.

1 mark for selecting B, which accurately describes the active pumping of protons into the cell wall and their facilitated diffusion back via a co-transporter carrying sucrose.

The constant region is identical in all antibodies of a given class and contains the binding site for phagocytes, facilitating opsonisation. The variable region has a unique, complementary shape that binds to a specific antigen. The hinge region provides flexibility, enabling the antibody to bind to multiple antigen molecules on the same or different pathogens.

1 mark for selecting A, showing correct mapping of constant, variable, and hinge region functions.

In double-stranded DNA, the percentage of thymine (T) equals the percentage of adenine (A). Therefore, T = 24% and A = 24%. Together, A + T = 48%. This leaves 100% - 48% = 52% for guanine (G) and cytosine (C). Since G and C pair together in equal amounts, the percentage of guanine is 52% / 2 = 26%.

1 mark for selecting B, demonstrating correct application of Chargaff's rules to calculate 26% guanine.

If the reaction rate can reach the same maximum velocity (\(V_{max}\)) as the uninhibited reaction when substrate concentration is high, the inhibitor is competitive. Competitive inhibitors compete with substrate molecules for the active site because they have a complementary shape to it. At high substrate concentrations, the substrate molecules outcompete the inhibitor molecules for the active sites, overcoming the inhibition.

1 mark for selecting B, correctly identifying that competitive inhibitors bind to the active site and can be outcompeted by high substrate concentrations.

Both glycogen and amylopectin are branched polymers composed of \(\alpha\)-glucose sub-units, which means they both contain both \(\alpha\)-1,4 and \(\alpha\)-1,6-glycosidic bonds. However, glycogen is much more highly branched than amylopectin because it contains a higher frequency of \(\alpha\)-1,6-glycosidic bonds (at branch points). Both molecules are completely insoluble in water, which prevents them from affecting the osmotic potential of cells.

1 mark for the correct answer B. 0 marks for any other response.

During Prophase I of meiosis, homologous chromosomes undergo synapsis to pair up and form bivalents. In Anaphase II, the centromeres divide, which allows sister chromatids to separate and be pulled to opposite poles. Homologous chromosomes separate during Anaphase I, not Anaphase II.

1 mark for the correct answer A. 0 marks for any other response.

Ethanol is an organic solvent that dissolves the non-polar tails of phospholipids, disrupting the highly ordered structure of the phospholipid bilayer. It also denatures membrane proteins by disrupting their tertiary structure. This causes the membrane to lose its integrity and become highly permeable, allowing betalain pigment molecules to leak out of the vacuole into the external solution.

1 mark for the correct answer C. 0 marks for any other response.

In actively respiring tissues, high levels of \(\text{CO}_2\) dissolve to form carbonic acid, which dissociates into \(\text{H}^+\) and \(\text{HCO}_3^-\). The accumulation of \(\text{H}^+\) ions lowers the pH. This acidic environment alters the tertiary structure of haemoglobin, reducing its affinity for oxygen. Consequently, the oxygen dissociation curve shifts to the right (the Bohr shift), causing haemoglobin to release oxygen more readily at any given partial pressure of oxygen.

1 mark for the correct answer A. 0 marks for any other response.

Active loading of sucrose is achieved by the active transport of hydrogen ions (\(\text{H}^+\)) out of the companion cells into the surrounding apoplast (cell wall) space using ATP-driven proton pumps. This establishes a high concentration of protons outside the cell. Protons then diffuse back down their electrochemical gradient into the companion cell through co-transporter proteins, bringing sucrose molecules with them against the sucrose concentration gradient.

1 mark for the correct answer C. 0 marks for any other response.

The constant region of an antibody is the same within a class of antibodies (e.g., IgG). Its primary role is to bind to receptors on immune cells like macrophages and neutrophils, which enhances phagocytosis (a process known as opsonisation). The variable region (Fab) contains the antigen-binding sites, which are highly variable to recognize different pathogens. Disulfide bridges are structural bonds and are found in both the constant and variable regions.

1 mark for the correct answer C. 0 marks for any other response.

According to Chargaff's rules for double-stranded DNA, the amount of cytosine (C) equals the amount of guanine (G), and the amount of adenine (A) equals the amount of thymine (T). Given that \(\text{C} = 22\%\), then \(\text{G} = 22\%\). Together, \(\text{C} + \text{G} = 44\%\). This leaves \(100\% - 44\% = 56\%\) for \(\text{A} + \text{T}\). Since \(\text{A} = \text{T}\), the percentage of adenine must be \(56\% / 2 = 28\%\).

1 mark for the correct answer B. 0 marks for any other response.

Magnification is defined as the number of times larger an image appears compared to the actual size of the specimen. Resolution is defined as the minimum distance between two objects at which they can still be clearly distinguished as separate entities. Resolution is limited by wavelength (e.g., of light or electrons), whereas magnification is not. Increasing magnification beyond the resolution limit merely produces a larger, blurrier image (empty magnification) without revealing extra detail.

1 mark for the correct answer B. 0 marks for any other response.

As temperature increases, phospholipid molecules gain more kinetic energy and move faster, increasing the fluidity of the membrane and creating temporary gaps in the bilayer. Concurrently, membrane proteins denature as their hydrogen and ionic bonds break at higher temperatures, creating large channels that allow the vacuolar pigment (betalain) to leak out easily. Therefore, option A is correct.

1 mark for the correct option A. Reject other options: cholesterol does not denature (B); active transport proteins denature and do not actively pump pigment out (C); the cell wall does not rupture under these conditions to cause this permeability change (D).

The aortic valve (a semi-lunar valve) opens when the blood pressure within the left ventricle exceeds the blood pressure within the aorta. At 0.25 s, the pressure in the left ventricle (10.2 kPa) is greater than the pressure in the aorta (10.1 kPa), which forces the valve to open. Prior to this (at 0.20 s), the ventricular pressure (5.2 kPa) is lower than the aortic pressure (10.0 kPa), so the valve remains shut.

1 mark for the correct option C. Only C provides the time at which ventricular pressure first exceeds aortic pressure in the dataset.

If increasing the substrate concentration can restore the maximum rate of reaction (Vmax), the inhibitor must be a competitive inhibitor. Competitive inhibitors have a similar shape to the substrate and bind reversibly to the active site. When substrate concentration is significantly increased, the substrate molecules outcompete the inhibitor molecules for the active sites, overcoming the inhibition.

1 mark for the correct option A. Reject: Non-competitive/allosteric inhibitors (B and C) which prevent the maximum rate from being restored regardless of substrate concentration. Reject: Inhibitors do not lower activation energy (D).

In double-stranded DNA, complementary base pairing rules state that the percentage of adenine (A) equals thymine (T), and the percentage of cytosine (C) equals guanine (G). Since A = 24%, then T = 24%. Together, A + T = 48%. The remaining bases must be C + G, which equals 100% - 48% = 52%. Since C and G are present in equal amounts, the percentage of cytosine is 52% / 2 = 26%.

1 mark for the correct option B. 26% calculated using Chargaff's rules where A = T and C = G.

Part (i): Standard glucose solutions of known concentrations are prepared using serial dilution (e.g., from a 1% stock). A fixed volume of quantitative Benedict's reagent is added to each solution, and they are heated in a water bath at 80-100 degrees Celsius for 5 minutes. The resulting precipitate of copper(I) oxide is removed by filtration or centrifugation. The absorbance or transmission of the remaining blue supernatant is measured using a colorimeter. A calibration curve is plotted with glucose concentration on the x-axis and absorbance/transmission on the y-axis. Part (ii): The unreacted Cu2+ ions in the supernatant are blue, which absorbs red light. Using a red filter increases the sensitivity of the measurements by maximizing light absorption. Part (iii): The Benedict's test is performed on the unknown sample using the identical procedure and volumes. Its absorbance is measured with the colorimeter using the red filter. The student locates this absorbance value on the y-axis of the calibration curve, reads across to the curve, and drops down to the x-axis to determine the glucose concentration.

1. Serial dilutions: describe making a range of known glucose concentrations from a stock solution. 2. Standardised test: add a constant volume of Benedict's reagent and heat at 80-100 degrees Celsius for a set time. 3. Filtration: describe removing the precipitate (by filtering or centrifuging) to isolate the supernatant. 4. Colorimetry: measure absorbance or transmission of the supernatant. 5. Graph: plot concentration on the x-axis and absorbance/transmission on the y-axis. 6. Absorption: explain that blue supernatant absorbs red light. 7. Sensitivity: explain that a red filter maximizes sensitivity / absorbance range. 8. Identical procedure: perform the exact same test/heating/filtering on the unknown sample. 9. Absorbance of unknown: measure the absorbance of the treated unknown sample. 10. Reading curve: locate the unknown's absorbance on the y-axis and read the corresponding concentration off the x-axis.

Part (i): Ethanol is a non-polar solvent that dissolves the lipids and phospholipids of the beetroot cell membrane and tonoplast, disrupting the bilayer structure and making it more fluid. It also denatures membrane proteins, forming gaps. This increases permeability, allowing the red pigment (betalain) to diffuse out of the vacuole and cell. Part (ii): Temperature must be controlled using a thermostatically controlled water bath because high temperatures also disrupt membranes. Surface area/size of beetroot must be controlled by using a cork borer for uniform diameter and cutting to equal lengths with a scalpel and ruler. Volume of ethanol must be controlled using a volumetric pipette to ensure the dilution of pigment is consistent. Part (iii): Use a blank cuvette containing distilled water or solvent to set the colorimeter absorbance to 0.00 (or transmission to 100%). Ensure the cuvette is dry and wiped clean of fingerprints. Use the same orientation of the cuvette for every reading, and select a green/blue-green filter (~520nm) as red pigment absorbs green light.

1. Lipids: state that ethanol dissolves phospholipids/lipids in membranes. 2. Fluidity: explain that this disrupts the bilayer / increases membrane fluidity. 3. Proteins: explain that ethanol denatures membrane proteins. 4. Diffusion: state that betalain leaks out of the vacuole via diffusion. 5. Control variable 1: Temperature controlled using a water bath. 6. Control variable 2: Surface area of beetroot controlled using cork borer and ruler/scalpel. 7. Control variable 3: Volume of ethanol/solvent controlled using a pipette/syringe. 8. Blank: use a blank (distilled water/solvent) to zero the absorbance (or set transmission to 100%). 9. Cuvette care: wipe cuvettes to remove dirt/fingerprints and orient them identically. 10. Filter: select an appropriate green/blue-green filter (~520nm) for the red pigment.

Part (i): Cut the terminal 1-2 mm of an onion root tip to capture the meristematic region. Heat the tip in dilute hydrochloric acid to break down the middle lamella/pectin, separating the cells (maceration). Rinse the tip with water and transfer to a slide. Add a drop of stain, such as acetic orcein, which binds to DNA and makes chromosomes visible. Lower a coverslip carefully with a mounted needle to prevent air bubbles, cover with a paper towel, and press down vertically to squash the tissue into a single thin layer of cells (without rolling the coverslip). Part (ii): Number of mitotic cells = 18 + 12 + 6 + 4 = 40. Total number of cells = 40 + 160 = 200. Mitotic Index = 40 / 200 = 0.20 (or 20%). Part (iii): In metaphase, spindle fibres attach to the centromeres of chromosomes and align them along the cell equator. In anaphase, the spindle fibres contract/shorten, pulling sister chromatids to opposite poles of the cell.

1. Meristem: cut the terminal 1-2 mm of the root tip. 2. Acid hydrolysis: use hydrochloric acid and heat to macerate the tissues / break middle lamella. 3. Staining: add a stain like acetic orcein or toluidine blue to dye the chromosomes. 4. Squash method: press down firmly/vertically with a paper towel on the coverslip without sliding. 5. Monolayer: explain that squashing creates a single layer of cells for light transmission. 6. Mitotic count: state total mitotic cells is 40 and total cells is 200. 7. Calculation: divide mitotic cells by total cells to get 0.20 / 20%. 8. Working: show clear intermediate step/working. 9. Metaphase role: spindle fibres attach to centromeres and align chromosomes along the equator. 10. Anaphase role: spindle fibres contract/shorten to pull sister chromatids to opposite poles.

Part (i): The P wave corresponds to the depolarisation of the atria, initiated by the sinoatrial node, leading to atrial systole. The QRS complex corresponds to the depolarisation of the ventricles, triggered as the wave of excitation spreads from the atrioventricular node down the bundle of His and Purkyne fibres, causing ventricular systole. The T wave represents the repolarisation of the ventricles, leading to ventricular diastole. Part (ii): Duration of cardiac cycle = 60 seconds / 75 bpm = 0.8 seconds. Part (iii): During ventricular systole, the muscular walls of the left ventricle contract, rapidly increasing intraventricular pressure. When the pressure in the left ventricle exceeds the pressure in the left atrium, blood is forced back against the cusps of the bicuspid (mitral) valve, closing it to prevent backflow. As ventricular contraction continues, ventricular pressure rises further until it exceeds the pressure within the aorta. This pressure gradient forces the aortic semilunar valve open, allowing blood to flow into the aorta. When the ventricle begins to relax (diastole), ventricular pressure drops below aortic pressure, and backflow of blood in the aorta fills the pockets of the semilunar valve, forcing it shut.

1. P wave: state that it represents atrial depolarisation / SAN firing. 2. QRS complex: state that it represents ventricular depolarisation / transmission via Purkyne tissue. 3. T wave: state that it represents ventricular repolarisation. 4. Formula: show the calculation 60 / 75. 5. Cycle length: state the correct answer of 0.8 seconds. 6. Bicuspid closure: ventricular pressure exceeds atrial pressure, forcing the bicuspid valve closed. 7. Atrial backflow: explain that bicuspid closure prevents backflow of blood into the left atrium. 8. Semilunar opening: ventricular pressure exceeds aortic pressure, opening the aortic semilunar valve. 9. Ventricular ejection: state that blood is ejected into the aorta. 10. Semilunar closure: ventricular pressure drops below aortic pressure, causing the semilunar valve to close to prevent backflow into the ventricle.

Part (i): The patient's serum sample containing the target viral antigen is added to a plastic microtiter well, allowing the antigens to bind (adsorb) to the well's surface. The well is then washed with a buffer solution to remove unbound proteins. Next, monoclonal antibodies specific to the viral antigen, which are conjugated to a detection enzyme, are added to the well to bind to the immobilized antigen. The well is washed again to remove any unbound antibody-enzyme conjugates. Finally, a colorless substrate is added; the enzyme catalyses a reaction that converts the substrate into a colored product, indicating a positive result. Part (ii): The washing steps are essential to remove any unbound, excess proteins and enzyme-linked antibodies. Without washing, unbound enzyme-linked antibodies would remain in the well even if no antigen were present. The addition of the substrate would still lead to a color reaction, creating a false-positive result. Part (iii): The primary immune response has a lag phase of several days while naive B-cells undergo clonal selection and expansion, producing low antibody levels that drop quickly. In contrast, the secondary response is faster (shorter lag phase), produces a much higher concentration of antibodies, and maintains high levels for longer. This is because memory B and T lymphocytes, produced during the primary response, remain in circulation and can immediately recognize the antigen, rapidly differentiating into plasma cells without delay.

1. Adsorption: patient sample containing antigen is bound to the well. 2. Wash 1: wash to remove unbound proteins/antigen. 3. Antibody: add enzyme-linked monoclonal antibodies specific to the target antigen. 4. Wash 2: wash to remove unbound antibody-enzyme conjugates. 5. Detection: add substrate and measure color intensity. 6. Wash role: washing removes unbound antibodies/enzymes. 7. False-positives: without washing, unbound enzymes would react with substrate, causing false-positive results. 8. Contrast: secondary response is faster, has a higher peak antibody concentration, and lasts longer than the primary response. 9. Memory cell presence: memory B and T lymphocytes persist in circulation from the primary response. 10. Clonal expansion: memory B-cells immediately undergo clonal expansion and differentiate into antibody-secreting plasma cells, bypassing the slow initial activation steps.

(a)
1. Make a range of known glucose concentrations using serial dilution.
2. Add a constant, excess volume of Benedict's reagent to each known concentration and the unknown sample, then heat in a water bath at \(80^\circ\text{C}-100^\circ\text{C}\) for a set time (e.g., 5 minutes).
3. Allow to cool and filter or centrifuge the samples to remove the red precipitate of copper(I) oxide.
4. Calibrate the colorimeter using a blank (distilled water) using a red filter (which is absorbed by the blue Benedict's solution).
5. Measure the absorbance or transmission of the remaining supernatant for each known concentration.
6. Plot a calibration curve of absorbance/transmission against glucose concentration. Test the unknown sample, measure its absorbance/transmission, and read its concentration from the calibration curve.

(b)
- Amylose is a polymer of \(\alpha\)-glucose, whereas cellulose is a polymer of \(\beta\)-glucose.
- Amylose molecules are coiled/helical due to 1,4-glycosidic bonds without rotation, making them highly compact and ideal for storage within cells without affecting osmotic potential.
- Cellulose molecules contain \(\beta\)-glucose monomers where alternating units are rotated by \(180^\circ\), resulting in straight, unbranched chains.
- These parallel cellulose chains are cross-linked by hydrogen bonds to form microfibrils, which aggregate into macrofibrils, providing immense tensile strength to plant cell walls.

(c)
- Phospholipids have a phosphate group substituting one of the three fatty acids found in a triglyceride (i.e., glycerol + 2 fatty acids + 1 phosphate vs glycerol + 3 fatty acids).
- The phosphate group is polar/charged and hydrophilic, while the fatty acid hydrocarbon chains are non-polar and hydrophobic.
- This amphipathic nature causes phospholipids in water to self-assemble into a bilayer, with hydrophobic tails pointing inwards (shielded from water) and hydrophilic heads facing outwards towards the aqueous cytoplasm and extracellular fluid, creating a selectively permeable barrier.

(a) Max 6 marks:
- MP1: Detail of creating a range of known glucose concentrations using serial dilution. (1)
- MP2: Addition of equal volumes of Benedict's reagent and heating in a water bath (temp \(>80^\circ\text{C}\)). (1)
- MP3: Centrifuging or filtering the mixtures to obtain a clear supernatant. (1)
- MP4: Calibrating colorimeter using a blank / water. (1)
- MP5: Use of a red filter (or wavelength around 680 nm) to measure absorbance or transmission of supernatant. (1)
- MP6: Plotting a calibration curve with concentration on x-axis and absorbance/transmission on y-axis. (1)
- MP7: Using the absorbance/transmission value of the unknown sample to find its concentration on the curve. (1)

(b) Max 5 marks:
- MP1: Monomer difference: Amylose contains \(\alpha\)-glucose, cellulose contains \(\beta\)-glucose. (1)
- MP2: Structural shape: Amylose is coiled/helical, cellulose is a straight/unbranched chain. (1)
- MP3: Rotation of monomer: In cellulose, alternate glucose monomers are rotated \(180^\circ\) (not in amylose). (1)
- MP4: Hydrogen bonding: Cellulose chains form hydrogen bonds between adjacent chains, amylose does not (it forms internal H-bonds). (1)
- MP5: Function link: Coiled amylose makes it compact for storage; cellulose microfibrils provide strength to cell walls. (1)

(c) Max 3.5 marks:
- MP1: Phospholipids have 2 fatty acids + 1 phosphate group; triglycerides have 3 fatty acids (both have glycerol). (1)
- MP2: Phospholipids have a hydrophilic/polar 'head' and hydrophobic/non-polar 'tails'. (1)
- MP3: Relates to bilayer: Hydrophobic tails orientate inwards away from water, while hydrophilic heads face outwards towards water, forming a stable membrane barrier. (1.5 marks for both parts of this explanation; award 0.5 for just saying 'bilayer forms' without description).

(a)
- Haemoglobin is a quaternary protein with four polypeptide chains, each containing a haem group that can bind one oxygen molecule.
- At low partial pressure of oxygen (\(pO_2\)), the shape of the haemoglobin molecule makes it difficult for the first oxygen molecule to bind, resulting in a shallow gradient initially.
- Once the first oxygen molecule binds, it induces a conformational change in the tertiary structure of the protein (cooperative binding).
- This makes it significantly easier for the second and third oxygen molecules to bind, causing the curve to steepen rapidly.
- The curve plateaus at high \(pO_2\) because most oxygen-binding sites are occupied, reducing the probability of a fourth oxygen molecule finding an empty site.
- The Bohr effect: Respiring tissues release carbon dioxide, increasing the partial pressure of carbon dioxide (\(pCO_2\)).
- This high \(pCO_2\) causes haemoglobin's affinity for oxygen to decrease, shifting the dissociation curve to the right. This allows oxygen to be released more readily to actively respiring tissues.

(b)
- In red blood cells, carbon dioxide (\(CO_2\)) reacts with water (\(H_2O\)) to form carbonic acid (\(H_2CO_3\)).
- This reaction is catalyzed by the enzyme carbonic anhydrase.
- Carbonic acid quickly dissociates into hydrogen ions (\(H^+\)) and hydrogencarbonate ions (\(HCO_3^-\)).
- Hydrogencarbonate ions diffuse out of the red blood cells into the blood plasma, where they are transported in solution to the lungs (accounting for ~85% of \(CO_2\) transport).
- To maintain electrical neutrality, chloride ions (\(Cl^-\)) diffuse into the erythrocyte when \(HCO_3^-\)` leaves (the chloride shift).
- Hydrogen ions bind to haemoglobin to form haemoglobonic acid, preventing a dangerous drop in blood pH (buffering effect).

(c)
1. Calculate Heart Rate (HR) in beats per minute:
\(5 \text{ cycles} = 4.0 \text{ seconds}\)
\(1 \text{ cycle} = \frac{4.0}{5} = 0.8 \text{ seconds per beat}\)
\(\text{HR} = \frac{60}{0.8} = 75 \text{ beats min}^{-1}\)
2. Calculate Cardiac Output in \(\text{cm}^3 \text{ min}^{-1}\):
\(\text{Cardiac Output} = \text{Stroke Volume} \times \text{Heart Rate}\)
\(\text{Cardiac Output} = 75 \text{ cm}^3 \times 75 \text{ beats min}^{-1} = 5625 \text{ cm}^3 \text{ min}^{-1}\)
3. Convert to \(\text{dm}^3 \text{ min}^{-1}\):
\(\frac{5625}{1000} = 5.625 \text{ dm}^3 \text{ min}^{-1}\)

(a) Max 6 marks:
- MP1: Reference to haemoglobin having quaternary structure / 4 haem groups. (1)
- MP2: Initial binding is difficult due to molecular conformation (causes flat initial curve). (1)
- MP3: Explanation of cooperative binding: binding of first \(O_2\) alters tertiary/quaternary shape. (1)
- MP4: This shape change uncovers other haem groups, making subsequent binding of 2nd/3rd \(O_2\) easier (causes steep curve). (1)
- MP5: Leveling off at high \(pO_2\) because probability of finding empty sites decreases. (1)
- MP6: Bohr effect description: increased \(pCO_2\) shifts curve to the right. (1)
- MP7: Explanation of shift to right: haemoglobin has lower affinity for oxygen at any given \(pO_2\), aiding oxygen unloading at respiring tissue. (1)

(b) Max 5 marks:
- MP1: Carbonic anhydrase catalyzes reaction of \(CO_2 + H_2O \rightarrow H_2CO_3\). (1)
- MP2: Carbonic acid (\(H_2CO_3\)) dissociates into \(H^+\) and \(HCO_3^-\). (1)
- MP3: \(HCO_3^-\)` ions diffuse out of the cytoplasm of erythrocytes into the plasma. (1)
- MP4: Mention of the chloride shift (chloride ions enter to balance charge). (1)
- MP5: Hydrogen ions (\(H^+\)) bind to haemoglobin to form haemoglobonic acid, acting as a buffer. (1)

(c) Max 3.5 marks:
- MP1: Correct heart rate calculation of 75 beats min⁻¹ (or showing work: \((5 / 4) \times 60\)). (1)
- MP2: Correct cardiac output formula: \(\text{Stroke Volume} \times \text{Heart Rate}\). (1)
- MP3: Final correct value: \(5.625\) (or \(5.63\) / \(5.6\)) with correct units of \(\text{dm}^3 \text{ min}^{-1}\). (1.5 marks; award 0.5 marks if calculation is correct but units are incorrect or omitted).

(a)
- 'Fluid': Phospholipids and proteins are free to move laterally/rotate within their layer, giving the membrane flexibility.
- 'Mosaic': Proteins of various sizes and shapes are scattered/embedded in the lipid bilayer like tiles in a mosaic.
- Glycoproteins: These consist of proteins with branching carbohydrate chains attached. They act as cell surface receptors (e.g., for hormones or neurotransmitters in cell signalling) or as cell markers (antigens) for cell-to-cell recognition.
- Cholesterol: Fits between phospholipid fatty acid tails. It binds to hydrophobic tails, stabilizing the membrane and regulating fluidity (preventing the membrane from becoming too fluid at high temperatures or too rigid/solidified at low temperatures).

(b)
- Independent variable: At least 5 different concentrations of ethanol (e.g., 0%, 20%, 40%, 60%, 80%).
- Dependent variable: Absorbance (or percentage transmission) of light of the bathing solution using a colorimeter (using a blue/green filter to detect the red betalain pigment).
- Key controlled variables (must state 3 for full marks):
1. Temperature of the water bath/environment.
2. Dimensions/surface area/volume of the beetroot discs (standardized using a cork borer and ruler).
3. Source/age/type of beetroot used.
4. Volume of ethanol solution in each tube.
5. Incubation time (time the discs are left in the ethanol).
- Brief method description: Beetroot discs must be thoroughly washed in running water before the experiment to remove any surface betalain pigment released during cutting.

(c)
- Energy requirement: Simple diffusion and facilitated diffusion are passive processes powered by the kinetic energy of the particles (moving down a concentration gradient, no ATP required). Active transport requires metabolic energy in the form of ATP to move molecules against their concentration gradient.
- Proteins involved:
- Simple diffusion occurs directly through the phospholipid bilayer; no membrane proteins are required.
- Facilitated diffusion requires specific transmembrane proteins, which can be either channel proteins (for ions/water-soluble substances) or carrier proteins.
- Active transport requires specific carrier proteins (often referred to as pumps) that undergo conformational changes using ATP.

(a) Max 5 marks:
- MP1: Explaining 'fluid' (phospholipids/proteins can move laterally). (1)
- MP2: Explaining 'mosaic' (proteins embedded in bilayer in random/scattered pattern). (1)
- MP3: Glycoprotein function: cell-signalling / cell recognition / receptors / adhesion. (1)
- MP4: Cholesterol function: fits between tails, regulates fluidity / stability / prevents leakage of water/ions. (1)
- MP5: Clarification that cholesterol stabilizes at high temperatures and prevents crystallisation/solidifying at low temperatures. (1)

(b) Max 6 marks:
- MP1: Explicit identification of Independent Variable: concentration of ethanol (must specify a range of at least 5 concentrations). (1)
- MP2: Explicit identification of Dependent Variable: light absorbance/transmission of solution containing betalain pigment using a colorimeter. (1)
- MP3: Controlled variable 1: Temperature of water bath. (1)
- MP4: Controlled variable 2: Surface area/volume/size/number of beetroot cylinders (e.g., cut with cork borer to same length). (1)
- MP5: Controlled variable 3: Volume of solution / time in solution. (1)
- MP6: Experimental detail: washing of beetroot discs prior to experiment to remove pigment released by cutting. (1)

(c) Max 3.5 marks:
- MP1: Passive vs Active: Simple and facilitated diffusion do not require ATP (passive), while active transport requires ATP. (1)
- MP2: Protein use in passive: Simple diffusion does not use proteins, while facilitated diffusion uses channel or carrier proteins. (1)
- MP3: Protein use in active: Active transport requires carrier proteins (pumps). (1.5 marks for complete comparison of protein roles across all three types; award 0.5 marks for partial protein comparison).

(a)
- Significance of mitosis:
- Growth: increases cell number in multicellular organisms.
- Repair of damaged tissues: replaces worn-out or damaged cells with genetically identical ones.
- Asexual reproduction: producing genetically identical offspring in some organisms.
- Events of metaphase:
- Chromosomes (consisting of two sister chromatids) align along the equator (metaphase plate) of the cell.
- Spindle fibres attach to the centromeres of each chromosome.
- Events of anaphase:
- The centromeres split, separating the sister chromatids.
- Spindle fibres contract, pulling chromatids (now chromosomes) to opposite poles of the spindle, centromere first.

(b)
(i) Calculate the Mitotic Index (MI):
- Total cells in mitosis = \(18 \text{ (prophase)} + 6 \text{ (metaphase)} + 4 \text{ (anaphase)} + 2 \text{ (telophase)} = 30\) cells.
- Total cells observed = \(150\) cells.
- \(\text{Mitotic Index} = \frac{\text{Number of cells in mitosis}}{\text{Total number of cells}} \times 100 = \frac{30}{150} \times 100 = 20.0\%\).

(ii) Calculate average duration of metaphase in minutes:
- Fraction of cells in metaphase = \(\frac{6}{150} = 0.04\).
- Total time of cell cycle in minutes = \(18 \text{ hours} \times 60 \text{ minutes/hour} = 1080 \text{ minutes}\).
- Duration of metaphase = \(0.04 \times 1080 \text{ minutes} = 43.2 \text{ minutes}\).
*(Alternative working: \(0.04 \times 18 \text{ hours} = 0.72 \text{ hours}\); \(0.72 \times 60 = 43.2 \text{ minutes}\))*

(c)
- Neutrophil specialization:
- Multi-lobed nucleus: allows flexibility so the cell can change shape and squeeze through narrow gaps in capillary walls (diapedesis) to reach infection sites.
- Many lysosomes/granules in cytoplasm: contain hydrolytic enzymes (such as lysozyme) to digest engulfed pathogens during phagocytosis.
- Well-developed cytoskeleton: helps with changing shape during phagocytosis (pseudopodia formation).
- Numerous mitochondria: provide ATP for active processes like engulfing pathogens.
- Contrast with Erythrocyte:
- Erythrocytes have no nucleus or organelles (ribosomes, mitochondria, rough endoplasmic reticulum) at maturity to maximize cytoplasmic volume for haemoglobin, enabling maximum oxygen transport capacity.
- Erythrocytes have a biconcave shape to increase surface area to volume ratio for diffusion of gases, whereas neutrophils have a flexible, variable shape.

(a) Max 5 marks:
- MP1: Significance: Produces genetically identical daughter cells for growth, tissue repair, or asexual reproduction (must state at least two). (1)
- MP2: Metaphase: Chromosomes line up along the equator/metaphase plate. (1)
- MP3: Metaphase: Spindle fibres attach to centromeres. (1)
- MP4: Anaphase: Centromeres divide/split separating sister chromatids. (1)
- MP5: Anaphase: Spindle fibres shorten/contract, pulling chromatids/chromosomes to opposite poles. (1)

(b) Max 4.5 marks:
- Part (i) [2 marks]:
- MI calculation: \((30 / 150) \times 100\) (1)
- Correct final answer: \(20\)% or \(20.0\)% (1)
- Part (ii) [2.5 marks]:
- Correct fraction: \(6 / 150\) or \(0.04\). (0.5)
- Correct conversion of 18 hours to minutes (1080 minutes) or showing appropriate working. (1)
- Correct final answer: \(43.2\) minutes (allow 43 minutes). (1)

(c) Max 5 marks:
- MP1: Neutrophil has a multi-lobed nucleus allowing it to squeeze through capillary walls. (1)
- MP2: Neutrophil has many lysosomes containing hydrolytic/digestive enzymes to destroy engulfed pathogens. (1)
- MP3: Neutrophil contains many mitochondria to generate ATP for phagocytosis. (1)
- MP4: Erythrocyte lacks a nucleus/organelles to maximize space for haemoglobin. (1)
- MP5: Erythrocyte has a biconcave disc shape to increase the SA:Vol ratio for gas exchange (contrast with dynamic shape of neutrophil). (1)

To achieve 5-6 marks (Level 3), students must clearly explain both the active loading mechanism in detail (including proton pumps, ATP, co-transporters, and plasmodesmata) and connect this loading to the physical changes that initiate mass flow (water potential, osmosis, hydrostatic pressure). Level 2 (3-4 marks) will describe either the active loading or mass flow in detail, or both with some missing steps. Level 1 (1-2 marks) will show a basic understanding of active transport or bulk flow without explaining the mechanism of co-transport.

Level 3 (5-6 marks): Detailed, logical explanation of both active loading (active H+ transport, co-transport of sucrose) and the physical process initiating mass flow (osmosis, hydrostatic pressure gradient). Key scientific terms used accurately throughout. | Level 2 (3-4 marks): Clear explanation of either active loading or mass flow, or a partial explanation of both. There may be minor omissions in the sequence. | Level 1 (1-2 marks): Basic description of sucrose movement or water movement, lacking specific mechanisms or correct sequence. | Indicative content: 1. Hydrogen ions/protons (H+) are actively pumped out of companion cells into the apoplast/cell wall. 2. This process requires ATP. 3. This establishes a proton concentration/electrochemical gradient. 4. Protons diffuse back into companion cells through co-transporter proteins. 5. Sucrose is carried into companion cells alongside protons, against its concentration gradient. 6. Sucrose diffuses into sieve tube elements via plasmodesmata. 7. The loading of sucrose lowers the water potential of the sieve tube element. 8. Water enters the sieve tube element from the xylem by osmosis. 9. This generates high hydrostatic pressure, initiating mass flow towards the sink.

To achieve 5-6 marks (Level 3), students must logically connect the cellular recognition phase (antigen presentation, clonal selection, helper T cell involvement) to the proliferation phase (clonal expansion, differentiation into plasma cells) and detail the exact mechanisms of antibody action (agglutination, opsonisation, neutralisation). Level 2 (3-4 marks) will explain the pathway of antibody production well but describe antibody functions superficially, or vice versa. Level 1 (1-2 marks) will provide a general description of white blood cells making antibodies with little detail on cellular interactions.

Level 3 (5-6 marks): Comprehensive, structured response detailing both the cellular activation pathway (presentation, selection, expansion, differentiation) and at least two distinct antibody defense mechanisms (neutralisation, opsonisation, agglutination). Biological terminology is precise. | Level 2 (3-4 marks): Explains the stages of the cellular response but with less detail on antibody function, or describes antibody actions well but misses key activation steps (e.g., role of T helper cells or interleukins). | Level 1 (1-2 marks): Simple description of B cells multiplying and producing antibodies to fight the virus, with minimal detail. | Indicative content: 1. Macrophages engulf virus and present antigens on MHC molecules to become APCs. 2. T helper cells with complementary receptors bind APCs and are activated. 3. Activated T helper cells release cytokines/interleukins. 4. Complementary B cells bind antigen (clonal selection). 5. Interleukins trigger B cells to undergo mitosis (clonal expansion). 6. B cells differentiate into plasma cells which secrete specific antibodies. 7. Antibodies cause agglutination (clumping pathogens to prevent spread). 8. Antibodies act as opsonins (enhancing phagocytosis). 9. Antibodies neutralise the virus (blocking host cell entry).