Cambridge IAL · Thinka-original Practice Paper

2024 Cambridge IAL Biology (9700) Practice Paper with Answers

Thinka Jun 2024 (V1) Cambridge International A Level-Style Mock — Biology (9700)

270 marks445 mins2024
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V1) Cambridge International A Level Biology (9700) paper. Not affiliated with or reproduced from Cambridge.

Paper 1 (Multiple Choice)

Answer all 40 multiple-choice questions. Each question has four options.
40 Question · 40 marks
Question 1 · multiple_choice
1 marks
Which sequence represents the correct pathway taken by a newly synthesized membrane glycoprotein from its site of synthesis to its final destination in the cell surface membrane?
  1. A.rough endoplasmic reticulum -> transport vesicle -> Golgi body -> secretory vesicle -> cell surface membrane
  2. B.smooth endoplasmic reticulum -> transport vesicle -> Golgi body -> secretory vesicle -> cell surface membrane
  3. C.rough endoplasmic reticulum -> Golgi body -> transport vesicle -> lysosome -> cell surface membrane
  4. D.nucleolus -> rough endoplasmic reticulum -> transport vesicle -> Golgi body -> cell surface membrane
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Worked solution

The protein component of a glycoprotein is synthesized by ribosomes bound to the rough endoplasmic reticulum (RER). The protein is co-translationally imported into the RER lumen where initial glycosylation may begin. It is then transported in transport vesicles to the cis face of the Golgi body. In the Golgi cisternae, the carbohydrate chains are modified and completed. The mature glycoprotein is then packaged into secretory vesicles at the trans face of the Golgi. These secretory vesicles move along the cytoskeleton and fuse with the cell surface membrane, delivering the membrane glycoprotein to its final destination.

Marking scheme

Award 1 mark for the correct option A. Reject options B, C, and D because they do not represent the correct sequence of organelles and transport intermediates involved in the secretory pathway of membrane glycoproteins.
Question 2 · multiple_choice
1 marks
A student placed cylinders of potato tissue in sucrose solutions of different concentrations. After 24 hours, the percentage change in mass was calculated and plotted. The concentration at which there was no change in mass was determined to be \(0.3\text{ mol dm}^{-3}\). Which statement about the potato cells at this concentration is correct?
  1. A.The pressure potential of the cells is equal to zero.
  2. B.The water potential of the cells is equal to the water potential of the sucrose solution.
  3. C.The solute potential of the cells is equal to the solute potential of the sucrose solution.
  4. D.Active transport of sucrose into the cells is equal to the rate of passive diffusion out of the cells.
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Worked solution

When there is no change in the mass of the potato cylinders, the net movement of water into or out of the cells is zero. This occurs because the water potential of the potato cells has reached dynamic equilibrium with the water potential of the external sucrose solution. The solute potential of the cells is not equal to that of the solution because the potato cells also possess a positive pressure potential (turgor pressure) exerted by their cell walls.

Marking scheme

Award 1 mark for the correct option B. Reject options A, C, and D because at equilibrium of mass, the water potentials must be equal, whereas pressure and solute potentials of the cells differ from those of the surrounding open solution.
Question 3 · multiple_choice
1 marks
Which statements correctly describe the differences between a triglyceride molecule and a phospholipid molecule? 1. Triglycerides have three fatty acid tails, while phospholipids have two fatty acid tails. 2. Triglycerides are completely hydrophobic, while phospholipids have a hydrophilic head and hydrophobic tails. 3. Triglycerides contain only carbon, hydrogen, and oxygen, while phospholipids contain carbon, hydrogen, oxygen, and phosphorus.
  1. A.1, 2 and 3
  2. B.1 and 2 only
  3. C.2 and 3 only
  4. D.1 and 3 only
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Worked solution

All three statements are correct. Triglycerides are formed from one glycerol molecule condensed with three fatty acids, making them entirely non-polar and hydrophobic. Phospholipids have one fatty acid tail replaced by a phosphate group, giving them a polar, hydrophilic phosphate head and two hydrophobic fatty acid tails (making them amphipathic). Because of the phosphate group, phospholipids contain phosphorus, which is absent in triglycerides.

Marking scheme

Award 1 mark for the correct option A. Reject options B, C, and D because they do not include all three correct statement numbers.
Question 4 · multiple_choice
1 marks
Which processes are involved in the loading of sucrose into the phloem sieve tube elements at a source? 1. Active transport of protons (\(\text{H}^+\)) out of the companion cells into the cell wall. 2. Co-transport of protons (\(\text{H}^+\)) and sucrose back into the companion cells down a proton gradient. 3. Diffusion of sucrose through plasmodesmata from the companion cells into the sieve tube elements.
  1. A.1, 2 and 3
  2. B.1 and 2 only
  3. C.2 and 3 only
  4. D.1 and 3 only
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Worked solution

Active loading of sucrose into companion cells begins with proton pumps actively transporting protons (\(\text{H}^+\)) out of the companion cell cytoplasm into the cell wall space. This builds up a high concentration of protons outside. Protons then diffuse back down their electrochemical gradient into the companion cells through a co-transporter protein, bringing sucrose along with them against its concentration gradient. Once high sucrose concentrations accumulate in the companion cells, sucrose diffuses symplastically through plasmodesmata into the neighboring sieve tube elements.

Marking scheme

Award 1 mark for the correct option A. Reject options B, C, and D because all three steps are necessary and correct parts of the phloem loading mechanism in sources.
Question 5 · multiple_choice
1 marks
A person consumes a meal with a very high salt (sodium chloride) content but does not drink any water. What are the short-term changes in the concentration of antidiuretic hormone (ADH) in the blood, the permeability of the collecting duct to water, and the volume of urine produced?
  1. A.Blood ADH concentration: increases; Permeability of collecting duct: increases; Urine volume: decreases
  2. B.Blood ADH concentration: decreases; Permeability of collecting duct: decreases; Urine volume: increases
  3. C.Blood ADH concentration: increases; Permeability of collecting duct: decreases; Urine volume: decreases
  4. D.Blood ADH concentration: decreases; Permeability of collecting duct: increases; Urine volume: increases
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Worked solution

A high salt intake without water intake decreases the water potential of the blood. Osmoreceptors in the hypothalamus detect this decrease, causing the posterior pituitary gland to secrete more antidiuretic hormone (ADH) into the blood. Increased blood ADH concentration stimulates cells of the collecting duct to insert more aquaporins into their luminal membranes, increasing the permeability of the collecting duct to water. Consequently, more water is reabsorbed from the filtrate into the medulla, resulting in a decreased volume of highly concentrated urine.

Marking scheme

Award 1 mark for the correct option A. Reject options B, C, and D because they do not correctly identify that low water potential causes an increase in ADH secretion, an increase in duct permeability, and a decrease in urine volume.
Question 6 · multiple_choice
1 marks
In an investigation of the light-independent stage of photosynthesis, algae were exposed to radioactive carbon dioxide (\(^{14}\text{CO}_2\)) in the light. The light was then turned off, but the supply of \(^{14}\text{CO}_2\) was maintained. What happens to the concentrations of ribulose bisphosphate (RuBP) and glycerate 3-phosphate (GP) immediately after the light is turned off?
  1. A.RuBP concentration decreases, GP concentration increases
  2. B.RuBP concentration increases, GP concentration decreases
  3. C.Both RuBP and GP concentrations increase
  4. D.Both RuBP and GP concentrations decrease
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Worked solution

In the dark, the light-dependent reactions cease, meaning no ATP and reduced NADP are synthesized. GP requires both ATP and reduced NADP to be reduced to triose phosphate (TP). In their absence, GP cannot be converted, and thus its concentration increases. Conversely, the regeneration of RuBP from TP requires ATP. Since no ATP is available, RuBP cannot be regenerated, but any remaining RuBP continues to react with the available \(\text{CO}_2\) to form GP, causing the RuBP concentration to decrease.

Marking scheme

Award 1 mark for the correct option A. Reject options B, C, and D because the absence of light prevents GP reduction (causing GP to rise) and prevents RuBP regeneration (causing RuBP to fall).
Question 7 · multiple_choice
1 marks
In a species of bird, individuals with intermediate wing lengths survive better during storms than individuals with very short or very long wings. What type of selection is operating on wing length in this bird population, and what is its effect on the variance of wing length over generations?
  1. A.Type of selection: Stabilizing; Effect on variance: Decreases
  2. B.Type of selection: Directional; Effect on variance: Decreases
  3. C.Type of selection: Stabilizing; Effect on variance: Increases
  4. D.Type of selection: Disruptive; Effect on variance: Increases
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Worked solution

Stabilizing selection occurs when environmental conditions favor intermediate phenotypes over extreme ones. Because both extreme phenotypes (very short and very long wings) are selected against, the extreme values are reduced in frequency in the gene pool, leading to a narrower distribution and a decrease in the phenotypic variance of wing length over generations.

Marking scheme

Award 1 mark for the correct option A. Reject options B, C, and D because they do not correctly identify both stabilizing selection and the decreasing effect on phenotypic variance.
Question 8 · multiple_choice
1 marks
The polymerase chain reaction (PCR) is used to amplify DNA. The mixture is cycled through three temperatures: \(95\,^\circ\text{C}\), \(55\,^\circ\text{C}\), and \(72\,^\circ\text{C}\). Which row correctly describes the purpose of each temperature in a cycle of PCR?
  1. A.\(95\,^\circ\text{C}\): Breaks hydrogen bonds to separate DNA strands; \(55\,^\circ\text{C}\): Allows primers to anneal to complementary sequences; \(72\,^\circ\text{C}\): Optimum temperature for Taq polymerase to extend the new DNA strands.
  2. B.\(95\,^\circ\text{C}\): Denatures Taq polymerase to initiate the reaction; \(55\,^\circ\text{C}\): Allows complementary base pairing of free nucleotides; \(72\,^\circ\text{C}\): Breaks hydrogen bonds between template strands.
  3. C.\(95\,^\circ\text{C}\): Breaks phosphodiester bonds to separate DNA strands; \(55\,^\circ\text{C}\): Allows primers to anneal to complementary sequences; \(72\,^\circ\text{C}\): Denatures the newly formed double-stranded DNA.
  4. D.\(95\,^\circ\text{C}\): Breaks hydrogen bonds to separate DNA strands; \(55\,^\circ\text{C}\): Allows Taq polymerase to bind to the template DNA; \(72\,^\circ\text{C}\): Allows primers to anneal to complementary sequences.
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Worked solution

At \(95\,^\circ\text{C}\), the high thermal energy denatures the double-stranded template DNA by breaking the hydrogen bonds between complementary base pairs to separate the strands. At \(55\,^\circ\text{C}\), the temperature is lowered to allow primers to bind (anneal) via complementary base pairing to the single-stranded template DNA. At \(72\,^\circ\text{C}\), the temperature is set to the optimum for the thermostable Taq polymerase to synthesize the complementary strands by adding free deoxynucleoside triphosphates (dNTPs) to extend the primers.

Marking scheme

Award 1 mark for the correct option A. Reject options B, C, and D because they contain errors regarding the biochemical mechanisms occurring at each temperature step.
Question 9 · multiple-choice
1 marks
A micrograph of a plant cell chloroplast has an image length of \(2.4\text{ cm}\). If the actual length of this chloroplast is \(6\\ \mu\text{m}\), what is the magnification of the micrograph?
  1. A.\(\times 400\)
  2. B.\(\times 4,000\)
  3. C.\(\times 40,000\)
  4. D.\(\times 400,000\)
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Worked solution

To calculate magnification, use the formula: \(M = \frac{I}{A}\). First, convert the image length and actual length to the same units. Image length = \(2.4\text{ cm} = 24\text{ mm} = 24,000\\ \mu\text{m}\). Actual length = \(6\\ \mu\text{m}\). Magnification = \(\frac{24,000\\ \mu\text{m}}{6\\ \mu\text{m}} = \times 4,000\).

Marking scheme

1 mark for correct calculation of magnification with consistent unit conversion.
Question 10 · multiple-choice
1 marks
A group of plant cells with a water potential of \(-500\text{ kPa}\) is placed in four different sucrose solutions, W, X, Y, and Z. The water potentials of the solutions are: W = \(-200\text{ kPa}\), X = \(-500\text{ kPa}\), Y = \(-800\text{ kPa}\), and Z = \(-1100\text{ kPa}\). In which of these solutions will the cell protoplasts pull away from the cell walls?
  1. A.W only
  2. B.W and X only
  3. C.Y and Z only
  4. D.Z only
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Worked solution

Plasmolysis occurs when water leaves the cell by osmosis down a water potential gradient (from a higher/less negative water potential to a lower/more negative water potential). Since the plant cells have a water potential of \(-500\text{ kPa}\), water will move out of the cells when they are placed in solutions with a more negative water potential. These solutions are Y (\(-800\text{ kPa}\\)) and Z (\(-1100\text{ kPa}\\)). This loss of water causes the protoplast to shrink and pull away from the cell wall.

Marking scheme

1 mark for identifying that plasmolysis occurs in solutions with a lower (more negative) water potential than the cell (Y and Z only).
Question 11 · multiple-choice
1 marks
Which row correctly describes the types of glycosidic bonds present in cellulose, amylose, and glycogen?
  1. A.cellulose: \(\alpha\)-1,4 only; amylose: \(\beta\)-1,4 only; glycogen: \(\alpha\)-1,4 and \(\alpha\)-1,6
  2. B.cellulose: \(\beta\)-1,4 only; amylose: \(\alpha\)-1,4 only; glycogen: \(\alpha\)-1,4 and \(\alpha\)-1,6
  3. C.cellulose: \(\beta\)-1,4 and \(\beta\)-1,6; amylose: \(\alpha\)-1,4 only; glycogen: \(\alpha\)-1,4 only
  4. D.cellulose: \(\beta\)-1,4 only; amylose: \(\alpha\)-1,4 and \(\alpha\)-1,6; glycogen: \(\alpha\)-1,4 only
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Worked solution

Cellulose is an unbranched polymer of \(\beta\)-glucose joined by \(\beta\)-1,4-glycosidic bonds. Amylose is an unbranched polymer of \(\alpha\)-glucose joined by \(\alpha\)-1,4-glycosidic bonds. Glycogen is a highly branched polymer of \(\alpha\)-glucose containing both \(\alpha\)-1,4-glycosidic bonds in the main chains and \(\alpha\)-1,6-glycosidic bonds at the branch points.

Marking scheme

1 mark for identifying the correct glycosidic bond types for cellulose, amylose, and glycogen.
Question 12 · multiple-choice
1 marks
Which features of companion cells are essential for their role in loading sucrose into phloem sieve tube elements? 1: Presence of numerous mitochondria to synthesize ATP. 2: Proton pumps in the cell surface membrane that actively transport hydrogen ions out of the cytoplasm. 3: Plasmodesmata connecting the cytoplasm of the companion cell to the sieve tube element.
  1. A.1 and 2 only
  2. B.2 and 3 only
  3. C.1 and 3 only
  4. D.1, 2 and 3
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Worked solution

All three statements describe essential features of companion cells during active loading. Mitochondria supply ATP (1) to power the active transport of hydrogen ions out of the cell via membrane proton pumps (2). This establishes an electrochemical gradient of hydrogen ions. The cotransport of hydrogen ions back into the companion cell brings sucrose along with them. This accumulated sucrose then diffuses into the sieve tube element via connecting plasmodesmata (3).

Marking scheme

1 mark for selecting the option that correctly identifies all three features as essential.
Question 13 · multiple-choice
1 marks
Which row correctly describes the physiological responses that occur in a human body when the water potential of the blood plasma is lower than normal?
  1. A.Osmoreceptors: swell; ADH secretion: decreases; Aquaporins: removed from luminal membrane; Urine concentration: decreases
  2. B.Osmoreceptors: shrink; ADH secretion: decreases; Aquaporins: inserted into luminal membrane; Urine concentration: decreases
  3. C.Osmoreceptors: shrink; ADH secretion: increases; Aquaporins: inserted into luminal membrane; Urine concentration: increases
  4. D.Osmoreceptors: swell; ADH secretion: increases; Aquaporins: removed from luminal membrane; Urine concentration: increases
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Worked solution

When the water potential of blood plasma decreases, water moves out of osmoreceptor cells in the hypothalamus by osmosis, causing them to shrink. This shrinkage stimulates the hypothalamus, which signals the posterior pituitary gland to increase the secretion of antidiuretic hormone (ADH). ADH binds to receptors on collecting duct cells, promoting the insertion of aquaporins into their luminal membranes. This increases water reabsorption back into the blood, producing highly concentrated urine.

Marking scheme

1 mark for correctly matching the state of osmoreceptors, ADH secretion, aquaporin movement, and urine concentration when blood water potential is low.
Question 14 · multiple-choice
1 marks
How many molecules of carbon dioxide, reduced NAD, and reduced FAD are produced in the Krebs cycle during the complete aerobic respiration of one molecule of glucose?
  1. A.Carbon dioxide: 2, Reduced NAD: 3, Reduced FAD: 1
  2. B.Carbon dioxide: 4, Reduced NAD: 6, Reduced FAD: 2
  3. C.Carbon dioxide: 6, Reduced NAD: 10, Reduced FAD: 2
  4. D.Carbon dioxide: 4, Reduced NAD: 8, Reduced FAD: 2
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Worked solution

One molecule of glucose produces two molecules of pyruvate via glycolysis, which are converted into two molecules of acetyl-CoA in the link reaction. Consequently, the Krebs cycle turns twice per glucose molecule. Each single turn of the Krebs cycle produces 2 molecules of carbon dioxide (CO2), 3 molecules of reduced NAD, and 1 molecule of reduced FAD. Thus, for one glucose molecule, the products are 4 CO2, 6 reduced NAD, and 2 reduced FAD.

Marking scheme

1 mark for correctly identifying the total Krebs cycle outputs per molecule of glucose (representing two complete turns of the cycle).
Question 15 · multiple-choice
1 marks
What is the correct sequence of temperatures and their primary purposes during one cycle of the polymerase chain reaction (PCR)?
  1. A.55 °C to break hydrogen bonds between complementary DNA strands; 95 °C to allow primers to bind; 72 °C for optimum Taq polymerase activity.
  2. B.95 °C to break covalent phosphodiester bonds; 55 °C to allow DNA polymerase to bind; 72 °C for optimum primer extension.
  3. C.72 °C to break hydrogen bonds between complementary DNA strands; 55 °C to allow primers to bind; 95 °C for optimum Taq polymerase activity.
  4. D.95 °C to break hydrogen bonds between complementary DNA strands; 55 °C to allow primers to bind; 72 °C for optimum Taq polymerase activity.
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Worked solution

During a PCR cycle: 1. Heat is applied at approx. 95 °C to break the hydrogen bonds between complementary base pairs, separating the double-stranded DNA (denaturation). 2. The temperature is lowered to approx. 55 °C to allow hydrogen bonds to form between primers and single-stranded templates (annealing). 3. The temperature is raised to approx. 72 °C, which is the optimum temperature for Taq DNA polymerase to synthesize a new complementary DNA strand (extension).

Marking scheme

1 mark for matching all three PCR steps (denaturation, annealing, and extension) to their correct temperatures and processes.
Question 16 · multiple-choice
1 marks
Which row correctly classifies the type of immunity acquired in each of the given scenarios?
  1. A.Injection of tetanus toxoid: active natural; Injection of anti-tetanus antibodies: passive artificial; Transfer of maternal antibodies via placenta: passive natural; Recovery from a measles infection: active artificial
  2. B.Injection of tetanus toxoid: passive artificial; Injection of anti-tetanus antibodies: active artificial; Transfer of maternal antibodies via placenta: active natural; Recovery from a measles infection: passive natural
  3. C.Injection of tetanus toxoid: active artificial; Injection of anti-tetanus antibodies: passive artificial; Transfer of maternal antibodies via placenta: passive natural; Recovery from a measles infection: active natural
  4. D.Injection of tetanus toxoid: active artificial; Injection of anti-tetanus antibodies: passive natural; Transfer of maternal antibodies via placenta: passive artificial; Recovery from a measles infection: active natural
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Worked solution

- Injection of tetanus toxoid (antigen) is active artificial immunity, as the host's body is artificially stimulated to make its own antibodies and memory cells. - Injection of anti-tetanus antibodies (antitoxin) is passive artificial immunity, as pre-formed antibodies are introduced artificially. - Transfer of maternal antibodies via the placenta is passive natural immunity, as pre-formed antibodies are transferred naturally. - Recovery from a measles infection is active natural immunity, as natural exposure to the pathogen induces the host's own immune response.

Marking scheme

1 mark for correctly classifying active artificial, passive artificial, passive natural, and active natural immunity.
Question 17 · multiple_choice
1 marks
An electron micrograph of a eukaryotic cell shows cross-sections of a cilium and a centriole. Which row in the table correctly describes the arrangement of microtubules in these two structures?
  1. A.Cilium: '9 + 2' arrangement of microtubule doublets; Centriole: '9 + 0' arrangement of microtubule triplets
  2. B.Cilium: '9 + 0' arrangement of microtubule triplets; Centriole: '9 + 2' arrangement of microtubule doublets
  3. C.Cilium: '9 + 2' arrangement of microtubule triplets; Centriole: '9 + 0' arrangement of microtubule doublets
  4. D.Cilium: '9 + 0' arrangement of microtubule doublets; Centriole: '9 + 2' arrangement of microtubule triplets
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Worked solution

Cilia have a characteristic '9 + 2' arrangement of microtubules consisting of nine outer doublets surrounding two central single microtubules. Centrioles, which organize spindle fibres during cell division, consist of nine triplets of microtubules arranged in a ring with no central microtubules ('9 + 0' arrangement).

Marking scheme

A is correct (1 mark). B, C, and D are incorrect arrangements.
Question 18 · multiple_choice
1 marks
A plant cell with a solute potential (\Psi_s) of -0.8 MPa and a pressure potential (\Psi_p) of +0.2 MPa is placed in a sucrose solution with a water potential (\Psi) of -0.4 MPa. Which statement correctly describes the net movement of water and the subsequent change in the cell?
  1. A.Water enters the cell by osmosis down a water potential gradient, and the pressure potential of the cell increases.
  2. B.Water leaves the cell by osmosis down a water potential gradient, and the protoplast pulls away from the cell wall.
  3. C.Water enters the cell by osmosis, causing the solute potential to become significantly more negative.
  4. D.There is no net movement of water because the water potential of the cell is already equal to that of the solution.
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Worked solution

The initial water potential of the cell (\Psi_w) is calculated as \Psi_w = \Psi_s + \Psi_p = -0.8\text{ MPa} + 0.2\text{ MPa} = -0.6\text{ MPa}. Since the surrounding solution has a higher (less negative) water potential of -0.4 MPa, there will be a net movement of water into the cell by osmosis. As water enters, the protoplast swells and exerts greater force against the cell wall, increasing the cell's pressure potential (\Psi_p) until equilibrium is reached.

Marking scheme

A is correct (1 mark). Candidates must calculate the initial cell water potential (-0.6 MPa) and determine that water moves from the higher water potential of the solution (-0.4 MPa) into the cell, which increases turgor (pressure potential).
Question 19 · multiple_choice
1 marks
Which statement correctly describes a structural or functional difference between amylose and cellulose?
  1. A.Amylose is a polymer of \alpha-glucose with 1,4-glycosidic bonds, whereas cellulose is a polymer of \beta-glucose with 1,4-glycosidic bonds in which alternate glucose monomers are rotated by 180 degrees.
  2. B.Amylose contains both 1,4- and 1,6-glycosidic bonds to form a branched structure, whereas cellulose contains only 1,4-glycosidic bonds to form unbranched chains.
  3. C.Amylose forms straight, unbranched chains held together by hydrogen bonds to form microfibrils, whereas cellulose forms helical structures held by intramolecular bonds.
  4. D.Amylose is a structural polysaccharide that reinforces plant cell walls, whereas cellulose is a metabolic storage carbohydrate found in chloroplasts.
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Worked solution

Amylose is an unbranched polymer made of \alpha-glucose molecules linked by 1,4-glycosidic bonds, coiled into a helix. Cellulose is a polymer of \beta-glucose where alternate monomers must be rotated 180 degrees to allow the 1,4-glycosidic bonds to form, resulting in a straight, unbranched chain. This straight structure allows hydrogen bonding between adjacent cellulose chains to form strong microfibrils.

Marking scheme

A is correct (1 mark). B is incorrect because amylose is unbranched (only 1,4-glycosidic bonds; amylopectin has 1,6-bonds). C and D describe the properties or locations of these molecules in reverse.
Question 20 · multiple_choice
1 marks
During the active loading of sucrose into companion cells, protons (\text{H}^+ ions) and sucrose molecules are transported. Which row correctly identifies the mechanisms and directions of transport for both protons and sucrose?
  1. A.Protons: active transport out of the companion cell; Sucrose: co-transport into the companion cell down the proton concentration gradient
  2. B.Protons: facilitated diffusion into the companion cell; Sucrose: active transport out of the companion cell against its concentration gradient
  3. C.Protons: active transport into the companion cell; Sucrose: co-transport out of the companion cell down the proton concentration gradient
  4. D.Protons: facilitated diffusion out of the companion cell; Sucrose: active transport into the companion cell against its concentration gradient
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Worked solution

Active loading of sucrose involves proton pumps actively transporting \text{H}^+ ions out of the companion cell cytoplasm into the apoplast (cell wall), establishing an electrochemical gradient. Protons then diffuse back into the companion cell down their concentration gradient via a co-transporter protein, which simultaneously transports sucrose molecules into the companion cell against their concentration gradient.

Marking scheme

A is correct (1 mark). Proton pumps actively transport protons out of the cell, and co-transporters bring them back in alongside sucrose.
Question 21 · multiple_choice
1 marks
An increase in the concentration of antidiuretic hormone (ADH) in the blood causes changes in the kidney nephrons. Which processes occur in the cells of the collecting duct in response to ADH binding to its cell-surface receptors? 1. Activation of adenylyl cyclase to produce cyclic AMP (cAMP) 2. Fusion of vesicles containing aquaporins with the luminal membrane 3. Facilitated diffusion of water molecules through aquaporins out of the tubule fluid 4. Active transport of sodium ions from the filtrate into the cytoplasm
  1. A.1, 2 and 3 only
  2. B.1 and 2 only
  3. C.2, 3 and 4 only
  4. D.1, 3 and 4 only
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Worked solution

When ADH binds to its receptors on the cell surface membrane of collecting duct cells, it activates adenylyl cyclase, which converts ATP to cyclic AMP (cAMP, a second messenger). This triggers a kinase cascade resulting in vesicles containing aquaporins fusing with the luminal (apical) membrane. Water then moves out of the tubule fluid down its water potential gradient via facilitated diffusion through these aquaporins. Sodium ions are actively transported out of the filtrate primarily in the loop of Henle, not in the collecting duct cells as a direct response to ADH.

Marking scheme

A is correct (1 mark). Statements 1, 2, and 3 are correct descriptions of the cell-signalling pathway of ADH in the collecting duct. Statement 4 is incorrect.
Question 22 · multiple_choice
1 marks
Which row correctly matches the infectious disease with its type of pathogen and its primary method of transmission?
  1. A.Disease: Cholera | Pathogen: Bacterium | Primary transmission: Ingestion of contaminated water or food
  2. B.Disease: Malaria | Pathogen: Virus | Primary transmission: Injection of saliva during a bite by an infected female mosquito
  3. C.Disease: Tuberculosis | Pathogen: Virus | Primary transmission: Inhalation of airborne droplets from coughing or sneezing
  4. D.Disease: Cholera | Pathogen: Protoctist | Primary transmission: Contact with contaminated faeces on hands
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Worked solution

Cholera is caused by the bacterium Vibrio cholerae and is transmitted via the faecal-oral route through contaminated food or water. Malaria is caused by a protoctist (Plasmodium), not a virus. Tuberculosis is caused by a bacterium (Mycobacterium tuberculosis), not a virus.

Marking scheme

A is correct (1 mark). Row A lists the correct pathogen type and transmission route for Cholera. Other rows contain factual errors about the pathogen classification.
Question 23 · multiple_choice
1 marks
What is the primary role of memory B-lymphocytes during a secondary immune response to a pathogen that has been encountered previously?
  1. A.They divide rapidly by mitosis to produce a clone of plasma cells that secrete large quantities of antibodies.
  2. B.They instantly secrete high concentrations of antibodies into the blood without the need for cell division.
  3. C.They act as antigen-presenting cells to engulf the pathogen and display its antigens to activate helper T-cells.
  4. D.They release cytokines to recruit and activate phagocytic neutrophils to the site of infection.
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Worked solution

During a secondary immune response, memory B-lymphocytes recognize the antigen, become activated, and divide rapidly by mitosis (clonal expansion). They differentiate into a large clone of antibody-secreting plasma cells and more memory cells. This produces a faster and much larger quantity of antibodies compared to the primary response.

Marking scheme

A is correct (1 mark). Memory B-lymphocytes must divide and differentiate into plasma cells to produce antibodies. Plasma cells, not memory cells, directly secrete antibodies (ruling out B). C is a primary function of macrophages/dendritic cells. D is a function of helper T-lymphocytes.
Question 24 · multiple_choice
1 marks
During the complete aerobic respiration of one molecule of glucose, how many molecules of reduced NAD (NADH) and reduced FAD (\text{FADH}_2) are produced combined in the link reaction and the Krebs cycle?
  1. A.8 reduced NAD and 2 reduced FAD
  2. B.6 reduced NAD and 2 reduced FAD
  3. C.10 reduced NAD and 2 reduced FAD
  4. D.2 reduced NAD and 0 reduced FAD
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Worked solution

For each glucose molecule, two molecules of pyruvate enter the link reaction. The link reaction produces 2 molecules of NADH (1 per pyruvate) and no \text{FADH}_2. The two acetyl-CoA molecules then enter the Krebs cycle, which produces 6 molecules of NADH (3 per acetyl-CoA) and 2 molecules of \text{FADH}_2 (1 per acetyl-CoA). Combining these gives: 2 + 6 = 8 NADH, and 0 + 2 = 2 \text{FADH}_2.

Marking scheme

A is correct (1 mark). 8 NADH and 2 FADH2 are produced across the link reaction and the Krebs cycle for one molecule of glucose.
Question 25 · multiple-choice
1 marks
Which feature of an organelle isolated from a plant leaf cell would confirm that it is a chloroplast and not a mitochondrion?
  1. A.The presence of a double outer membrane
  2. B.The presence of circular DNA
  3. C.The presence of 70S ribosomes
  4. D.The presence of thylakoid membranes
Show answer & marking scheme

Worked solution

Both chloroplasts and mitochondria are double-membraned organelles that contain circular DNA and 70S ribosomes due to their endosymbiotic origin. However, only chloroplasts contain internal thylakoid membranes, which are organized into grana and contain photosynthetic pigments.

Marking scheme

1 mark for selecting option D. Options A, B, and C are incorrect because these features are present in both mitochondria and chloroplasts.
Question 26 · multiple-choice
1 marks
The fluid mosaic model describes the structure of cell membranes. Which change to the membrane components would increase the fluidity of a mammalian cell surface membrane at low temperatures?
  1. A.An increase in the proportion of phospholipids with saturated fatty acid tails
  2. B.An increase in the proportion of phospholipids with unsaturated fatty acid tails
  3. C.A decrease in the cholesterol content of the membrane
  4. D.An increase in the concentration of extrinsic membrane proteins
Show answer & marking scheme

Worked solution

At low temperatures, phospholipids tend to pack closely together, decreasing membrane fluidity. Phospholipids with unsaturated fatty acid tails contain double bonds that introduce kinks, preventing close packing and maintaining fluidity. Decreasing cholesterol would reduce the membrane's ability to resist freezing, thereby decreasing fluidity at low temperatures.

Marking scheme

1 mark for identifying that increasing the proportion of unsaturated fatty acids increases fluidity (B). Reject A (saturated lipids pack tightly and decrease fluidity), C (reducing cholesterol decreases fluidity at low temps), and D (extrinsic proteins do not regulate primary lipid bilayer fluidity in this way).
Question 27 · multiple-choice
1 marks
Which level of protein structure is maintained by hydrophobic interactions, hydrogen bonds, ionic bonds, and disulfide bonds between the R-groups of a single polypeptide chain?
  1. A.Primary structure
  2. B.Secondary structure
  3. C.Tertiary structure
  4. D.Quaternary structure
Show answer & marking scheme

Worked solution

The tertiary structure of a protein is the folding of a single polypeptide chain into a complex three-dimensional shape, stabilized by interactions between the R-groups of the amino acids. Secondary structure is stabilized by hydrogen bonds within the polypeptide backbone, not the R-groups. Quaternary structure involves interactions between multiple polypeptide chains.

Marking scheme

1 mark for selecting tertiary structure (C). Reject A (primary structure involves only peptide bonds), B (secondary structure involves backbone hydrogen bonds), and D (quaternary structure involves multiple chains).
Question 28 · multiple-choice
1 marks
Which statements about the transport of sucrose in the phloem of a plant are correct? 1. Sucrose is loaded into companion cells by active transport of hydrogen ions out of the cell followed by co-transport of sucrose with hydrogen ions back in. 2. Hydrostatic pressure increases at the source due to water entering the sieve tube elements by osmosis. 3. Sucrose moves through the sieve tube elements by diffusion down a concentration gradient.
  1. A.1, 2 and 3
  2. B.1 and 2 only
  3. C.2 and 3 only
  4. D.1 and 3 only
Show answer & marking scheme

Worked solution

Statement 1 is correct: proton pumps actively pump hydrogen ions out, creating a gradient that drives the co-transport of sucrose back into companion cells. Statement 2 is correct: high solute concentration at the source decreases water potential, causing water to enter by osmosis, raising hydrostatic pressure. Statement 3 is incorrect: sucrose moves through the sieve tubes by mass flow driven by a hydrostatic pressure gradient, not by simple diffusion.

Marking scheme

1 mark for identifying statements 1 and 2 as correct (B). Reject options containing statement 3 (A, C, D).
Question 29 · multiple-choice
1 marks
The concentration of glucose in the glomerular filtrate of a healthy human is 1.0 g dm^-3, while its concentration in the final urine is 0.0 g dm^-3. Which process explains this change in glucose concentration?
  1. A.Active transport in the proximal convoluted tubule
  2. B.Facilitated diffusion in the collecting duct
  3. C.Ultrafiltration in the Bowman’s capsule
  4. D.Active transport in the loop of Henle
Show answer & marking scheme

Worked solution

In a healthy human, all of the glucose filtered into the Bowman's capsule is reabsorbed back into the blood from the glomerular filtrate in the proximal convoluted tubule (PCT). This selective reabsorption is an active process involving secondary active transport (co-transport with sodium ions across the apical membrane).

Marking scheme

1 mark for active transport in the proximal convoluted tubule (A). Reject B, C, and D as glucose is not reabsorbed in these regions.
Question 30 · multiple-choice
1 marks
Which of the following correctly pairs an infectious disease with its causative pathogen type and its primary method of transmission?
  1. A.Cholera | Bacterium | Vector-borne (insect bite)
  2. B.Malaria | Virus | Vector-borne (mosquito bite)
  3. C.Tuberculosis | Bacterium | Airborne droplets
  4. D.HIV/AIDS | Protoctist | Direct contact with bodily fluids
Show answer & marking scheme

Worked solution

Tuberculosis is caused by the bacterium Mycobacterium tuberculosis (or M. bovis) and is transmitted via airborne droplets. Cholera is a water-borne bacterial disease. Malaria is caused by a protoctist (Plasmodium) via a vector. HIV is caused by a virus (HIV) via direct contact with bodily fluids.

Marking scheme

1 mark for identifying the correct pathogen type and transmission for tuberculosis (C). Reject A (cholera is water-borne), B (malaria is caused by a protoctist, not a virus), and D (HIV is a virus, not a protoctist).
Question 31 · multiple-choice
1 marks
In the production of monoclonal antibodies using hybridoma technology, which cells are fused together, and what is the function of the fusing agent?
  1. A.B-lymphocytes and T-lymphocytes are fused using detergent to make the membranes permeable to antibodies.
  2. B.Plasma cells and myeloma cells are fused using polyethylene glycol (PEG) to allow the cell membranes to merge.
  3. C.T-lymphocytes and myeloma cells are fused using calcium ions to stimulate rapid mitotic cell division.
  4. D.Macrophages and plasma cells are fused using polyethylene glycol (PEG) to increase antibody secretion rate.
Show answer & marking scheme

Worked solution

To produce monoclonal antibodies, antibody-producing plasma cells are isolated from an immunized animal and fused with cancer cells (myeloma cells). This fusion is assisted by polyethylene glycol (PEG), which destabilizes cell membranes and allows them to fuse to form hybridoma cells.

Marking scheme

1 mark for selecting option B. Reject A, C, and D because they contain incorrect cell types or incorrect fusing mechanisms/agents.
Question 32 · multiple-choice
1 marks
For every molecule of glucose that undergoes aerobic respiration, what are the total net yields of reduced NAD and reduced FAD produced specifically during the Krebs cycle?
  1. A.2 reduced NAD and 2 reduced FAD
  2. B.6 reduced NAD and 2 reduced FAD
  3. C.8 reduced NAD and 2 reduced FAD
  4. D.10 reduced NAD and 2 reduced FAD
Show answer & marking scheme

Worked solution

One molecule of glucose yields two molecules of pyruvate via glycolysis, which are converted into two molecules of acetyl-CoA in the link reaction. Each acetyl-CoA enters the Krebs cycle once. One turn of the Krebs cycle produces 3 reduced NAD and 1 reduced FAD. Therefore, for two turns (corresponding to one glucose molecule), the Krebs cycle produces 6 reduced NAD and 2 reduced FAD.

Marking scheme

1 mark for the correct yields specifically for the Krebs cycle (B). Reject A (only 1 turn), C (includes link reaction or glycolysis), and D (total reduced NAD across all aerobic respiration stages combined).
Question 33 · multiple-choice
1 marks
A student uses a light microscope to measure the size of a stomatal pore. At a magnification of \(\times 400\), the image of the stomatal pore is \(8.0\text{ mm}\) wide. What is the actual width of the stomatal pore?
  1. A.0.02 \(\mu\)m
  2. B.2.0 \(\mu\)m
  3. C.20 \(\mu\)m
  4. D.200 \(\mu\)m
Show answer & marking scheme

Worked solution

The formula for magnification is \(M = I / A\), where \(M\) is magnification, \(I\) is image size, and \(A\) is actual size. Rearranging the formula: \(A = I / M\). Given \(I = 8.0\text{ mm}\) and \(M = 400\), we get \(A = 8.0\text{ mm} / 400 = 0.02\text{ mm}\). Converting millimetres to micrometres: \(0.02\text{ mm} \times 1000 = 20\text{ \mu m}\). Thus, the actual width of the stomatal pore is \(20\text{ \mu m}\).

Marking scheme

1 mark for the correct calculation of 20 \(\mu\)m (C).
Question 34 · multiple-choice
1 marks
A plant cell is placed in a solution with a higher solute concentration than the cytoplasm of the cell. Which row correctly describes the changes in the water potential of the cytoplasm and the state of the cell?
  1. A.cytoplasm water potential decreases | cell becomes plasmolysed
  2. B.cytoplasm water potential decreases | cell becomes turgid
  3. C.cytoplasm water potential increases | cell becomes plasmolysed
  4. D.cytoplasm water potential increases | cell becomes turgid
Show answer & marking scheme

Worked solution

The external solution has a higher solute concentration, meaning it has a lower water potential than the cytoplasm. Water leaves the cell by osmosis down a water potential gradient. As water leaves, the cytoplasm becomes more concentrated, so its water potential decreases. The protoplast shrinks and pulls away from the cell wall, causing the cell to become plasmolysed.

Marking scheme

1 mark for identifying that the cytoplasm water potential decreases and the cell becomes plasmolysed.
Question 35 · multiple-choice
1 marks
Which statement correctly describes a difference between amylose and cellulose?
  1. A.Amylose contains \(\beta\)-1,4-glycosidic bonds, whereas cellulose contains \(\alpha\)-1,4-glycosidic bonds.
  2. B.Amylose has a branched structure, whereas cellulose is a straight, unbranched chain.
  3. C.Amylose is formed by the condensation of \(\alpha\)-glucose, whereas cellulose is formed by the condensation of \(\beta\)-glucose.
  4. D.Amylose forms hydrogen bonds between parallel chains, whereas cellulose does not.
Show answer & marking scheme

Worked solution

Amylose is a polymer of \(\alpha\)-glucose joined by \(\alpha\)-1,4-glycosidic bonds, forming a coiled, unbranched structure. Cellulose is a polymer of \(\beta\)-glucose joined by \(\beta\)-1,4-glycosidic bonds, forming straight, unbranched chains. Parallel cellulose chains form hydrogen bonds with one another to produce microfibrils, which amylose does not do.

Marking scheme

1 mark for identifying that amylose is formed from \(\alpha\)-glucose, while cellulose is formed from \(\beta\)-glucose.
Question 36 · multiple-choice
1 marks
In a plant, what drives the upward movement of water through xylem vessel elements from the roots to the leaves?
  1. A.Active transport of water molecules into the xylem vessels at the roots
  2. B.The hydrostatic pressure gradient created by the active loading of sucrose into sieve tubes
  3. C.The tension created by the evaporation of water from the cell walls of mesophyll cells
  4. D.The active pumping of hydrogen ions out of the xylem vessels by companion cells
Show answer & marking scheme

Worked solution

The evaporation of water from the cell walls of mesophyll cells into the air spaces (transpiration) reduces the hydrostatic pressure at the top of the xylem. This creates a tension (negative pressure) that pulls water up through the xylem vessels, aided by the cohesive properties of water molecules.

Marking scheme

1 mark for identifying that tension created by the evaporation of water drives the water upward.
Question 37 · multiple-choice
1 marks
During aerobic respiration, how many molecules of carbon dioxide are produced from the complete oxidation of one molecule of pyruvate?
  1. A.1
  2. B.2
  3. C.3
  4. D.6
Show answer & marking scheme

Worked solution

The complete oxidation of a single 3-carbon pyruvate molecule produces 3 molecules of carbon dioxide. The first molecule of carbon dioxide is released during the link reaction when pyruvate is decarboxylated to form acetyl-coenzyme A (a 2-carbon compound). The remaining two molecules of carbon dioxide are released during the subsequent Krebs cycle reactions.

Marking scheme

1 mark for calculating that 3 molecules of carbon dioxide are released per pyruvate molecule.
Question 38 · multiple-choice
1 marks
Which of the following occurs in the kidneys of a human when the water potential of the blood decreases below the set point?
  1. A.Osmoreceptors in the hypothalamus swell and stop secreting ADH.
  2. B.The cells of the collecting duct become less permeable to water due to the internalization of aquaporins.
  3. C.ADH binds to receptors on the cell surface membrane of collecting duct cells, initiating a second messenger cascade.
  4. D.More water is excreted in the urine, making the urine less concentrated (dilute).
Show answer & marking scheme

Worked solution

A decrease in the water potential of the blood is detected by osmoreceptors in the hypothalamus, which shrink and stimulate the release of ADH from the posterior pituitary. ADH travels in the bloodstream to the kidneys, where it binds to specific cell surface receptors on collecting duct cells. This binding initiates a second messenger cascade (via cyclic AMP) that leads to the exocytosis of vesicles containing aquaporins into the luminal membranes, increasing water reabsorption.

Marking scheme

1 mark for identifying that ADH binding initiates a second messenger cascade in the collecting duct cells.
Question 39 · multiple-choice
1 marks
A population of snails exhibits a wide range of shell colorations, from very light yellow to very dark brown. A new predator is introduced to the habitat. This predator hunts by sight and is highly efficient at detecting and consuming snails with intermediate (medium-brown) shell colors against the background of light sand and dark rocks. Which type of natural selection is most likely to occur in this snail population over time?
  1. A.Directional selection towards lighter shell colors
  2. B.Directional selection towards darker shell colors
  3. C.Disruptive selection
  4. D.Stabilising selection
Show answer & marking scheme

Worked solution

Disruptive selection occurs when environmental factors select against the intermediate phenotype and favor both extreme phenotypes. In this case, the predator preferentially preys on snails with intermediate medium-brown shell colors, while snails with extreme phenotypes (very light yellow, which blend into the sand, and very dark brown, which blend into the rocks) survive and reproduce at a higher rate.

Marking scheme

1 mark for identifying disruptive selection.
Question 40 · multiple-choice
1 marks
In genetic engineering, which enzyme is correctly matched with its role?
  1. A.DNA ligase | Synthesises a complementary DNA strand from a single-stranded mRNA template.
  2. B.Restriction endonuclease | Cleaves phosphodiester bonds at specific nucleotide sequences.
  3. C.Reverse transcriptase | Joins the sugar-phosphate backbones of two DNA fragments.
  4. D.Taq polymerase | Unwinds double-stranded DNA during the denaturation step of PCR.
Show answer & marking scheme

Worked solution

Restriction endonucleases cleave phosphodiester bonds at specific target nucleotide sequences. DNA ligase is used to join sugar-phosphate backbones (not synthesise cDNA). Reverse transcriptase synthesises cDNA from an mRNA template. Taq polymerase synthesises DNA strands during PCR, while denaturation is achieved by heating.

Marking scheme

1 mark for selecting the correct enzyme and role combination.

Paper 2 (AS Structured Questions)

Answer all structured questions in the spaces provided.
6 Question · 60 marks
Question 1 · Structured Short Answer
10 marks
The epithelial lining of the trachea contains goblet cells and ciliated epithelial cells. Goblet cells synthesize and secrete mucin, which is a large glycoprotein.

(a) Describe the roles of the rough endoplasmic reticulum and the Golgi body in the synthesis and processing of mucin. [4]

(b) Explain how the structure of a ciliated epithelial cell is adapted to its function of clearing mucus from the airways. [3]

(c) Suggest why goblet cells contain a large number of mitochondria. [3]
Show answer & marking scheme

Worked solution

(a)
1. Ribosomes on the rough endoplasmic reticulum (rER) translate mRNA to synthesize the polypeptide chains of mucin.
2. The newly synthesized polypeptides enter the lumen of the rER, where they undergo folding into their initial tertiary structure.
3. The proteins are packaged into transport vesicles that bud off the rER and fuse with the Golgi body.
4. In the Golgi body, the proteins are modified by the addition of carbohydrate chains (glycosylation) to form glycoproteins (mucin).
5. The Golgi body packages the mature mucin into secretory vesicles.

(b)
1. Ciliated epithelial cells possess numerous tiny hair-like projections called cilia on their apical membrane.
2. The cilia beat in a coordinated, synchronized, rhythmic wave-like motion.
3. This beating action sweeps the mucus (containing trapped dust and pathogens) upwards and away from the lungs towards the throat to be swallowed or coughed out.

(c)
1. Mitochondria are the sites of aerobic respiration, which produces ATP.
2. ATP is required to provide energy for anabolic reactions, such as the synthesis of proteins (mucin polypeptide chains) on ribosomes.
3. ATP is required for active transport and the movement of secretory vesicles along the cytoskeleton (microtubules) to the cell surface membrane for exocytosis.

Marking scheme

(a) [Max 4 marks]
- Ribosomes on rER translate mRNA to synthesize proteins/polypeptides. [1]
- Polypeptides enter the rER lumen / undergo folding. [1]
- Transport vesicles move proteins from rER to the Golgi body. [1]
- In the Golgi, carbohydrates are added / glycosylation occurs (to form glycoprotein). [1]
- Golgi packages mucin into secretory vesicles. [1]

(b) [Max 3 marks]
- Presence of cilia on the apical surface of the cell. [1]
- Cilia beat in a synchronized / coordinated / wave-like fashion. [1]
- Sweeps / moves mucus containing pathogens/dust upwards / away from lungs / towards throat. [1]

(c) [Max 3 marks]
- Mitochondria perform aerobic respiration to produce ATP. [1]
- ATP is required for protein / mucin synthesis / translation. [1]
- ATP is required for vesicle transport / movement of vesicles along microtubules / exocytosis. [1]
Question 2 · Structured Short Answer
10 marks
The movement of substances across cell membranes is essential for cell function. In the small intestine, glucose is absorbed into epithelial cells against its concentration gradient.

(a) Explain why glucose cannot pass directly through the phospholipid bilayer by simple diffusion. [2]

(b) Describe how the co-transport of sodium ions (\(\text{Na}^+\)) and glucose takes place across the apical membrane of these epithelial cells. [5]

(c) State how the active transport of sodium ions across the basolateral membrane into the blood maintains this absorption process. [3]
Show answer & marking scheme

Worked solution

(a)
1. Glucose is a relatively large, polar (hydrophilic) molecule.
2. The non-polar hydrophobic core of the phospholipid bilayer (formed by fatty acid tails) repels polar substances, preventing them from diffusing through the lipid matrix.

(b)
1. Co-transporter (symport) proteins are present in the apical membrane of the epithelial cells.
2. Sodium ions (\(\text{Na}^+\)) and glucose molecules bind to specific binding sites on the same co-transporter protein.
3. There is a higher concentration of \(\text{Na}^+\) in the lumen of the intestine than inside the epithelial cell.
4. \(\text{Na}^+\) moves down its electrochemical/concentration gradient into the cell through the co-transporter.
5. The downhill movement of \(\text{Na}^+\) provides the energy to drive glucose into the cell against its concentration gradient.
6. This causes a conformational (shape) change in the co-transporter protein, releasing both \(\text{Na}^+\) and glucose into the cytoplasm.

(c)
1. Active transport of \(\text{Na}^+\) occurs via the sodium-potassium pump (\(\text{Na}^+/\text{K}^+\) ATPase) on the basolateral membrane.
2. This pump actively moves \(\text{Na}^+\) out of the cytoplasm and into the blood using ATP.
3. This keeps the intracellular concentration of \(\text{Na}^+\) low, maintaining the steep concentration gradient for \(\text{Na}^+\) between the lumen and the cell, allowing continuous co-transport of glucose.

Marking scheme

(a) [Max 2 marks]
- Glucose is a polar / hydrophilic / large molecule. [1]
- It cannot pass through the non-polar / hydrophobic core of the phospholipid bilayer / is repelled by fatty acid tails. [1]

(b) [Max 5 marks]
- Co-transporter / symport protein is located in the apical membrane. [1]
- Both \(\text{Na}^+\) and glucose bind to the co-transporter protein. [1]
- \(\text{Na}^+\) moves down its electrochemical / concentration gradient (from lumen into cell). [1]
- The entry of \(\text{Na}^+\) drives / provides energy for the movement of glucose against its concentration gradient. [1]
- Co-transporter protein undergoes a conformational / shape change. [1]
- Both substances are released into the cytoplasm. [1]

(c) [Max 3 marks]
- Sodium-potassium pump (on basolateral membrane) uses ATP to pump \(\text{Na}^+\) out of the cell. [1]
- Keeps intracellular concentration of \(\text{Na}^+\) low. [1]
- Maintains the concentration / electrochemical gradient of \(\text{Na}^+\) from lumen to cytoplasm (so co-transport continues). [1]
Question 3 · Structured Short Answer
10 marks
Collagen is the most abundant structural protein in the mammalian body, found in skin, tendons, and bones.

(a) Describe the primary and secondary structures of a single collagen polypeptide chain. [3]

(b) Explain how the tertiary and quaternary structures of collagen are adapted to its structural role. [4]

(c) Describe the structural differences between collagen and a globular protein such as hemoglobin. [3]
Show answer & marking scheme

Worked solution

(a)
1. The primary structure is a repeating sequence of three amino acids, typically Glycine-X-Y, where glycine is found at every third position (X is often proline and Y is often hydroxyproline).
2. The secondary structure of a single polypeptide chain is a tight, left-handed helix (collagen helix).
3. Unlike an alpha-helix, it is looser and stabilized by steric repulsion of proline rings rather than internal hydrogen bonds.

(b)
1. In the quaternary structure, three of these left-handed polypeptide chains twist around each other to form a tight, right-handed triple helix called tropocollagen.
2. Glycine is the smallest amino acid, containing only a single hydrogen atom in its R-group. Its small size allows it to fit into the tight center of the triple helix, enabling close packing.
3. Hydrogen bonds form between the peptide groups of adjacent chains, stabilizing the triple helix structure.
4. Many tropocollagen molecules lie parallel to one another and are joined by covalent cross-links to form fibrils, which associate into thick collagen fibers, providing extremely high tensile strength.

(c)
1. Collagen is a fibrous protein with an elongated, repeating structure, whereas hemoglobin is a globular protein folded into a compact, spherical shape.
2. Collagen is insoluble in water due to its large size and hydrophobic outer amino acids, whereas hemoglobin is soluble because hydrophilic R-groups are exposed on its outer surface.
3. Collagen does not contain a prosthetic group, whereas hemoglobin contains four iron-containing prosthetic heme groups that bind oxygen.
4. Collagen's quaternary structure consists of three helices wound together, whereas hemoglobin consists of four folded polypeptide chains (two alpha and two beta globins).

Marking scheme

(a) [Max 3 marks]
- Primary structure has a repeating sequence of three amino acids / Gly-X-Y. [1]
- Glycine is located at every third position. [1]
- Secondary structure is a tight, left-handed helix. [1]
- Accept: No hydrogen bonds within the single chain (stabilized by proline repulsion). [1]

(b) [Max 4 marks]
- Three polypeptide chains wind together to form a triple helix / tropocollagen. [1]
- Glycine has a very small R-group (hydrogen atom) allowing tight packing in the center of the helix. [1]
- Hydrogen bonds form between the chains to stabilize the triple helix. [1]
- Covalent cross-links form between adjacent tropocollagen molecules. [1]
- Staggered arrangement of tropocollagen molecules prevents weak spots / forms strong fibrils/fibers. [1]

(c) [Max 3 marks - must be comparative differences]
- Collagen is fibrous / elongated whereas hemoglobin is globular / spherical. [1]
- Collagen is insoluble in water whereas hemoglobin is soluble. [1]
- Collagen has no prosthetic group whereas hemoglobin contains four prosthetic heme groups. [1]
- Collagen has three polypeptide chains in a triple helix whereas hemoglobin has four polypeptide chains (two alpha, two beta). [1]
Question 4 · Structured Short Answer
10 marks
Water and assimilates are transported through specialized vascular tissues in plants.

(a) Explain how companion cells load sucrose into phloem sieve tube elements. [4]

(b) Distinguish between the symplastic pathway and the apoplastic pathway of water transport through the root cortex. [3]

(c) Explain how the structure of xylem vessels is adapted to prevent the collapse of the vessel under tension. [3]
Show answer & marking scheme

Worked solution

(a)
1. Proton pumps (H+-ATPases) in the plasma membrane of companion cells actively pump hydrogen ions (protons) out of the companion cell cytoplasm into the cell wall/apoplast.
2. This process requires ATP as an energy source.
3. This creates a high concentration of protons outside the cell, establishing an electrochemical / proton gradient.
4. Protons diffuse back into the companion cell down their gradient through a specific co-transporter protein.
5. Sucrose molecules are transported into the companion cell simultaneously against their concentration gradient, carried along with the protons via the co-transporter.
6. Sucrose then diffuses into the phloem sieve tube element from the companion cell through connecting plasmodesmata.

(b)
1. In the symplastic pathway, water moves through the living cytoplasm and vacuoles of the cells.
2. Water passes from cell to cell via plasmodesmata (cytoplasmic connections through cell walls).
3. In the apoplastic pathway, water moves through the non-living cell walls and intercellular spaces.
4. Water moves by cohesion and tension through the apoplast until it reaches the endodermis, where the Casparian strip blocks it, forcing it into the symplast.

(c)
1. Xylem vessel walls are impregnated with a hard, waterproof substance called lignin.
2. Lignin provides immense mechanical strength and rigidity to the cellulose cell walls.
3. This prevents the xylem vessels from collapsing inwards when water is pulled up the stem under extremely high negative pressure (tension) created by transpiration.

Marking scheme

(a) [Max 4 marks]
- Protons / hydrogen ions (\(\text{H}^+\)) are actively pumped out of companion cells into the apoplast / cell wall. [1]
- This process requires ATP / active transport. [1]
- Creates a proton / electrochemical gradient (higher concentration outside). [1]
- Protons diffuse back into the companion cell down their gradient via a co-transporter / carrier protein. [1]
- Sucrose is co-transported / carried alongside the protons against its concentration gradient. [1]
- Sucrose diffuses from the companion cell to the sieve tube element via plasmodesmata. [1]

(b) [Max 3 marks]
- Symplast pathway involves water moving through the cytoplasm / vacuole, whereas apoplast involves movement through cell walls / intercellular spaces. [1]
- In symplast, water passes between cells via plasmodesmata. [1]
- Apoplast pathway is non-living, whereas symplast pathway is living. [1]
- Apoplast pathway is blocked by the Casparian strip (suberin) at the endodermis, forcing water into the symplast. [1]

(c) [Max 3 marks]
- Xylem walls contain lignin / are lignified. [1]
- Lignin is very strong / rigid. [1]
- Prevents the vessel walls from collapsing inward. [1]
- Under the influence of tension / negative pressure / suction created by transpiration. [1]
Question 5 · Structured Short Answer
10 marks
Vaccination is an effective method of controlling the spread of infectious diseases such as measles.

(a) Explain how a vaccine stimulates active immunity against a specific pathogen. [4]

(b) Distinguish between natural active immunity and artificial passive immunity. [3]

(c) Suggest why vaccination programs fail to eradicate some infectious diseases, such as malaria or cholera. [3]
Show answer & marking scheme

Worked solution

(a)
1. A vaccine containing harmless antigens, weakened (attenuated) pathogens, or dead pathogens is injected or ingested into the body.
2. Antigen-presenting cells (APCs), such as macrophages, engulf the pathogen/antigens and present them on their surface.
3. Specific T-helper cells with complementary receptors bind to the presented antigens and release cytokines.
4. Cytokines stimulate specific B-lymphocytes with complementary receptors to undergo clonal selection, dividing rapidly by mitosis (clonal expansion).
5. B-lymphocytes differentiate into plasma cells, which secrete specific, complementary antibodies, and memory B-cells.
6. Memory cells remain in the circulation for a long time, providing a rapid and large antibody response if the real pathogen is encountered later.

(b)
1. Natural active immunity occurs following exposure to a live pathogen (infection), where the individual's immune system produces its own antibodies and memory cells.
2. Artificial passive immunity involves injecting ready-made antibodies (e.g., antitoxins or immunoglobulins) into the individual, providing immediate but short-term protection.
3. In natural active immunity, memory cells are produced, leading to long-term protection, whereas artificial passive immunity produces no memory cells, so protection is temporary.

(c)
1. Some pathogens, such as the influenza virus or Plasmodium (malaria), undergo frequent mutations, leading to antigenic variation (changing surface antigens so old antibodies no longer recognize them).
2. Eukaryotic pathogens like Plasmodium have complex life cycles with different stages in different hosts (mosquito and human) and hide inside host cells (liver and red blood cells), escaping antibodies.
3. Some pathogens (like Vibrio cholerae) remain in the gut lumen, where antibodies from the blood cannot easily reach them.
4. Poor sanitation, crowded living conditions, and difficulties in delivering/storing vaccines in remote, developing regions make complete eradication hard to achieve.

Marking scheme

(a) [Max 4 marks]
- Vaccine contains antigens / dead / attenuated / harmless pathogens. [1]
- Macrophages / APCs present the antigens. [1]
- T-helper cells release cytokines to activate specific B-lymphocytes. [1]
- B-lymphocytes undergo clonal selection / clonal expansion / mitosis. [1]
- B-lymphocytes differentiate into plasma cells (which secrete antibodies) and memory cells. [1]
- Memory cells persist in the body to allow a rapid secondary response. [1]

(b) [Max 3 marks - must be comparative differences]
- Natural active is caused by infection / natural exposure to pathogen, whereas artificial passive involves injection of pre-formed antibodies. [1]
- Natural active leads to the production of antibodies by the host's own immune system, whereas passive does not. [1]
- Natural active leads to the production of memory cells / long-term protection, whereas artificial passive does not produce memory cells / is short-term. [1]

(c) [Max 3 marks]
- Pathogens undergo antigenic variation / mutations that change surface antigens. [1]
- Pathogens hide inside host cells (e.g., inside red blood cells) / away from antibodies. [1]
- Complex life cycle of eukaryotic pathogens (different antigens at different stages). [1]
- Pathogen remains in the gut lumen (e.g., cholera) where systemic antibodies are less effective. [1]
- Poor infrastructure / lack of cold chain storage / high transmission rates make mass vaccination difficult. [1]
Question 6 · Structured Short Answer
10 marks
Mitosis is part of the cell cycle that ensures genetic stability in eukaryotic cells.

(a) Describe the events that occur during interphase that prepare a cell for division. [3]

(b) Describe and explain the behaviour of chromosomes during metaphase and anaphase. [4]

(c) Explain the significance of mitosis in producing genetically identical daughter cells. [3]
Show answer & marking scheme

Worked solution

(a)
1. During the G1 phase of interphase, the cell undergoes growth, protein synthesis, and replication of organelles (e.g., mitochondria and centrioles).
2. During the S (synthesis) phase, semi-conservative DNA replication occurs, doubling the DNA content so each chromosome now consists of two genetically identical sister chromatids joined at a centromere.
3. During the G2 phase, the cell continues to grow, synthesizes tubulin (spindle proteins), and stores ATP required for chromosome movement.

(b)
1. In metaphase, chromosomes align individually along the equator (metaphase plate) of the cell.
2. Spindle fibers (microtubules) attach to the centromere (specifically the kinetochores) of each chromosome.
3. In anaphase, the centromeres divide / split.
4. The spindle fibers contract and shorten, pulling the sister chromatids apart towards opposite poles of the cell.
5. The chromatids move centromere first, appearing in a characteristic V-shape.

(c)
1. Semi-conservative replication in S phase ensures that the two sister chromatids are exact genetic copies of each other.
2. The equal division and separation of these sister chromatids during mitosis ensures that each of the two new daughter nuclei receives exactly one copy of every chromosome.
3. This maintains the diploid chromosome number and ensures no genetic variation is introduced (excluding rare mutations), which is vital for growth, tissue repair / replacement of damaged cells, and asexual reproduction.

Marking scheme

(a) [Max 3 marks]
- G1 phase: cell growth / protein synthesis / replication of organelles. [1]
- S phase: DNA replicates / semi-conservative replication of DNA (to form sister chromatids). [1]
- G2 phase: tubulin / spindle protein synthesis / ATP storage / further growth. [1]

(b) [Max 4 marks]
- Metaphase: Chromosomes line up individually along the equator / metaphase plate. [1]
- Spindle fibers attach to the centromeres / kinetochores. [1]
- Anaphase: Centromeres divide / split. [1]
- Spindle fibers shorten / contract. [1]
- Chromatids / sister chromosomes are pulled to opposite poles (centromere first). [1]

(c) [Max 3 marks]
- DNA replication in S phase produces identical sister chromatids. [1]
- Mitosis ensures sister chromatids are separated equally into separate nuclei. [1]
- Maintains chromosome number / diploid number / ensures no loss of genetic information. [1]
- Essential for growth / replacing damaged cells / tissue repair / asexual reproduction. [1]

Paper 3 (Advanced Practical Skills)

Carry out practical investigations, record results in tables, plot graphs, and make microscope drawings.
2 Question · 40 marks
Question 1 · Practical and Analysis Tasks
20 marks
You are required to investigate the effect of sucrose concentration on the water potential of potato tissue.

You are provided with:
* \(1.0\text{ mol dm}^{-3}\) sucrose solution, labeled **S**
* Distilled water, labeled **W**
* Five identical potato cylinders, cut to approximately \(50\text{ mm}\) in length

### (a) Preparation of dilutions and investigation

(i) You need to prepare \(20\text{ cm}^3\) of five different concentrations of sucrose solution: \(0.8\text{ mol dm}^{-3}\), \(0.6\text{ mol dm}^{-3}\), \(0.4\text{ mol dm}^{-3}\), \(0.2\text{ mol dm}^{-3}\), and \(0.0\text{ mol dm}^{-3}\) (pure distilled water), using the stock solution **S** and distilled water **W**.

Complete Table 1.1 to show the volumes of stock sucrose solution **S** and distilled water **W** needed to prepare each concentration.

**Table 1.1**
| Concentration of sucrose / \(\text{mol dm}^{-3}\) | Volume of stock solution S / \(\text{cm}^3\) | Volume of distilled water W / \(\text{cm}^3\) | Total volume / \(\text{cm}^3\) |
| :---: | :---: | :---: | :---: |
| \(0.8\) | | | \(20\) |
| \(0.6\) | | | \(20\) |
| \(0.4\) | | | \(20\) |
| \(0.2\) | | | \(20\) |
| \(0.0\) | | | \(20\) |

[3 marks]

(ii) Carry out the investigation as follows:
1. Label five test-tubes with the concentrations of sucrose prepared.
2. Transfer \(20\text{ cm}^3\) of the corresponding concentration into each tube.
3. Measure the initial length of each potato cylinder to the nearest \(0.5\text{ mm}\).
4. Place one potato cylinder into each of the test-tubes.
5. Leave the potato cylinders in the solutions for 30 minutes.
6. After 30 minutes, remove the potato cylinders from the test-tubes.
7. Gently blot them dry with a paper towel.
8. Measure the final length of each potato cylinder to the nearest \(0.5\text{ mm}\).
9. Calculate the change in length of each cylinder.

Record all your results in a single, appropriate table.

[6 marks]

(iii) Explain the change in length of the potato cylinder in the \(0.0\text{ mol dm}^{-3}\) solution and in the \(0.8\text{ mol dm}^{-3}\) solution in terms of water potential.

[3 marks]

### (b) Graph plotting and interpretation

In a separate investigation, a student measured the percentage change in mass of sweet potato discs in a range of sucrose concentrations. Their results are shown in Table 1.2.

**Table 1.2**
| Sucrose concentration / \(\text{mol dm}^{-3}\) | Mean percentage change in mass (%) |
| :---: | :---: |
| \(0.0\) | \(+14.5\) |
| \(0.2\) | \(+8.0\) |
| \(0.4\) | \(+1.5\) |
| \(0.6\) | \(-4.0\) |
| \(0.8\) | \(-9.5\) |
| \(1.0\) | \(-13.0\) |

(i) Plot a graph of the data in Table 1.2 on suitable graph axes.

[4 marks]

(ii) Use your graph to find the concentration of sucrose solution that is isotonic to the sweet potato tissue.

[1 mark]

### (c) Evaluation

Identify two sources of error in the practical procedure described in (a)(ii) that could affect the accuracy of the measurements of change in length. For each error, suggest an improvement to reduce its impact.

[3 marks]
Show answer & marking scheme

Worked solution

### Part (a)(i)
To find the volume of stock solution S needed for each dilution, the formula \(C_1 V_1 = C_2 V_2\) is used, where:
* \(C_1 = 1.0\text{ mol dm}^{-3}\) (stock concentration)
* \(V_2 = 20\text{ cm}^3\) (desired total volume)
* \(C_2\) is the desired concentration.

For \(0.8\text{ mol dm}^{-3}\):
\(V_1 = \frac{0.8 \times 20}{1.0} = 16.0\text{ cm}^3\) of S. The volume of distilled water is \(20.0 - 16.0 = 4.0\text{ cm}^3\).
This calculation is repeated systematically for all concentrations.

### Part (a)(ii)
Practical results tables must follow standard conventions: independent variable in the first column, dependent variables following. No units should be included within the body cells of the table; they must be stated in the column headers separated by a slash (/). All measurements of length using a ruler must be recorded to a consistent level of precision, typically \(0.5\text{ mm}\).

### Part (a)(iii)
Water potential (\(\psi\)) dictates the direction of water movement. Pure water has a \(\psi\) of \(0\text{ kPa}\), which is higher than the negative water potential inside plant cells. Therefore, water moves into the potato. Conversely, highly concentrated sucrose solutions (like \(0.8\text{ mol dm}^{-3}\)) have a lower (more negative) water potential than the cell sap, driving water out.

### Part (b)(i) & (ii)
The graph must be plotted carefully. The independent variable (sucrose concentration) goes on the horizontal x-axis and the dependent variable (mean percentage change in mass) on the vertical y-axis. The point of zero net mass change represents the isotonic point where the water potential of the tissue is equal to the water potential of the surrounding solution. From the plotted points, this intersection point is approximately \(0.44\text{ mol dm}^{-3}\).

### Part (c)
Common errors in this experiment arise from scale calibration and environmental control. Minimizing evaporation ensures concentrations remain constant. Consistent physical treatment (blotting) ensures external liquid film thickness is minimized and uniform.

Marking scheme

### (a)(i) [3 marks total]
* **1 mark:** Correctly calculates volumes for all concentrations (16, 12, 8, 4, 0 \(\text{cm}^3\) of S).
* **1 mark:** Correctly calculates matching volumes of distilled water W (4, 8, 12, 16, 20 \(\text{cm}^3\)).
* **1 mark:** All volumes recorded to \(1\text{ decimal place}\) (e.g. 16.0, 4.0) to reflect the precision of the measuring cylinder/pipette.

### (a)(ii) [6 marks total]
* **1 mark:** Table structured with clear borders and logical layout.
* **1 mark:** Correct column headings with units: 'Concentration of sucrose / \(\text{mol dm}^{-3}\)', 'Initial length / \(\text{mm}\)', 'Final length / \(\text{mm}\)', 'Change in length / \(\text{mm}\' (or other acceptable units like \)\text{cm}\)).
* **1 mark:** Units restricted solely to the table headers; no units (like 'mm') written next to numbers in the data rows.
* **1 mark:** All 5 sucrose concentrations represented in the table.
* **1 mark:** Raw lengths recorded consistently to the nearest \(0.5\text{ mm}\) (all values ending in .0 or .5).
* **1 mark:** Correctly calculated 'Change in length' values with appropriate operational signs (+ or -).

### (a)(iii) [3 marks total]
* **1 mark:** Identifies osmosis as the process driven by water potential differences.
* **1 mark:** For \(0.0\text{ mol dm}^{-3}\): Net movement of water is *into* the cells because external \(\psi\) is higher, leading to increased turgor and increased length.
* **1 mark:** For \(0.8\text{ mol dm}^{-3}\): Net movement of water is *out* of the cells because external \(\psi\) is lower, leading to plasmolysis/flaccidity and decreased length.

### (b)(i) [4 marks total]
* **1 mark:** Axes labeled correctly with units: x-axis: 'Sucrose concentration / \(\text{mol dm}^{-3}\)', y-axis: 'Mean percentage change in mass (%)'.
* **1 mark:** Linear scale chosen so that the plotted data covers more than 50% of the grid in both horizontal and vertical directions. (No awkward scale divisions like 3s or 7s).
* **1 mark:** All 6 points plotted accurately to within half a small square.
* **1 mark:** Points joined with ruled, clean straight lines from point to point, OR a smooth curve of best fit. (Line must not be thicker than half a small square, no double-lines).

### (b)(ii) [1 mark total]
* **1 mark:** Isotonic concentration estimated correctly based on student's graph intersection with the x-axis (accept any value in range \(0.43 - 0.45\text{ mol dm}^{-3}\)).

### (c) [3 marks total]
* **1 mark:** Identifies one valid source of error (e.g. evaporation changing solution volume/concentration, uneven surface blotting, differences in starting cylinder dimensions).
* **1 mark:** Identifies second valid source of error.
* **1 mark:** Suggests a correct, corresponding improvement for at least one of the identified errors.
Question 2 · Practical and Analysis Tasks
20 marks
Marram grass (*Ammophila arenaria*) is a xerophytic plant commonly found on sand dunes.

### (a) Drawings of Plant Structure

(i) **Fig. 2.1** is a photomicrograph showing a transverse section through the rolled leaf of *Ammophila arenaria*.

Draw a large, clear, low-power plan diagram of the transverse section of the leaf shown in **Fig. 2.1** to show the overall shape of the rolled leaf and the arrangement of its tissues.

*Do not draw any individual cells.*

On your plan diagram, use clear label lines to identify:
1. Cuticle
2. A vascular bundle

[5 marks]

(ii) **Fig. 2.2** is a high-power view showing details of the inner epidermis and hinge cells of the leaf of *Ammophila arenaria*, showing guard cells, stomata, and adjacent epidermal cells.

Make a high-power drawing of three adjacent epidermal cells, including one guard cell flanking a stoma.

On your drawing, use a clear label line to identify:
1. Cell wall

[4 marks]

### (b) Calibration and Measurement

A student calibrated an eyepiece graticule under a light microscope using a stage micrometer.
* The stage micrometer scale has divisions spaced at \(0.1\text{ mm}\) intervals.
* Using the \(\times 10\) objective lens, 40 divisions of the eyepiece graticule coincided with exactly 12 divisions of the stage micrometer.

(i) Calculate the actual distance represented by one eyepiece graticule division (epu) in micrometers (\(\mu\text{m}\)). Show all your working.

[3 marks]

(ii) The student then used the same \(\times 10\) objective lens to measure the total leaf thickness at the midrib in a slide of *Ammophila*. The thickness was measured as 32 eyepiece graticule divisions.

Calculate the actual thickness of the leaf in micrometers (\(\mu\text{m}\)). Show your working and state your answer to an appropriate number of significant figures.

[4 marks]

### (c) Comparison

Table 2.1 is an incomplete table comparing the leaf anatomy of *Ammophila arenaria* (a xerophyte) with that of *Ligustrum* (a mesophyte).

Complete Table 2.1 by describing three anatomical differences between the two leaves.

**Table 2.1**
| Anatomical Feature | *Ammophila arenaria* (Xerophyte) | *Ligustrum* (Mesophyte) |
| :--- | :--- | :--- |
| **Leaf shape / rolling** | | |
| **Location of stomata** | | |
| **Cuticle thickness** | | |

[4 marks]
Show answer & marking scheme

Worked solution

### Part (a)(i)
When drawing a plan diagram, the rule of 'no cells' must be strictly followed. The overall rolled structure of the leaf is represented by outline boundaries of the tissue layers (epidermis, mesophyll, vascular bundles). The outer boundary of the leaf is smooth and rounded, while the inner boundary has deep folds (grooves/ridges).

### Part (a)(ii)
High-power drawings require the detail of individual cells. Cell walls in plant cells must always be drawn as double lines to represent their 3D physical thickness. Shading or sketchy, feathered lines are rejected.

### Part (b)(i)
To calibrate, we use the fact that the stage micrometer has divisions spaced at \(0.1\text{ mm}\):
\(12\text{ divisions} = 12 \times 0.1\text{ mm} = 1.2\text{ mm} = 1200\ \mu\text{m}\).
Since \(40\text{ eyepiece divisions} = 1200\ \mu\text{m}\),
\(1\text{ eyepiece division} = 1200 / 40 = 30\ \mu\text{m}\).

### Part (b)(ii)
To calculate the actual thickness at the midrib, we multiply the measured eyepiece divisions by the calibration factor:
\(\text{Thickness} = 32\text{ divisions} \times 30\ \mu\text{m/division} = 960\ \mu\text{m}\).
Since the measurements are precise to 2 significant figures, the final value should be written to 2 or 3 significant figures, which is \(960\ \mu\text{m}\).

### Part (c)
Xerophytes are highly adapted to minimize transpirational water loss. In *Ammophila arenaria*, the rolling of the leaf creates a microenvironment with high humidity around the stomata. Stomata are located only on the inner surface inside protective grooves, shielded by hairs. The outer surface has a very thick, impermeable waxy cuticle to limit cuticular transpiration. In contrast, *Ligustrum* is a mesophyte with a flat leaf shape, stomata open to the general environment on its lower surface, and a much thinner cuticle.

Marking scheme

### (a)(i) [5 marks total]
* **1 mark:** Large plan diagram occupying at least 50% of the available box space.
* **1 mark:** Clear, single, continuous lines showing no sketching or overlapping pencil marks; no shading included.
* **1 mark:** Correct proportions of the rolled leaf shape (showing deep inner ridges and a smooth curved outer outline).
* **1 mark:** No individual cells drawn (vascular bundles shown strictly as circular or oval outlines).
* **1 mark:** Correctly labels 'Cuticle' (pointing to outer boundary) and 'Vascular bundle' (pointing to a circular region in the ridges).

### (a)(ii) [4 marks total]
* **1 mark:** High-power drawing of exactly three adjacent cells, including one recognizable guard cell (smaller, curved structure flanking a gap/stoma).
* **1 mark:** All cell walls drawn with double lines to show thickness; walls meet at realistic angles with no gaps in corners.
* **1 mark:** Cells drawn in correct relative sizes (guard cell smaller than the flanking epidermal cells).
* **1 mark:** Correct label 'Cell wall' with a straight, uncrossed line pointing precisely to a cell wall.

### (b)(i) [3 marks total]
* **1 mark:** Recognizes that \(1.2\text{ mm}\) is equivalent to \(1200\ \mu\text{m}\).
* **1 mark:** Shows division of distance by the number of eyepiece units (\(1200 / 40\)).
* **1 mark:** Correct answer of \(30\ \mu\text{m}\) with complete working.

### (b)(ii) [4 marks total]
* **1 mark:** Shows working multiplying 32 by the calibration value from (b)(i).
* **1 mark:** Obtains numerical value \(960\).
* **1 mark:** Correct units (\(\mu\text{m}\)) given in the final answer.
* **1 mark:** Expresses the final answer to an appropriate number of significant figures (2 or 3 sig figs: \(960\) or \(960.\)).

### (c) [4 marks total]
* **1 mark:** Correctly contrasts leaf shape/rolling (Ammophila: rolled / incurved vs. Ligustrum: flat / unrolled).
* **1 mark:** Correctly contrasts stomatal location (Ammophila: inner surface only / inside grooves / sunken vs. Ligustrum: lower surface / not sunken).
* **1 mark:** Correctly contrasts cuticle thickness (Ammophila: thick waxy outer cuticle vs. Ligustrum: thin waxy cuticle).
* **1 mark:** Table completed with clear comparative language on each row.

Paper 4 (A Level Structured Questions)

Answer all structured and data analysis questions based on core A Level content.
10 Question · 100 marks
Question 1 · structured
10 marks
A biotechnology company aims to produce human antithrombin III, a therapeutic protein, in the milk of transgenic goats.

(a) Explain why the human gene must be paired with a promoter sequence from a mammary gland-specific gene, such as beta-casein, before microinjection. [3]

(b) Outline the steps involved in using a PCR-based method to confirm that the human antithrombin gene has integrated successfully into the goat's genome. [4]

(c) Discuss one social or ethical concern associated with producing pharmaceutical proteins in GM animals. [3]
Show answer & marking scheme

Worked solution

Part (a):
- Promoters act as binding sites for RNA polymerase to initiate transcription.
- A tissue-specific promoter (e.g., from beta-casein) ensures that transcription of the human antithrombin gene only occurs in the mammary gland cells.
- This prevents expression of the protein in other organs where it might disrupt normal physiological functions (systemic toxicity).
- It ensures high-yield secretion of the recombinant protein directly into the milk for easy harvesting.

Part (b):
- DNA extraction: Isolate genomic DNA from cells of the transgenic goat (e.g., from skin or blood).
- Primer design: Synthesize forward and reverse primers complementary to specific sequences of the human antithrombin gene.
- PCR cycles: Run the mixture through temperature cycles (95 degrees C to denature DNA strands, 55-60 degrees C to allow primers to anneal, 72 degrees C for Taq polymerase extension).
- Analysis: Perform gel electrophoresis alongside DNA markers to check for the presence and correct size of the amplified target DNA band.

Part (c):
- Animal welfare: The transgene insertion or microinjection process may cause unintended birth defects, physiological stress, or suffering.
- Environmental/Ecological: Potential risk of transgenic animals escaping and crossbreeding with wild/non-GM populations, altering the gene pool.
- Socio-ethical: Opposition to genetic modification on moral/religious grounds ('playing God') or concerns about the commercialization and ownership of animal life.

Marking scheme

Part (a) (Max 3):
1. Promoters are binding sites for RNA polymerase / transcription factors (initiating transcription) [1].
2. Specific promoter ensures transcription occurs only in mammary gland tissues / secretory cells [1].
3. Prevents harmful systemic expression / toxic effects in other organs of the goat [1].
4. Ensures the target protein is secreted into milk for non-invasive extraction [1].

Part (b) (Max 4):
1. Extract genomic DNA from transgenic goat tissue / blood sample [1].
2. Design and add primers specific / complementary to the human antithrombin gene sequence [1].
3. Detail PCR cycle steps: denaturation at 90-95 degrees C, annealing at 50-65 degrees C, extension at 70-75 degrees C (accept Taq polymerase) [1].
4. Perform agarose gel electrophoresis [1].
5. Check for the presence of a band of the predicted size / molecular weight matching the target gene [1].

Part (c) (Max 3):
1. Concern over animal welfare / potential physical distress or suffering during cloning or insertion [1].
2. Unintended target effects where gene insertion disrupts other essential genes in the host genome [1].
3. Religious or moral objections to modifying the genetic identity of animals / species boundaries [1].
4. Potential risks of escape and ecological impacts (though limited for domestic goats) [1].
Question 2 · structured
10 marks
Xenozin is a chemical compound that acts as an uncoupler of oxidative phosphorylation in mammalian mitochondria. It makes the inner mitochondrial membrane highly permeable to hydrogen ions (\(H^+\)).

(a) Define the term 'oxidative phosphorylation' and state its precise location within a mitochondrion. [2]

(b) An experiment showed that adding Xenozin to respiring liver cells increased oxygen consumption but stopped ATP synthesis. Explain how Xenozin causes these effects. [5]

(c) State the main pathway of respiration that continues to produce ATP when oxidative phosphorylation is blocked, and name its net ATP yield per glucose molecule. [2]

(d) State one physiological hazard of uncoupling agents to mammalian homeostasis. [1]
Show answer & marking scheme

Worked solution

Part (a):
- Definition: The synthesis of ATP from ADP and inorganic phosphate (\(P_i\)) using energy derived from electron transport, with oxygen acting as the terminal electron acceptor.
- Location: Cristae / inner mitochondrial membrane.

Part (b):
- Normally, protons are pumped from the matrix to the intermembrane space by proteins in the electron transport chain (ETC), creating an electrochemical proton gradient.
- Protons flow back into the matrix down this gradient exclusively through the hydrophilic channel of ATP synthase (chemiosmosis), driving ATP synthesis.
- Xenozin makes the inner membrane permeable to \(H^+\), allowing protons to leak directly back into the matrix, bypassing ATP synthase.
- This dissipates / destroys the proton gradient and the proton motive force, so ATP synthesis stops.
- Since there is no high proton concentration in the intermembrane space to resist pumping, the ETC accelerates, consuming more oxygen (which acts as the final electron acceptor) to try and pump more protons.

Part (c):
- Pathway: Glycolysis.
- Net ATP yield: 2 ATP molecules (per glucose molecule).

Part (d):
- Excessive heat generation / hyperthermia. Because the chemical energy of the proton gradient is not captured as ATP, it is entirely dissipated as heat energy, which can denature proteins.

Marking scheme

Part (a) (Max 2):
1. Synthesis of ATP from ADP and \(P_i\) driven by the transfer of electrons along the ETC / oxygen as terminal acceptor [1].
2. Located on the cristae / inner mitochondrial membrane [1].

Part (b) (Max 5):
1. Protons (\(H^+\)) are pumped into the intermembrane space by electron transport chain carriers [1].
2. Creates a proton gradient / electrochemical gradient / proton motive force [1].
3. Xenozin allows protons to diffuse back / leak across the lipid bilayer directly into the matrix [1].
4. Protons bypass ATP synthase [1].
5. No ATP is synthesized as chemiosmotic flow through ATP synthase is lost [1].
6. Electron transport chain runs at maximum rate / faster, leading to increased consumption of oxygen (accept oxygen reduced to water at terminal end) [1].

Part (c) (Max 2):
1. Glycolysis [1].
2. Net 2 ATP (per glucose) [1] (Reject: gross 4 ATP, must state net).

Part (d) (Max 1):
1. Hyperthermia / dangerous increase in body temperature / heat exhaustion [1].
Question 3 · structured
10 marks
Rubisco (ribulose-1,5-bisphosphate carboxylase-oxygenase) is a key enzyme in the light-independent stage of photosynthesis. It can catalyze both the carboxylation and oxygenation of RuBP.

(a) Describe how the first stable product of carbon fixation is converted to triose phosphate (TP) in the Calvin cycle. [4]

(b) Under high temperatures and dry conditions, \(C_3\) crops exhibit high rates of photorespiration.
(i) Suggest why high temperature increases the rate of photorespiration relative to photosynthesis. [2]
(ii) Explain why photorespiration reduces the overall productivity of \(C_3\) crops. [2]

(c) State two structural or anatomical features of \(C_4\) plants that minimize photorespiration. [2]
Show answer & marking scheme

Worked solution

Part (a):
- Carbon dioxide is fixed to RuBP to form an unstable 6C compound, which splits into two molecules of glycerate 3-phosphate (GP).
- GP is phosphorylated by ATP (derived from the light-dependent stage) to form glycerate 1,3-bisphosphate.
- This intermediate is then reduced using hydrogen/electrons from reduced NADP (also from the light-dependent stage) to yield triose phosphate (TP).
- Inorganic phosphate is released during this process.

Part (b):
(i)
- At elevated temperatures, the solubility of carbon dioxide in water decreases more rapidly than the solubility of oxygen, leading to a lower ratio of dissolved \(CO_2\) to \(O_2\) in the chloroplast.
- High temperatures also cause conformational changes in the Rubisco enzyme, increasing its relative affinity for oxygen.
(ii)
- Photorespiration involves Rubisco binding to oxygen instead of \(CO_2\), leading to the production of phosphoglycolate, which must be metabolized in a costly salvage pathway.
- This pathway releases carbon dioxide (losing previously fixed carbon) and consumes ATP, decreasing net primary productivity.

Part (c):
- Kranz anatomy / tight ring of mesophyll cells surrounding bundle sheath cells.
- Spatial separation of initial carbon fixation (in mesophyll cells) and the Calvin cycle (in bundle sheath cells).
- High density of chloroplasts in bundle sheath cells, which are impermeable to gases, keeping \(CO_2\) concentrations high.

Marking scheme

Part (a) (Max 4):
1. Glycerate 3-phosphate (GP) is the first stable product [1].
2. GP is phosphorylated by ATP (to form 1,3-bisphosphoglycerate) [1].
3. ATP provides energy / phosphate group [1].
4. Reduced NADP provides hydrogen / electrons to reduce the molecule [1].
5. Triose phosphate (TP) and NADP / ADP are produced [1].

Part (b)(i) (Max 2):
1. Solubility of \(CO_2\) decreases more than that of \(O_2\) at high temperatures, raising the oxygen-to-carbon dioxide ratio [1].
2. High temperature changes the shape of Rubisco's active site to increase its relative affinity for oxygen [1].

Part (b)(ii) (Max 2):
1. Carbon is lost / wasted as carbon dioxide is released (reducing net carbon fixation) [1].
2. ATP / metabolic energy is consumed to salvage carbon skeleton without yielding sugars [1].

Part (c) (Max 2):
1. Kranz anatomy / presence of specialized bundle sheath cells [1].
2. Spatial separation of carbon dioxide fixation (mesophyll) and Calvin cycle (bundle sheath) [1].
3. Thick / gas-impermeable cell walls of bundle sheath cells that trap \(CO_2\) [1].
Question 4 · structured
10 marks
Antidiuretic hormone (ADH) plays a crucial role in maintaining blood water potential in mammals.

(a) Describe how ADH increases the water permeability of the collecting duct cells. [4]

(b) Nephrogenic diabetes insipidus (NDI) is a genetic disorder where the kidneys fail to respond to ADH due to a mutation in the ADH receptor gene.
(i) Explain why an individual with NDI produces a large volume of dilute urine, even when severely dehydrated. [3]
(ii) Explain why injections of synthetic ADH (desmopressin) are ineffective as a treatment for this patient. [1]

(c) State two physiological responses in a healthy mammal that occur when blood water potential increases above the set point. [2]
Show answer & marking scheme

Worked solution

Part (a):
- ADH is transported in the blood and binds to specific V2 receptors on the basolateral membrane of the collecting duct cells.
- This binding activates a G-protein, which stimulates the enzyme adenyl cyclase to convert ATP into cyclic AMP (cAMP) (secondary messenger).
- cAMP initiates a kinase cascade that causes vesicles containing aquaporins (water channel proteins) to move towards and fuse with the luminal (apical) membrane of the cell.
- This increases the number of aquaporins in the membrane, vastly increasing water permeability.

Part (b):
(i)
- Due to the mutated receptor, ADH cannot bind, so the G-protein is not activated and cAMP is not produced.
- Intracellular vesicles containing aquaporins do not fuse with the luminal membrane of the collecting duct.
- As a result, water cannot leave the collecting duct lumen by osmosis down the water potential gradient into the hypertonic medulla; instead, it is lost in the urine.
(ii)
- Desmopressin acts as an agonist for the ADH receptor. Since the receptor itself is structurally defective and non-functional, adding more hormone has no effect.

Part (c):
- Decrease in ADH release from the posterior pituitary gland.
- Reduction in the permeability of the collecting ducts to water.
- Excretion of a larger volume of dilute urine to restore normal water potential.

Marking scheme

Part (a) (Max 4):
1. ADH binds to specific V2 receptors on the basolateral membrane [1].
2. Activates G-protein / adenyl cyclase, converting ATP to cyclic AMP (cAMP) [1].
3. cAMP acts as a second messenger to trigger an intracellular kinase cascade [1].
4. Vesicles containing aquaporins move to and fuse with the luminal / apical membrane [1].
5. Aquaporins act as water channels increasing permeability [1].

Part (b)(i) (Max 3):
1. Mutated receptor cannot bind ADH / signal transduction fails [1].
2. Aquaporins are not inserted into the luminal membrane / remain in cytoplasm [1].
3. Collecting duct remains impermeable to water [1].
4. Water cannot be reabsorbed by osmosis into the hypertonic medulla, leading to dilute urine [1].

Part (b)(ii) (Max 1):
1. The mutation is in the receptor, not the hormone itself, so target cells cannot respond to desmopressin [1].

Part (c) (Max 2):
1. Osmoreceptors in hypothalamus detect high water potential and decrease stimulation of pituitary [1].
2. Posterior pituitary secretes less ADH into the blood [1].
3. Collecting ducts become less permeable, leading to excretion of large volumes of dilute urine [1].
Question 5 · structured
10 marks
Sarin is a highly toxic organophosphate compound that acts as an irreversible inhibitor of the enzyme acetylcholinesterase.

(a) Describe the events occurring at the presynaptic membrane of a neuromuscular junction upon the arrival of an action potential, leading to the release of acetylcholine (ACh). [4]

(b) In the presence of Sarin, acetylcholinesterase is inactivated.
(i) Describe the normal role of acetylcholinesterase at the synapse. [2]
(ii) Explain the consequences of Sarin poisoning on post-synaptic transmission and muscle contraction. [4]
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Worked solution

Part (a):
- The arrival of an action potential depolarizes the presynaptic membrane.
- This depolarization causes voltage-gated calcium ion channels to open.
- Calcium ions (\(Ca^{2+}\)) diffuse down their electrochemical gradient from the synaptic cleft into the presynaptic knob.
- The rise in intracellular calcium causes synaptic vesicles containing acetylcholine (ACh) to move towards and fuse with the presynaptic membrane.
- ACh is released into the synaptic cleft via exocytosis.

Part (b):
(i)
- Acetylcholinesterase rapidly hydrolyzes acetylcholine in the synaptic cleft into choline and ethanoic acid (acetate).
- This stops the stimulation of the postsynaptic receptors and allows the postsynaptic membrane to repolarize and recover for the next signal.
(ii)
- If acetylcholinesterase is inhibited by Sarin, acetylcholine is not broken down and remains accumulated in the cleft.
- ACh remains continuously bound to receptors on the postsynaptic membrane (sarcolemma).
- Sodium channels stay open, causing continuous depolarization of the postsynaptic membrane and repeated generation of action potentials along the sarcolemma.
- Calcium ions remain continuously released from the sarcoplasmic reticulum, resulting in continuous muscle contraction (spastic paralysis / tetany) and inability of the muscle to relax.

Marking scheme

Part (a) (Max 4):
1. Action potential arrives and depolarizes the presynaptic membrane [1].
2. Voltage-gated calcium ion channels open [1].
3. Calcium ions (\(Ca^{2+}\)) diffuse into the presynaptic knob [1].
4. Calcium ions stimulate synaptic vesicles containing ACh to move to and fuse with the presynaptic membrane [1].
5. ACh is released into the synaptic cleft via exocytosis [1].

Part (b)(i) (Max 2):
1. Hydrolyzes acetylcholine into choline and acetate / ethanoic acid [1].
2. Terminates synaptic transmission / stops depolarization of post-synaptic membrane, allowing it to repolarize [1].

Part (b)(ii) (Max 4):
1. ACh is not broken down and remains in the synaptic cleft [1].
2. ACh remains bound to receptor proteins on the postsynaptic membrane [1].
3. Sodium ion channels remain open, keeping membrane continuously depolarized [1].
4. Continuous action potentials are sent along the muscle fibre / sarcolemma [1].
5. Calcium ions remain high in the muscle sarcoplasm, resulting in continuous contraction / spastic paralysis / failure to relax [1].
Question 6 · structured
10 marks
The Great Rift Valley lakes of East Africa contain hundreds of species of cichlid fish, which have evolved through rapid adaptive radiation.

(a) Define the term 'speciation' and distinguish between allopatric and sympatric speciation. [3]

(b) Describe how natural selection can lead to the evolution of a new species of cichlid fish feeding on different food sources within the same lake. [5]

(c) Explain the role of reproductive isolation in maintaining separate gene pools after speciation has occurred. [2]
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Worked solution

Part (a):
- Speciation: The evolutionary process by which new, distinct species arise from a single ancestral species.
- Allopatric speciation occurs when populations are geographically isolated (e.g., by physical barriers like mountains or rivers), preventing gene flow.
- Sympatric speciation occurs without geographic isolation; populations live in the same geographic area but become reproductively isolated due to ecological, behavioral, or genetic differences.

Part (b):
- Within the cichlid population in the lake, genetic variation exists due to random mutations (e.g., affecting jaw morphology or feeding behavior).
- Overpopulation leads to intraspecific competition for limited food resources.
- Individuals with mutations that allow them to exploit an underutilized food source (e.g., scraping algae from rocks vs. crushing snail shells) experience less competition.
- These individuals have a selective advantage, survive longer, and are more likely to reproduce (natural selection).
- They pass on these advantageous alleles to their offspring.
- Over generations, directional selection changes the allele frequencies in different ecological niches.
- Assortative mating (preferring mates with similar feeding traits or color patterns) reinforces reproductive isolation, leading to sympatric speciation.

Part (c):
- Reproductive isolation (prezygotic or postzygotic barriers) prevents interbreeding between the newly formed species.
- This blocks gene flow between their gene pools.
- It preserves the distinct adaptive gene complexes of each species, preventing them from fusing back into a single species.

Marking scheme

Part (a) (Max 3):
1. Speciation: Formation of new / distinct species (that cannot interbreed to produce fertile offspring) [1].
2. Allopatric speciation requires physical / geographic isolation [1].
3. Sympatric speciation occurs in the same geographic location / without physical barriers [1].

Part (b) (Max 5):
1. Genetic variation exists in the population due to mutation [1].
2. Intraspecific competition for food / resources occurs [1].
3. Fish with adaptive traits / variations exploit new niches / food sources [1].
4. Individuals with beneficial alleles have higher survival and reproductive success (natural selection) [1].
5. Beneficial alleles are passed on to offspring, changing allele frequencies over generations [1].
6. Behavioral / ecological isolation (e.g. microhabitat preference or assortative mating) prevents cross-breeding [1].

Part (c) (Max 2):
1. Prevents gene flow / genetic mixing between the two emerging populations [1].
2. Maintains distinct allele frequencies / genetic integrity of each species [1].
Question 7 · structured
10 marks
The regulation of the operon system in Escherichia coli is a classic model of prokaryotic gene control.

(a) Outline the function of the following parts of the lac operon:
(i) Promoter [1]
(ii) Operator [1]
(iii) lacZ gene [1]

(b) Explain the sequence of molecular events that occurs in E. coli cells when lactose is present but glucose is absent. [5]

(c) State why the lac operon is described as an inducible operon, and distinguish this from a repressible operon. [2]
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Worked solution

Part (a):
- (i) Promoter: The DNA sequence where RNA polymerase binds to initiate transcription of the structural genes.
- (ii) Operator: The DNA sequence located between the promoter and structural genes where the repressor protein binds to block transcription.
- (iii) lacZ gene: A structural gene that codes for the enzyme \(\beta\)-galactosidase, which hydrolyzes lactose into glucose and galactose.

Part (b):
- In the absence of glucose, cAMP levels are high, which activates the CAP activator protein, facilitating RNA polymerase binding.
- Lactose enters the cell and some is converted into allolactose (the inducer).
- Allolactose binds to the allosteric site of the active lac repressor protein.
- This binding causes a conformational (shape) change in the repressor protein.
- The repressor protein can no longer bind to the operator site and detaches from the DNA.
- The operator is now unoccupied, allowing RNA polymerase to bind to the promoter and transcribe the structural genes (lacZ, lacY, and lacA) into a single polycistronic mRNA.

Part (c):
- The lac operon is 'inducible' because it is normally switched off (repressed) and is only switched on (induced) in the presence of its substrate (lactose/allolactose).
- In contrast, a repressible operon (like the trp operon) is normally switched on and is turned off when the concentration of a specific end-product (like tryptophan) increases and binds to the repressor, activating it.

Marking scheme

Part (a) (Max 3):
1. (i) Promoter: Binding site for RNA polymerase to start transcription [1].
2. (ii) Operator: Binding site for the active repressor protein to prevent transcription [1].
3. (iii) lacZ gene: Codes for \(\beta\)-galactosidase (which hydrolyzes lactose to glucose and galactose) [1].

Part (b) (Max 5):
1. Lactose enters the cell and is converted to allolactose (accept lactose acts as the inducer) [1].
2. Allolactose binds to the regulatory repressor protein [1].
3. Repressor protein undergoes a conformational change / change in tertiary structure [1].
4. Repressor detaches from / cannot bind to the operator [1].
5. RNA polymerase is able to bind to the promoter and transcribe the structural genes (lacZ, lacY, lacA) [1].
6. High cAMP levels / CAP binding enhances transcription when glucose is absent [1].

Part (c) (Max 2):
1. Inducible operon is default 'off' and turned 'on' in the presence of the substrate (inducer) [1].
2. Repressible operon is default 'on' and turned 'off' by the product (co-repressor) [1].
Question 8 · structured
10 marks
Auxins, such as indole-3-acetic acid (IAA), control plant growth responses including phototropism.

(a) Describe the distribution of IAA in a coleoptile tip exposed to unilateral light and explain how this distribution is achieved. [3]

(b) Explain how IAA stimulates cell elongation in the growing region of a plant shoot according to the acid growth hypothesis. [5]

(c) State the role of gibberellins (GA) in the germination of barley seeds. [2]
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Worked solution

Part (a):
- IAA is synthesized in the cells of the coleoptile tip.
- When unilateral light is applied, photoreceptors called phototropins detect the light gradient.
- This triggers the lateral active transport of IAA from the illuminated side to the shaded side of the tip.
- As a result, there is a higher concentration of IAA on the shaded side of the shoot compared to the illuminated side.

Part (b):
- IAA binds to specific receptors on the cell surface membrane of cells in the elongation zone.
- This binding stimulates proton pumps (\(H^+\)-ATPases) in the cell membrane to actively pump hydrogen ions (\(H^+\)) from the cytoplasm into the cell wall matrix.
- The pH of the cell wall decreases (becomes more acidic, around pH 4.5).
- The acidic environment activates wall-loosening proteins called expansins.
- Expansins break the non-covalent hydrogen bonds holding cellulose microfibrils and hemicellulose together.
- This weakens and loosens the cell wall matrix.
- Potassium ions (\(K^+\)) enter the cell, lowering its water potential so water enters by osmosis.
- The resulting high turgor pressure stretches the weakened cell wall, causing irreversible cell elongation.

Part (c):
- Water absorption by the seed triggers the embryo to synthesize and secrete gibberellins (GA).
- GA diffuses to the aleurone layer.
- In the aleurone layer, GA stimulates the transcription of genes encoding hydrolytic enzymes, specifically \(\alpha\)-amylase.
- Amylase is secreted into the endosperm, where it hydrolyzes stored starch into soluble maltose / glucose, providing respiratory substrates for the growing embryo.

Marking scheme

Part (a) (Max 3):
1. IAA accumulates on the shaded / dark side of the coleoptile [1].
2. Unilateral light is detected by phototropins / photoreceptors [1].
3. Causes active lateral transport of auxin from the illuminated side to the shaded side [1].

Part (b) (Max 5):
1. IAA binds to receptors on the cell membrane, activating proton (\(H^+\)) pumps [1].
2. Protons are actively pumped into the cell wall, lowering its pH [1].
3. Expansins are activated by low pH [1].
4. Expansins break hydrogen bonds / cross-links between cellulose microfibrils [1].
5. Ions (e.g. \(K^+\)) enter cell, lowering water potential, causing water uptake by osmosis [1].
6. High turgor pressure stretches the loosened / flexible cell wall [1].

Part (c) (Max 2):
1. GA is secreted by the embryo and diffuses to the aleurone layer [1].
2. Stimulates the synthesis / transcription of \(\alpha\)-amylase [1].
3. Amylase digests starch in the endosperm to maltose / glucose to support embryo growth [1].
Question 9 · structured
10 marks
An investigation was carried out to study the effect of a synthetic peptide, AMP-4, on the osmoregulatory response of healthy volunteers. Two groups of volunteers, Group A (control) and Group B (experimental), fasted overnight and then drank 500 cm3 of pure water at time = 0 minutes. At time = 0 minutes: Group A received an injection of sterile saline solution; Group B received an injection of AMP-4 dissolved in saline. The urine flow rate and urine osmolarity (concentration of solutes) were monitored over 180 minutes. The results showed: In Group A, the mean urine flow rate increased rapidly from 1.5 cm3 min-1 to a peak of 12.0 cm3 min-1 at 60 minutes, before decreasing back to 1.5 cm3 min-1 by 180 minutes. Urine osmolarity decreased to a minimum of 100 mOsmol kg-1 at 60 minutes. In Group B, the mean urine flow rate remained constant at approximately 1.5 to 2.0 cm3 min-1 throughout the 180 minutes. Urine osmolarity remained high at approximately 800 mOsmol kg-1. (a) Describe the physiological homeostatic mechanism by which an increase in blood water potential leads to the production of dilute urine, as observed in Group A. [4] (b) Suggest and explain how the synthetic peptide AMP-4 could prevent the response seen in Group A, keeping urine flow rate low and osmolarity high. [4] (c) State the precise location of the osmoreceptors that detect changes in blood water potential. [1] (d) Name one hormone, other than ADH, involved in homeostasis in mammals that acts via a second messenger system. [1]
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Worked solution

(a) An increase in blood water potential is detected by hypothalamic osmoreceptors, causing fewer impulses to be sent to the posterior pituitary gland. This results in less ADH being released into the bloodstream. Consequently, the walls of the collecting ducts become less permeable to water as aquaporins are removed from the luminal membranes. Less water is reabsorbed by osmosis into the hypertonic renal medulla, leading to the excretion of a larger volume of dilute urine. (b) AMP-4 likely acts as an ADH agonist by binding to ADH receptors on the cell surface membrane of collecting duct epithelial cells. This binding stimulates a second messenger pathway (such as cAMP), leading to the fusion of vesicles containing aquaporins with the luminal membrane. Therefore, the collecting duct remains highly permeable to water even in the absence of endogenous ADH, causing continuous water reabsorption and maintaining a low volume of concentrated urine. (c) Hypothalamus. (d) Glucagon (or adrenaline).

Marking scheme

(a) [Max 4 marks] 1. Increase in blood water potential is detected by osmoreceptors in the hypothalamus; 2. Fewer nerve impulses are sent to the posterior pituitary gland; 3. Less ADH (antidiuretic hormone) is secreted/released into the blood; 4. Collecting duct walls (and DCT) become less permeable to water / fewer aquaporins in the luminal/apical membrane; 5. Less water is reabsorbed by osmosis into the tissue fluid/blood of the renal medulla; 6. A larger volume of more dilute urine is produced. (b) [Max 4 marks] 1. AMP-4 acts as an ADH agonist / mimics the molecular structure of ADH; 2. AMP-4 binds to ADH receptors on the cell surface membrane of collecting duct cells; 3. This activates G-proteins and triggers a second messenger pathway / cAMP production; 4. Intracellular vesicles containing aquaporins fuse with the luminal cell surface membrane; 5. Collecting duct walls remain highly permeable to water even when endogenous ADH is low; 6. High volumes of water continue to be reabsorbed down the water potential gradient into the hypertonic medulla, keeping urine concentrated and of low volume. (c) [1 mark] Hypothalamus (accept: supraoptic or paraventricular nuclei). (d) [1 mark] Glucagon / Adrenaline.
Question 10 · structured
10 marks
Promoters are essential components of vectors used in genetic engineering. An investigation compared the efficiency of two different eukaryotic promoters, Prom-X and Prom-Y, in driving the expression of a human therapeutic protein (HTP) gene in transgenic host mammalian cells. To monitor expression, a reporter gene encoding Green Fluorescent Protein (GFP) was fused downstream of each promoter inside identical plasmids. Cells transfected with the Prom-X-GFP construct showed a mean fluorescence intensity of 120 arbitrary units (au). Cells transfected with the Prom-Y-GFP construct showed a mean fluorescence intensity of 450 au. To confirm transcription levels, quantitative RT-PCR (reverse transcription polymerase chain reaction) was performed on isolated mRNA using primers specific to the HTP gene sequence. The cycle threshold (Ct) value was determined: Cells with Prom-X had a mean Ct value of 24.5; Cells with Prom-Y had a mean Ct value of 19.1. (a) Explain why a reporter gene, such as the gene for GFP, is useful when transforming cells with a gene of interest. [2] (b) State the role of a promoter in transcription. [2] (c) With reference to the quantitative RT-PCR results, explain which promoter is more efficient at driving the transcription of the HTP gene. [3] (d) PCR involves a repeating cycle of three temperature steps. Outline the temperature and purpose of each step. [3]
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Worked solution

(a) A reporter gene allows researchers to easily identify which host cells have successfully taken up the plasmid vector (transformants) and are actively expressing the transgene, as GFP fluoresces green under UV light without destroying the host cells. (b) The promoter is the sequence of DNA to which RNA polymerase binds to initiate transcription, determining the starting point and frequency of mRNA synthesis for the downstream gene. (c) Prom-Y is more efficient because it has a lower Ct value (19.1 compared to 24.5 for Prom-X). A lower Ct value means the threshold of fluorescent detection was reached in fewer PCR cycles, indicating a higher initial concentration of target HTP mRNA in the cell extract. This is also supported by the significantly higher GFP fluorescence intensity of 450 au compared to 120 au. (d) Step 1: Denaturation at 92 to 95 degrees C to break hydrogen bonds and separate double-stranded DNA into single strands. Step 2: Annealing at 50 to 65 degrees C to allow primers to bind to their complementary target sequences. Step 3: Extension at 70 to 75 degrees C to allow Taq/thermostable DNA polymerase to synthesize new complementary strands by adding free nucleotides.

Marking scheme

(a) [Max 2 marks] 1. Identifies which host cells have successfully taken up the plasmid/vector/gene of interest (transformants); 2. GFP fluoresces under UV light, allowing easy, non-invasive detection without killing the cells; 3. Indicates that the promoter is functional and the gene of interest is likely being expressed. (b) [Max 2 marks] 1. Binding site for RNA polymerase; 2. Initiates/starts transcription of the target gene; 3. Directs the synthesis of mRNA from the correct template strand. (c) [Max 3 marks] 1. Prom-Y is more efficient; 2. Prom-Y has a lower Ct value (19.1 vs 24.5 for Prom-X), meaning its mRNA threshold was reached in fewer cycles; 3. A lower Ct value indicates a higher initial quantity/concentration of HTP mRNA template in cells with Prom-Y; 4. This correlates with higher GFP expression / higher fluorescence intensity (450 au vs 120 au), indicating greater transcription rates. (d) [Max 3 marks] 1. Denaturation: heating to 92-95 degrees C to break hydrogen bonds between complementary bases and separate double-stranded DNA; 2. Annealing: cooling to 50-65 degrees C to allow primers to bind/hybridize to complementary sequences on the single DNA strands; 3. Extension: heating to 70-75 degrees C to provide the optimum temperature for Taq / thermostable DNA polymerase to synthesize new complementary DNA strands.

Paper 5 (Planning, Analysis and Evaluation)

Answer planning design, statistical evaluation, and risk assessment questions.
3 Question · 30 marks
Question 1 · Experimental Design and Statistical Analysis
10 marks
A student wants to investigate the effect of different concentrations of copper sulfate (acting as a heavy metal respiratory inhibitor) on the rate of anaerobic respiration in yeast. The student is provided with: 10% yeast suspension, 5% glucose solution, 1.0% methylene blue indicator, 0.10 mol dm^{-3} copper sulfate stock solution, distilled water, and standard laboratory glassware with a thermostatically controlled water bath. Describe a method the student could use to carry out this investigation to obtain quantitative, reliable results. Your description should include: (1) how to vary the independent variable to obtain five different concentrations, (2) how to control two key confounding variables, (3) how to measure the dependent variable, (4) how to ensure reliability of the results, and (5) a safety hazard and corresponding precaution.
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Worked solution

1. Varying the Independent Variable: Prepare a simple dilution series of five copper sulfate concentrations from the 0.10 mol dm^{-3} stock solution. For example, to make 10 cm^{3} of each: (a) 0.10 mol dm^{-3}: 10 cm^{3} stock, 0 cm^{3} water; (b) 0.08 mol dm^{-3}: 8 cm^{3} stock, 2 cm^{3} water; (c) 0.06 mol dm^{-3}: 6 cm^{3} stock, 4 cm^{3} water; (d) 0.04 mol dm^{-3}: 4 cm^{3} stock, 6 cm^{3} water; (e) 0.02 mol dm^{-3}: 2 cm^{3} stock, 8 cm^{3} water. Distilled water alone (0.00 mol dm^{-3}) serves as the negative control. 2. Control of Confounding Variables: (a) Temperature: Keep constant using a thermostatically controlled water bath set at 35 degrees Celsius, as respiration is enzyme-controlled. Allow reactants to equilibrate in the water bath for 5 minutes before mixing. (b) Concentration and Volume of Reactants: Use exactly 5 cm^{3} of 10% yeast suspension and 5 cm^{3} of 5% glucose solution in each test. 3. Measuring the Dependent Variable: Add 1 cm^{3} of 1.0% methylene blue indicator. Start a stopwatch immediately. Measure the time taken in seconds for the blue color to completely disappear (decolorise to a creamy pink/white end-point), comparing against a fully decolorised standard tube. Calculate the rate of respiration as \( 1 / \text{time} \) (s^{-1}). 4. Reliability: Perform at least three replicates for each concentration of copper sulfate, calculate the mean rate, and identify any anomalous results. 5. Safety: Copper sulfate is toxic and a skin/eye irritant. Wear safety goggles and protective nitrile gloves, and wash hands immediately if skin contact occurs.

Marking scheme

[Method - maximum 6 marks] 1. Correct volumes of stock solution and distilled water specified to make 5 different concentrations. 2. Standardisation of temperature using a water bath at a stated optimum temperature (e.g. 30 to 40 degrees Celsius). 3. Yeast and glucose volumes and concentrations kept constant (e.g. 5 cm^{3} of 10% yeast and 5 cm^{3} of 5% glucose). 4. Allowing tubes to equilibrate to the water bath temperature before starting the reaction. 5. Standardising the endpoint of decolorisation (using a blank/color standard tube for comparison). 6. Use of a negative control (distilled water instead of copper sulfate) to show respiration without inhibitor. [Reliability and Analysis - maximum 3 marks] 7. Replicates: Repeating the experiment at least three times at each concentration. 8. State how rate is calculated: Rate = 1 / time taken (s^{-1}). 9. Detail of calculating a mean and identifying anomalies. [Safety - maximum 1 mark] 10. Risk: Copper sulfate is toxic/irritant AND Precaution: wear safety goggles/gloves to prevent contact.
Question 2 · Experimental Design and Statistical Analysis
10 marks
A student investigated the stomatal density on the lower epidermis of leaves from a single species of tree grown in two different environments: high light exposure (sun leaves) and shaded exposure (shade leaves). The results are: Sun leaves: mean = 184 stomata mm^{-2}, standard deviation = 15.4, sample size = 15. Shade leaves: mean = 142 stomata mm^{-2}, standard deviation = 12.8, sample size = 15. (a) State a suitable null hypothesis for this investigation. (b) Explain why a Student's t-test is appropriate for analyzing these data. (c) Calculate the t-value using the formula: \( t = \frac{|\bar{x}_1 - \bar{x}_2|}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \). Show your working and give your answer to two decimal places. (d) State the degrees of freedom for this test. (e) The critical value of t at p = 0.05 for the calculated degrees of freedom is 2.048. Explain what conclusion can be drawn from the calculated t-value.
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Worked solution

(a) Null Hypothesis: There is no significant difference between the mean stomatal density of sun leaves and shade leaves. (b) Why t-test is appropriate: The data is continuous, follows a normal distribution, and compares the means of two distinct, independent groups. (c) Calculation: \( \bar{x}_1 = 184 \), \( s_1 = 15.4 \), \( n_1 = 15 \); \( \bar{x}_2 = 142 \), \( s_2 = 12.8 \), \( n_2 = 15 \). Numerator: \( |184 - 142| = 42 \). Denominator: \( \sqrt{\frac{15.4^2}{15} + \frac{12.8^2}{15}} = \sqrt{\frac{237.16}{15} + \frac{163.84}{15}} = \sqrt{15.811 + 10.923} = \sqrt{26.734} = 5.1705 \). Therefore, \( t = 42 / 5.1705 = 8.123 \), which rounds to 8.12. (d) Degrees of freedom: \( (n_1 - 1) + (n_2 - 1) = (15 - 1) + (15 - 1) = 28 \). (e) Conclusion: Since the calculated t-value of 8.12 is greater than the critical value of 2.048 at the p = 0.05 significance level, the null hypothesis is rejected. The difference in stomatal density between sun leaves and shade leaves is statistically significant, meaning there is less than a 5% probability that this difference is due to random chance.

Marking scheme

[Part a - 1 mark] 1. Correct null hypothesis stating 'no significant difference between the means'. [Part b - 2 marks] 2. Mentioning comparison of two means / independent groups. 3. Mentioning data is continuous / normally distributed. [Part c - 3 marks] 4. Correct calculation of variance over sample sizes: 15.81 and 10.92 (allow 15.8 and 10.9). 5. Correct calculation of the denominator: sqrt(26.73) = 5.17. 6. Final calculated t-value of 8.12 (accept 8.12 or 8.13; reject other values). [Part d - 1 mark] 7. Correct degrees of freedom = 28. [Part e - 3 marks] 8. Comparison: States that calculated t-value is greater than the critical value (8.12 > 2.048). 9. Reject null hypothesis. 10. Conclude that the difference in mean stomatal density is statistically significant / probability of being due to chance is less than 5% (p < 0.05).
Question 3 · Experimental Design and Statistical Analysis
10 marks
An ecological survey investigated soil biodiversity near a heavy-metal smelting factory. Species richness of soil macroinvertebrates was measured at six sites at increasing distances from the factory. The raw data and ranks are as follows: Site A (1.0 km, Rank 1; Species richness 4, Rank 1), Site B (2.5 km, Rank 2; Species richness 9, Rank 3), Site C (5.0 km, Rank 3; Species richness 8, Rank 2), Site D (7.5 km, Rank 4; Species richness 14, Rank 4), Site E (10.0 km, Rank 5; Species richness 15, Rank 5), Site F (15.0 km, Rank 6; Species richness 23, Rank 6). (a) Calculate the Spearman's rank correlation coefficient (\( r_s \)) using the formula: \( r_s = 1 - \left( \frac{6 \sum d^2}{n(n^2 - 1)} \right) \). Show your working. (b) State what your calculated \( r_s \) value indicates about the relationship. (c) At p = 0.05 and n = 6, the critical value for \( r_s \) is 0.886. State the conclusion that can be drawn from this statistical test. (d) Identify two abiotic soil factors that should have been controlled or measured at each site, and explain why each factor should be controlled.
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Worked solution

(a) Calculation: Step 1: Find differences (d) and d^2. Site A: d = 1 - 1 = 0, d^2 = 0; Site B: d = 2 - 3 = -1, d^2 = 1; Site C: d = 3 - 2 = 1, d^2 = 1; Site D: d = 4 - 4 = 0, d^2 = 0; Site E: d = 5 - 5 = 0, d^2 = 0; Site F: d = 6 - 6 = 0, d^2 = 0. Sum of d^2 = 0 + 1 + 1 + 0 + 0 + 0 = 2. Step 2: Denominator = n(n^2 - 1) = 6(36 - 1) = 6 * 35 = 210. Step 3: \( r_s = 1 - (6 * 2 / 210) = 1 - (12 / 210) = 1 - 0.0571 = 0.9429 \). Rounded to two decimal places, \( r_s = 0.94 \). (b) Relationship: There is a strong, positive correlation between distance from the factory and the species richness of soil macroinvertebrates (as distance increases, species richness increases). (c) Conclusion: Since the calculated \( r_s \) value of 0.94 is greater than the critical value of 0.886, the correlation is statistically significant at the 5% level (p < 0.05). The null hypothesis is rejected, and there is a genuine positive relationship between the two variables. (d) Abiotic factors: 1. Soil pH: Heavy metal solubility and toxicity change drastically with pH, and soil pH directly affects macroinvertebrate survival. 2. Soil moisture content: Water availability is a primary limiting factor for soil organisms, and differences in dampness across sites would confound the effect of distance/pollution.

Marking scheme

[Part a - 3 marks] 1. Correct calculation of sum of d^2 = 2. 2. Correct substitution into the formula: rs = 1 - (12 / 210). 3. Final calculated rs value of 0.94 (accept 0.94 to 0.943). [Part b - 2 marks] 4. State that it is a positive correlation. 5. State that the correlation is strong/highly correlated. [Part c - 2 marks] 6. Correct comparison: calculated rs value (0.94) is greater than the critical value (0.886). 7. Conclude that the correlation is statistically significant (null hypothesis rejected). [Part d - 3 marks] 8. Identify a first valid abiotic factor (e.g., soil pH, soil moisture, organic matter, soil temperature). 9. Identify a second valid abiotic factor. 10. Provide a correct matching explanation for why at least one factor must be controlled (e.g., soil pH affects metal toxicity; moisture directly affects hydration and survival of soft-bodied invertebrates).

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