Cambridge IAL · Thinka-original Practice Paper

2024 Cambridge IAL Biology (9700) Practice Paper with Answers

Thinka Jun 2024 (V2) Cambridge International A Level-Style Mock — Biology (9700)

270 marks465 mins2024

An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V2) Cambridge International A Level Biology (9700) paper. Not affiliated with or reproduced from Cambridge.

Paper 12 - Multiple Choice

Answer all 40 multiple-choice questions on the answer sheet. Each correct answer scores 1 mark.

40 Question · 40 marks

Question 1 · Multiple Choice

1 marks

A mitochondrion in an electron micrograph has an image length of \(45\text{ mm}\). The magnification of the micrograph is \(\times 30\,000\). What is the actual length of the mitochondrion in micrometers (\(\mu\text{m}\))?

A.0.15
B.1.5
C.15
D.150

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Worked solution

Actual size \((A) = I / M\). Here, the image length \((I)\) is \(45\text{ mm}\), which is equal to \(45\,000\text{ }\mu\text{m}\). The magnification \((M)\) is \(30\,000\). Therefore, \(A = 45\,000 / 30\,000 = 1.5\text{ }\mu\text{m}\).

Marking scheme

1 mark for the correct answer B (1.5). Correct conversion of mm to \(\mu\text{m}\) and application of the magnification formula.

Question 2 · Multiple Choice

1 marks

Which statement describes a feature that is common to both amylose and cellulose?

A.Both are unbranched polysaccharides containing only \(\beta\)-glucose monomers.
B.Both contain 1,4-glycosidic bonds formed by condensation reactions.
C.Both form helical structures stabilized by hydrogen bonds between hydroxyl groups.
D.Both can be hydrolyzed by the same enzyme to produce maltose.

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Worked solution

Amylose is an unbranched polymer of \(\alpha\)-glucose, while cellulose is an unbranched polymer of \(\beta\)-glucose. Both polymers consist of monomers linked by 1,4-glycosidic bonds formed through condensation reactions. Amylose forms a helix, whereas cellulose forms straight, flat chains. They require different enzymes (amylase and cellulase, respectively) for hydrolysis.

Marking scheme

1 mark for the correct option B. Distractors are incorrect because: A is incorrect as amylose contains \(\alpha\)-glucose; C is incorrect because cellulose does not form a helix; D is incorrect because different enzymes hydrolyze them.

Question 3 · Multiple Choice

1 marks

An experiment was carried out to investigate the effect of an inhibitor on an enzyme-catalyzed reaction. The rate of reaction was measured at different substrate concentrations in the absence and in the presence of the inhibitor. The maximum rate of reaction (\(V_{\max}\)) was unchanged, but the Michaelis-Menten constant (\(K_{\text{m}}\)) was increased.

Which statement is correct about this inhibitor?

A.It is a non-competitive inhibitor that binds to an allosteric site on the enzyme.
B.It is a non-competitive inhibitor that binds to the active site of the enzyme.
C.It is a competitive inhibitor that binds to an allosteric site on the enzyme.
D.It is a competitive inhibitor that binds to the active site of the enzyme.

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Worked solution

A competitive inhibitor has a similar structure to the substrate and binds to the active site of the enzyme, competing with the substrate. Because it can be outcompeted by high substrate concentrations, the maximum rate of reaction (\(V_{\max}\)) remains unchanged. However, a higher concentration of substrate is needed to reach half of \(V_{\max}\), which means the Michaelis-Menten constant (\(K_{\text{m}}\)) is increased.

Marking scheme

1 mark for the correct option D. Non-competitive inhibitors decrease \(V_{\max}\) and typically bind to an allosteric site. Competitive inhibitors bind to the active site and increase \(K_{\text{m}}\) while keeping \(V_{\max}\) constant.

Question 4 · Multiple Choice

1 marks

Plant cells with a water potential of \(-600\text{ kPa}\) were placed in four different sucrose solutions: P, Q, R, and S. The water potentials of the sucrose solutions are shown in the list:

- Solution P: \(-200\text{ kPa}\)
- Solution Q: \(-400\text{ kPa}\)
- Solution R: \(-600\text{ kPa}\)
- Solution S: \(-800\text{ kPa}\)

In which solution(s) would net movement of water be out of the plant cells, potentially leading to plasmolysis?

A.P and Q only
B.S only
C.P, Q and R
D.Q and S only

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Worked solution

Water moves by osmosis down a water potential gradient, from a higher (less negative) water potential to a lower (more negative) water potential. For there to be a net movement of water out of the cell, the external solution must have a lower (more negative) water potential than the cell. The cell water potential is \(-600\text{ kPa}\), so water will move out of the cell only into solution S (\(-800\text{ kPa}\)), which is hypertonic relative to the cell, leading to plasmolysis.

Marking scheme

1 mark for B. Correct application of water potential gradients and osmosis principles.

Question 5 · Multiple Choice

1 marks

During water transport across the root cortex of a plant, water can travel via different pathways. Which statement about these pathways is correct?

A.In the apoplast pathway, water moves through the cytoplasm via plasmodesmata.
B.In the symplast pathway, water moves through the cellulose cell walls and intercellular spaces.
C.The Casparian strip in the endodermis blocks the apoplast pathway, forcing water to enter the symplast pathway.
D.Water movement through the symplast pathway is entirely passive and never crosses any cell membranes.

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Worked solution

The Casparian strip is a suberized, waterproof barrier located in the cell walls of endodermal cells. It blocks the apoplast pathway (which is the movement through non-living cell walls and intercellular spaces), forcing water and dissolved ions to cross the selectively permeable cell membrane of the endodermal cell and enter the symplast pathway (movement through living cytoplasm and plasmodesmata).

Marking scheme

1 mark for C. Distractor analysis: A is incorrect because apoplast is cell wall pathway; B is incorrect because symplast is cytoplasm pathway; D is incorrect because water must cross at least one cell membrane to enter the symplast.

Question 6 · Multiple Choice

1 marks

What is the role of the enzyme carbonic anhydrase in the transport of carbon dioxide by the blood?

A.It catalyzes the dissociation of carbonic acid into hydrogen ions and hydrogencarbonate ions.
B.It catalyzes the reaction between carbon dioxide and water to form carbonic acid inside red blood cells.
C.It binds directly to carbon dioxide to transport it through the plasma as carbaminohaemoglobin.
D.It actively transports hydrogencarbonate ions out of red blood cells in exchange for chloride ions.

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Worked solution

Carbonic anhydrase is an enzyme found inside red blood cells that catalyzes the reversible hydration reaction: \(\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3\). This allows carbon dioxide to be quickly converted into carbonic acid, which then spontaneously dissociates into hydrogen ions and hydrogencarbonate ions.

Marking scheme

1 mark for B. Correct identification of the role of carbonic anhydrase. A is incorrect because the dissociation of carbonic acid is spontaneous. C describes haemoglobin, not carbonic anhydrase. D describes the chloride shift, which is facilitated by an anion exchange protein, not carbonic anhydrase.

Question 7 · Multiple Choice

1 marks

Which row correctly identifies the type of immunity acquired by a newborn baby receiving antibodies from breast milk, and the type of immunity acquired by an individual receiving a tetanus toxoid injection?

A.Antibodies from breast milk: natural passive; Tetanus toxoid injection: artificial active
B.Antibodies from breast milk: natural active; Tetanus toxoid injection: artificial passive
C.Antibodies from breast milk: artificial passive; Tetanus toxoid injection: natural active
D.Antibodies from breast milk: artificial active; Tetanus toxoid injection: natural passive

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Worked solution

Antibodies from breast milk provide natural passive immunity because the antibodies are transferred naturally from mother to baby, and the baby does not produce its own antibodies or memory cells. A tetanus toxoid injection contains inactivated toxin (an antigen) which stimulates the recipient's own immune system to produce antibodies and memory cells, representing artificial active immunity.

Marking scheme

1 mark for A. Distractors correctly identified as incorrect based on active/passive and natural/artificial classification definitions.

Question 8 · Multiple Choice

1 marks

Which processes of aerobic respiration produce both carbon dioxide and reduced NAD?

A.glycolysis and the Link reaction
B.the Link reaction and the Krebs cycle
C.the Krebs cycle and oxidative phosphorylation
D.glycolysis, the Link reaction and the Krebs cycle

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Worked solution

The Link reaction produces carbon dioxide (decarboxylation) and reduced NAD (oxidation). The Krebs cycle also produces both carbon dioxide (during decarboxylation steps) and reduced NAD (during dehydrogenase steps). Glycolysis produces reduced NAD but no carbon dioxide. Oxidative phosphorylation utilizes reduced NAD and does not produce carbon dioxide.

Marking scheme

1 mark for B. Correctly identifying the products of glycolysis (reduced NAD, ATP, pyruvate), Link reaction (acetyl CoA, CO2, reduced NAD), Krebs cycle (CO2, reduced NAD, reduced FAD, ATP), and oxidative phosphorylation (ATP, H2O).

Question 9 · multiple_choice

1 marks

Cells of a photosynthetic eukaryote were disrupted and centrifuged to separate their organelles. The centrifugal force was increased step-wise, producing different fractions.

Four fractions (1, 2, 3, and 4) were collected:
- Fraction 1 contained the largest, heaviest structures, including nuclei.
- Fraction 2 contained structures containing green pigments.
- Fraction 3 contained structures capable of producing large amounts of ATP under aerobic conditions.
- Fraction 4 contained very small structures consisting only of RNA and proteins.

Which statement about these fractions is correct?

A.Fraction 2 organelles possess a double membrane, circular DNA, and 70S ribosomes.
B.Fraction 3 contains organelles that are the primary site of transcription of ribosomal RNA in the cell.
C.Fraction 4 contains organelles surrounded by a single phospholipid bilayer.
D.The structures in Fraction 1 contain the only double-stranded DNA in the cell.

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Worked solution

During cell fractionation:
- Fraction 1 contains nuclei (largest/heaviest).
- Fraction 2 contains chloroplasts (possess green chlorophyll pigments).
- Fraction 3 contains mitochondria (produce ATP via aerobic respiration).
- Fraction 4 contains ribosomes (small, non-membrane bound, consisting of rRNA and proteins).

Chloroplasts (Fraction 2) are double-membrane bound organelles containing circular DNA and 70S ribosomes, supporting the endosymbiotic theory. This makes statement A correct.

Statement B is incorrect because ribosomal RNA is primarily transcribed in the nucleolus of the nucleus (Fraction 1).
Statement C is incorrect because ribosomes are not membrane-bound.
Statement D is incorrect because mitochondria (Fraction 3) and chloroplasts (Fraction 2) also contain double-stranded circular DNA.

Marking scheme

1 mark for the correct option A.

Question 10 · multiple_choice

1 marks

A linear oligosaccharide consists of five hexose residues: three \(\alpha\)-glucose molecules and two \(\beta\)-fructose molecules.

During complete acid hydrolysis of this oligosaccharide, how many water molecules are used, and what are the products?

A.4 water molecules used; producing 3 \(\alpha\)-glucose and 2 \(\beta\)-fructose molecules
B.4 water molecules used; producing 5 glucose molecules
C.5 water molecules used; producing 3 \(\alpha\)-glucose and 2 \(\beta\)-fructose molecules
D.5 water molecules used; producing 5 glucose molecules

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Worked solution

A linear oligosaccharide consisting of five monomer units has four glycosidic bonds linking them together (number of bonds = \(n - 1 = 4\)).

Hydrolysis is the chemical breakdown of a compound due to reaction with water. To completely break down four glycosidic bonds, four molecules of water must be added.

The hydrolysis of these bonds yields the original monomer units: three \(\alpha\)-glucose molecules and two \(\beta\)-fructose molecules.

Marking scheme

1 mark for the correct option A.

Question 11 · multiple_choice

1 marks

An enzyme-catalysed reaction was studied under three conditions:
1. Enzyme + substrate (control)
2. Enzyme + substrate + Inhibitor X
3. Enzyme + substrate + Inhibitor Y

The experimental results showed:
- For Inhibitor X, the maximum rate of reaction (\(V_{\max}\)) was unchanged, but a higher substrate concentration was required to reach \(\frac{1}{2} V_{\max}\).
- For Inhibitor Y, the maximum rate of reaction (\(V_{\max}\)) was reduced, but the substrate concentration required to reach \(\frac{1}{2} V_{\max}\) was unchanged.

Which statements about the inhibitors are correct?
1. Inhibitor X binds to the active site of the enzyme.
2. Inhibitor Y binds to an allosteric site on the enzyme.
3. The effect of Inhibitor X can be overcome by increasing substrate concentration.
4. Inhibitor Y decreases the affinity of the enzyme for its substrate.

A.1, 2, 3 and 4
B.1, 2 and 3 only
C.1 and 3 only
D.2 and 4 only

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Worked solution

Analysis of the inhibitors:
- Inhibitor X does not alter \(V_{\max}\) but increases \(K_{\mathrm{m}}\) (the substrate concentration at \(\frac{1}{2} V_{\max}\)). This is characteristic of a competitive inhibitor. Competitive inhibitors bind to the active site (Statement 1) and can be outcompeted by high substrate concentrations (Statement 3).
- Inhibitor Y decreases \(V_{\max}\) but leaves \(K_{\mathrm{m}}\) unchanged. This is characteristic of a non-competitive inhibitor. Non-competitive inhibitors bind to an allosteric site away from the active site (Statement 2). Since \(K_{\mathrm{m}}\) is unchanged, the enzyme's affinity for the substrate is not affected, making Statement 4 incorrect.

Marking scheme

1 mark for the correct option B.

Question 12 · multiple_choice

1 marks

Four identical cylinders of potato tissue were placed in different concentrations of sucrose solution for three hours. The initial mass of each cylinder was 5.0 g.

The table shows the final mass of the cylinders.

| Cylinder | Sucrose concentration / \(\text{mol dm}^{-3}\) | Final mass / g |
| :---: | :---: | :---: |
| 1 | 0.0 | 5.8 |
| 2 | 0.2 | 5.3 |
| 3 | 0.4 | 4.8 |
| 4 | 0.6 | 4.4 |

Which statement is correct?

A.The solute potential of the potato cells before the experiment is equivalent to a sucrose solution between 0.0 and 0.2 \(\text{mol dm}^{-3}\).
B.In cylinder 3, the water potential of the solution was higher than the water potential inside the potato cells.
C.At a sucrose concentration of approximately 0.32 \(\text{mol dm}^{-3}\), there is no net movement of water into or out of the potato cells.
D.The water potential of the potato cells in cylinder 4 became less negative during the experiment.

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Worked solution

The point of no net movement of water (where the water potential of the cells equals the water potential of the external sucrose solution) occurs when there is no change in mass (final mass = 5.0 g).

Looking at the table:
- At 0.2 \(\text{mol dm}^{-3}\), the mass increases to 5.3 g (net water gain of +0.3 g).
- At 0.4 \(\text{mol dm}^{-3}\), the mass decreases to 4.8 g (net water loss of -0.2 g).

The point of zero mass change lies between 0.2 and 0.4 \(\text{mol dm}^{-3}\). By linear interpolation:
\(0.2 + \left(\frac{0.3}{0.3 - (-0.2)}\right) \times (0.4 - 0.2) = 0.2 + 0.12 = 0.32\) \(\text{mol dm}^{-3}\).
Therefore, statement C is correct.

- Statement A is incorrect because the concentration of equivalent solute potential is between 0.2 and 0.4 \(\text{mol dm}^{-3}\).
- Statement B is incorrect because the potato lost mass in cylinder 3, meaning water moved out down a water potential gradient, so the solution water potential was lower (more negative) than inside the cells.
- Statement D is incorrect because losing water concentrates the vacuolar sap, making the water potential more negative.

Marking scheme

1 mark for the correct option C.

Question 13 · multiple_choice

1 marks

Which sequence correctly describes the pathway of a water molecule moving through a leaf during transpiration?

A.xylem vessels \(\rightarrow\) cytoplasm of mesophyll cells \(\rightarrow\) vacuole \(\rightarrow\) stomatal pore
B.xylem vessels \(\rightarrow\) cell walls of mesophyll cells \(\rightarrow\) evaporation into air spaces \(\rightarrow\) diffusion of water vapour through stomata
C.xylem vessels \(\rightarrow\) symplast of mesophyll cells \(\rightarrow\) intercellular air spaces as liquid water \(\rightarrow\) diffusion through stomata
D.xylem vessels \(\rightarrow\) companion cells \(\rightarrow\) phloem sieve tubes \(\rightarrow\) evaporation to air spaces \(\rightarrow\) stomatal pore

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Worked solution

Water is transported to the leaves in xylem vessels. It then moves into the hydrophilic cell walls of the mesophyll cells (the apoplastic pathway). At the surface of these cell walls, the liquid water evaporates into the intercellular air spaces as water vapour. Finally, water vapour diffuses down its water potential gradient out of the leaf through the stomata into the atmosphere.

Marking scheme

1 mark for the correct option B.

Question 14 · multiple_choice

1 marks

Which reactions and processes occur in the red blood cells of a mammal as they pass through actively respiring muscle tissue?

1. Carbonic anhydrase catalyses the hydration of carbon dioxide to form carbonic acid.
2. Hydrogencarbonate ions (\(\text{HCO}_3^-\)) diffuse out of the red blood cells into the plasma.
3. Chloride ions (\(\text{Cl}^-\)) diffuse out of the red blood cells into the plasma to maintain electrical neutrality.
4. Oxyhaemoglobin dissociates more readily because of the decrease in pH.

A.1, 2, 3 and 4
B.1, 2 and 4 only
C.2 and 3 only
D.1 and 4 only

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Worked solution

In actively respiring muscle tissue, high levels of carbon dioxide diffuse into the red blood cells:
1. Carbonic anhydrase catalyses: \(\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3\) (carbonic acid). (Statement 1 is correct).
2. Carbonic acid dissociates into hydrogen ions (\(\text{H}^+\)) and hydrogencarbonate ions (\(\text{HCO}_3^-\)).
3. \(\text{HCO}_3^-\) ions diffuse out of the red blood cells into the plasma via a transport protein. (Statement 2 is correct).
4. To maintain electrical neutrality, chloride ions (\(\text{Cl}^-\)) diffuse INTO the red blood cells from the plasma (the chloride shift). Thus, statement 3 is incorrect because it says chloride ions diffuse out.
5. The accumulated \(\text{H}^+\) ions bind to haemoglobin (forming haemoglobinic acid), which buffers the cell and lowers the affinity of haemoglobin for oxygen (Bohr effect), causing oxyhaemoglobin to release oxygen more readily in acidic conditions. (Statement 4 is correct).

Marking scheme

1 mark for the correct option B.

Question 15 · multiple_choice

1 marks

During non-cyclic photophosphorylation in photosynthesis, what is the correct sequence of electron transfer and the immediate source of electrons for replacing those lost by photosystem II (PSII)?

A.Sequence of electron transfer: \(\text{water} \rightarrow \text{PSII} \rightarrow \text{electron transport chain} \rightarrow \text{PSI} \rightarrow \text{NADP}\)

Source of electrons to replace those lost by PSII: photolysis of water
B.Sequence of electron transfer: \(\text{water} \rightarrow \text{PSI} \rightarrow \text{electron transport chain} \rightarrow \text{PSII} \rightarrow \text{NADP}\)

Source of electrons to replace those lost by PSII: reduced NADP
C.Sequence of electron transfer: \(\text{PSI} \rightarrow \text{electron transport chain} \rightarrow \text{PSII} \rightarrow \text{NADP}\)

Source of electrons to replace those lost by PSII: photolysis of water
D.Sequence of electron transfer: \(\text{PSII} \rightarrow \text{electron transport chain} \rightarrow \text{PSI} \rightarrow \text{electron transport chain} \rightarrow \text{water}\)

Source of electrons to replace those lost by PSII: reduced NADP

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Worked solution

In non-cyclic photophosphorylation (the Z-scheme):
1. Photolysis of water splits water molecules into oxygen, protons (\(\text{H}^+\)), and electrons: \(2\text{H}_2\text{O} \rightarrow 4\text{H}^+ + 4\text{e}^- + \text{O}_2\). These electrons replace those lost by the reaction centre of PSII (chlorophyll \(a\), P680) when it is photoactivated.
2. Photoactivated electrons from PSII pass along an electron transport chain to PSI.
3. Photoactivated electrons from PSI are used along with protons to reduce \(\text{NADP}^+\) to reduced NADP (NADPH).
Thus, the correct sequence is: \(\text{water} \rightarrow \text{PSII} \rightarrow \text{electron transport chain} \rightarrow \text{PSI} \rightarrow \text{NADP}\), and the electron source for PSII is the photolysis of water.

Marking scheme

1 mark for the correct option A.

Question 16 · multiple_choice

1 marks

Which row correctly identifies the concentrations of glucose, urea, and proteins in different fluids of a healthy mammalian kidney?

A.Glucose: concentration in glomerular filtrate is the same as in blood plasma; concentration in urine is zero.
B.Urea: concentration in glomerular filtrate is lower than in blood plasma; concentration in urine is the same as in glomerular filtrate.
C.Proteins: concentration in glomerular filtrate is the same as in blood plasma; concentration in urine is zero.
D.Glucose: concentration in glomerular filtrate is lower than in blood plasma; concentration in urine is higher than in glomerular filtrate.

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Worked solution

- **Glucose**: Glucose is a small monomer that is freely filtered through the basement membrane into Bowman's capsule, so its concentration in the glomerular filtrate is the **same** as in blood plasma. In a healthy individual, 100% of the filtered glucose is selectively reabsorbed in the proximal convoluted tubule (PCT) via active co-transport, meaning its concentration in the urine is **zero**.
- **Urea**: Being a small molecule, urea is also freely filtered, so its concentration in the glomerular filtrate is the **same** as in blood plasma. It is not fully reabsorbed, and as water is reabsorbed along the nephron, the concentration of urea in the urine becomes **much higher** than in the glomerular filtrate.
- **Proteins**: Most plasma proteins are too large to pass through the filtration barriers (basement membrane), so their concentration in the glomerular filtrate is **extremely low or absent**.

Marking scheme

1 mark for the correct option A.

Question 17 · Multiple Choice

1 marks

Which row correctly matches the cellular structures with their membrane systems?

A.Single membrane: lysosome, tonoplast; Double membrane: nucleus, mitochondrion; No membrane: ribosome, centriole
B.Single membrane: lysosome, ribosome; Double membrane: nucleus, chloroplast; No membrane: microtubule, nucleolus
C.Single membrane: Golgi body, tonoplast; Double membrane: nucleus, lysosome; No membrane: ribosome, centriole
D.Single membrane: mitochondrion, tonoplast; Double membrane: nucleus, chloroplast; No membrane: ribosome, nucleolus

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Worked solution

Lysosomes and the tonoplast (the membrane of the vacuole) are surrounded by a single membrane. The nucleus and mitochondria are bounded by double membranes (envelopes). Ribosomes and centrioles do not have any membrane. Therefore, row A is correct.

Marking scheme

Correct option is A (1 mark). Any other option is incorrect (0 marks).

Question 18 · Multiple Choice

1 marks

Four statements are made about glycogen, amylose, and amylopectin:

1. Amylose and amylopectin both contain \(\alpha\)-1,4-glycosidic bonds.
2. Glycogen contains more \(\alpha\)-1,6-glycosidic bonds per unit mass than amylopectin.
3. Amylose is a highly branched polysaccharide, whereas amylopectin is an unbranched helical chain.
4. Glycogen and amylopectin are both major storage polysaccharides found within plant starch granules.

Which statements are correct?

A.1 and 2 only
B.1 and 3 only
C.2 and 4 only
D.3 and 4 only

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Worked solution

Statement 1 is correct: both amylose (linear unbranched) and amylopectin (branched) contain \(\alpha\)-1,4-glycosidic bonds. Statement 2 is correct: glycogen is highly branched with a higher frequency of \(\alpha\)-1,6-glycosidic bonds than amylopectin. Statement 3 is incorrect: amylose is unbranched and helical, while amylopectin is branched. Statement 4 is incorrect: glycogen is an animal and fungal storage polysaccharide, not found in plant starch granules.

Marking scheme

Correct option is A (1 mark). Any other option is incorrect (0 marks).

Question 19 · Multiple Choice

1 marks

An experiment was carried out to investigate the effect of an inhibitor on an enzyme-catalysed reaction.

In the presence of this inhibitor, the Michaelis-Menten constant (\(K_m\)) of the enzyme increased, but the maximum velocity (\(V_{max}\)) of the reaction remained unchanged.

Which type of inhibitor was used, and what is its effect on the apparent affinity of the enzyme for its substrate?

A.Competitive inhibitor; decreases the apparent affinity
B.Competitive inhibitor; increases the apparent affinity
C.Non-competitive inhibitor; decreases the apparent affinity
D.Non-competitive inhibitor; has no effect on the apparent affinity

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Worked solution

A competitive inhibitor competes with the substrate for the active site, which increases the substrate concentration needed to reach half of the maximum velocity, thereby increasing \(K_m\). Because the active site is still functionally identical and high substrate concentrations can overcome the inhibitor, \(V_{max}\) remains unchanged. An increased \(K_m\) value represents a decreased apparent affinity of the enzyme for its substrate.

Marking scheme

Correct option is A (1 mark). Any other option is incorrect (0 marks).

Question 20 · Multiple Choice

1 marks

Four adjacent plant cells (P, Q, R, and S) have the water potentials (\(\psi\)) and solute potentials (\(\psi_s\)) shown below:

- Cell P: \(\psi = -600\text{ kPa}\); \(\psi_s = -800\text{ kPa}\)
- Cell Q: \(\psi = -400\text{ kPa}\); \(\psi_s = -700\text{ kPa}\)
- Cell R: \(\psi = -500\text{ kPa}\); \(\psi_s = -900\text{ kPa}\)
- Cell S: \(\psi = -700\text{ kPa}\); \(\psi_s = -1000\text{ kPa}\)

Which option correctly identifies a direction of net water movement between these cells?

A.from Cell Q to Cell R
B.from Cell S to Cell P
C.from Cell P to Cell Q
D.from Cell R to Cell Q

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Worked solution

Net water movement always occurs from a cell with a higher (less negative) water potential (\(\psi\)) to a cell with a lower (more negative) water potential. Solute potential (\(\psi_s\)) is already factored into the net water potential value given, so it should not be calculated again. Comparing water potentials: Cell Q has \(\psi = -400\text{ kPa}\), and Cell R has \(\psi = -500\text{ kPa}\). Since \(-400\text{ kPa} > -500\text{ kPa}\), there will be a net movement of water from Cell Q to Cell R.

Marking scheme

Correct option is A (1 mark). Any other option is incorrect (0 marks).

Question 21 · Multiple Choice

1 marks

Which event causes the aortic semi-lunar valve to open during the cardiac cycle?

A.Pressure in the left ventricle becomes greater than the pressure in the aorta.
B.Pressure in the left ventricle becomes greater than the pressure in the left atrium.
C.Pressure in the left atrium becomes greater than the pressure in the left ventricle.
D.Pressure in the aorta becomes greater than the pressure in the left ventricle.

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Worked solution

Valves open and close solely due to relative differences in hydrostatic pressure. The aortic semi-lunar valve opens when the pressure inside the left ventricle exceeds the pressure inside the aorta, allowing blood to flow from the ventricle into the systemic circulation.

Marking scheme

Correct option is A (1 mark). Any other option is incorrect (0 marks).

Question 22 · Multiple Choice

1 marks

Which option correctly classifies the types of immunity described?

- Type 1: A baby receiving maternal antibodies via breast milk.
- Type 2: An individual receiving an injection of monoclonal antibodies after a venomous snake bite.
- Type 3: A child recovering from an infection of measles.
- Type 4: A person receiving an injection of a weakened pathogen (vaccination).

A.Type 1: natural passive; Type 2: artificial passive; Type 3: natural active; Type 4: artificial active
B.Type 1: artificial passive; Type 2: natural passive; Type 3: artificial active; Type 4: natural active
C.Type 1: natural passive; Type 2: artificial active; Type 3: natural active; Type 4: artificial passive
D.Type 1: natural active; Type 2: artificial passive; Type 3: natural passive; Type 4: artificial active

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Worked solution

Type 1 is natural passive immunity because maternal antibodies are transferred naturally without the baby making them. Type 2 is artificial passive immunity because pre-made antibodies are injected medically. Type 3 is natural active immunity because the body naturally encounters the live pathogen and initiates its own immune response. Type 4 is artificial active immunity because a harmless form of the antigen is introduced medically, triggering the body's own immune response.

Marking scheme

Correct option is A (1 mark). Any other option is incorrect (0 marks).

Question 23 · Multiple Choice

1 marks

For every molecule of glucose entering aerobic respiration, how many molecules of carbon dioxide (\(\text{CO}_2\)), reduced NAD, and reduced FAD are produced in the link reaction and the Krebs cycle combined?

A.6 \(\text{CO}_2\), 8 reduced NAD, 2 reduced FAD
B.4 \(\text{CO}_2\), 6 reduced NAD, 2 reduced FAD
C.6 \(\text{CO}_2\), 6 reduced NAD, 2 reduced FAD
D.4 \(\text{CO}_2\), 8 reduced NAD, 4 reduced FAD

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Worked solution

One molecule of glucose yields two molecules of pyruvate during glycolysis. In the link reaction, 2 pyruvates produce 2 \(\text{CO}_2\) and 2 reduced NAD. In the Krebs cycle, the 2 resulting acetyl-CoA molecules produce 4 \(\text{CO}_2\), 6 reduced NAD, and 2 reduced FAD. Thus, combined, they yield 6 \(\text{CO}_2\) (2 from link, 4 from Krebs), 8 reduced NAD (2 from link, 6 from Krebs), and 2 reduced FAD (0 from link, 2 from Krebs).

Marking scheme

Correct option is A (1 mark). Any other option is incorrect (0 marks).

Question 24 · Multiple Choice

1 marks

In the *lac* operon of *Escherichia coli*, which molecule binds to the operator in the absence of lactose, and what is the consequence of this binding?

A.The active repressor protein binds; transcription of the structural genes *lacZ* and *lacY* is prevented.
B.The active repressor protein binds; transcription of the structural genes *lacZ* and *lacY* is activated.
C.RNA polymerase binds; transcription of the structural genes *lacZ* and *lacY* is prevented.
D.Lactose binds; transcription of the structural genes *lacZ* and *lacY* is activated.

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Worked solution

In the absence of lactose, the active repressor protein (encoded by the regulatory gene *lacI*) binds directly to the operator. This blocks RNA polymerase from moving along the DNA from the promoter, preventing transcription of the structural genes *lacZ* and *lacY*.

Marking scheme

Correct option is A (1 mark). Any other option is incorrect (0 marks).

Question 25 · multiple-choice

1 marks

An enzyme-catalysed reaction is carried out in the presence of a competitive inhibitor. How do the Michaelis-Menten constant (\(K_m\)) and the maximum velocity (\(V_{max}\)) of the enzyme change compared to the reaction without the inhibitor?

A.The \(K_m\) increases and the \(V_{max}\) remains unchanged.
B.The \(K_m\) decreases and the \(V_{max}\) remains unchanged.
C.The \(K_m\) remains unchanged and the \(V_{max}\) decreases.
D.The \(K_m\) increases and the \(V_{max}\) decreases.

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Worked solution

A competitive inhibitor competes with the substrate for binding to the active site of the enzyme. This decreases the apparent affinity of the enzyme for its substrate, which increases the value of the Michaelis-Menten constant (\(K_m\)). However, because the inhibition can be overcome by high concentrations of substrate, the maximum rate of reaction (\(V_{max}\)) remains unchanged.

Marking scheme

[1 mark] A is the correct option.
- Reject B: Competitive inhibitors do not decrease \(K_m\).
- Reject C: This describes non-competitive inhibition.
- Reject D: Competitive inhibitors do not reduce the maximum possible velocity.

Question 26 · multiple-choice

1 marks

A scientist measures the rate of entry of a specific polar solute into animal cells across a range of external concentrations.

Initially, the rate of entry increases linearly with concentration, but then reaches a maximum level (plateaus) at high concentrations. When a respiratory inhibitor is added to the cells, the rate of entry of the solute remains completely unaffected.

Which mechanism of transport is responsible for the entry of this solute?

A.active transport
B.facilitated diffusion
C.simple diffusion
D.osmosis

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Worked solution

The plateau indicates that the transport mechanism is saturable, meaning that transport proteins (either carrier or channel proteins) are involved and become fully occupied at high concentrations. This rules out simple diffusion (which does not plateau). Because adding a respiratory inhibitor (which blocks ATP synthesis) has no effect on the rate, the mechanism must be passive and does not require ATP. Therefore, the process is facilitated diffusion.

Marking scheme

[1 mark] B is the correct option.
- Reject A: Active transport requires ATP, so its rate would be significantly reduced by a respiratory inhibitor.
- Reject C: Simple diffusion does not involve transport proteins, so its rate would continue to increase linearly without reaching a plateau.
- Reject D: Osmosis refers specifically to the net movement of water molecules, not a polar solute.

Question 27 · multiple-choice

1 marks

What is the correct sequence of structures and pathways that a water molecule passes through as it moves from the soil to the lumen of a root xylem vessel via the apoplast pathway?

A.root hair cell wall \(\rightarrow\) cell walls of cortex cells \(\rightarrow\) cytoplasm of endodermal cells \(\rightarrow\) xylem vessel elements
B.root hair cell membrane \(\rightarrow\) cell walls of cortex cells \(\rightarrow\) Casparian strip of endodermis \(\rightarrow\) xylem vessel elements
C.root hair cell wall \(\rightarrow\) cytoplasm of cortex cells \(\rightarrow\) Casparian strip of endodermis \(\rightarrow\) xylem vessel elements
D.root hair cell membrane \(\rightarrow\) cytoplasm of cortex cells \(\rightarrow\) cytoplasm of endodermal cells \(\rightarrow\) xylem vessel elements

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Worked solution

In the apoplast pathway, water moves through the cell walls of the root hair and cortex cells without crossing any plasma membranes. However, upon reaching the endodermis, the Casparian strip (made of suberin) blocks the apoplastic movement. Water is forced to cross the plasma membrane of the endodermal cell into its cytoplasm (switching to the symplast pathway) before it can enter the lumen of the xylem vessels.

Marking scheme

[1 mark] A is the correct option.
- Reject B: Water cannot pass through the hydrophobic Casparian strip in the cell wall.
- Reject C: Movement through cytoplasm represents the symplast pathway, not the apoplast pathway.
- Reject D: This is the symplast pathway entirely, and it starts with crossing the cell membrane.

Question 28 · multiple-choice

1 marks

Which statement correctly describes what occurs inside a red blood cell as it passes through actively respiring tissues?

A.Carbon dioxide diffuses into the cell and is converted to carbonic acid by carbonic anhydrase; hydrogencarbonate ions then diffuse out of the cell in exchange for chloride ions entering.
B.Carbon dioxide diffuses into the cell and is converted to carbonic acid by carbonic anhydrase; chloride ions then diffuse out of the cell in exchange for hydrogencarbonate ions entering.
C.Carbonic acid is formed from carbon dioxide and water in the plasma; hydrogencarbonate ions then diffuse into the cell in exchange for chloride ions leaving.
D.Oxygen and carbon dioxide compete for the same binding site on the haem group, forming carbaminohemoglobin and oxyhaemoglobin.

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Worked solution

In actively respiring tissues, carbon dioxide (\(\text{CO}_2\)) diffuses down its concentration gradient into the red blood cells. Inside, the enzyme carbonic anhydrase catalyses its reaction with water to form carbonic acid (\(\text{H}_2\text{CO}_3\)), which quickly dissociates into hydrogen ions (\(\text{H}^+\)) and hydrogencarbonate ions (\(\text{HCO}_3^-\)). The hydrogencarbonate ions diffuse out of the red blood cell down their concentration gradient into the plasma, and chloride ions (\(\text{Cl}^-\)) diffuse into the cell to maintain electrical neutrality (the chloride shift).

Marking scheme

[1 mark] A is the correct option.
- Reject B: Chloride ions move into the cell, and hydrogencarbonate ions move out of the cell, not vice versa.
- Reject C: Carbonic acid is formed primarily inside red blood cells due to the presence of carbonic anhydrase, not in the plasma.
- Reject D: Carbon dioxide binds to amine groups on the globin polypeptide chains of haemoglobin, not to the haem group where oxygen binds.

Question 29 · multiple-choice

1 marks

Monoclonal antibodies are produced using hybridoma cells. Why are antibody-producing plasma cells fused with myeloma (cancer) cells during this process?

A.To allow the resulting cells to divide repeatedly in culture and secrete a single, specific type of antibody.
B.To increase the affinity of the antibodies for the target antigen by inducing rapid mutations.
C.To cause the cancer cells to recognize and bind to the specific antigen.
D.To enable the plasma cells to synthesize and secrete several different types of antibodies simultaneously.

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Worked solution

Plasma cells are differentiated B-lymphocytes that produce a single type of specific antibody, but they have a short lifespan and cannot divide in cell culture. Myeloma (cancer) cells are immortal and divide rapidly. Fusing a plasma cell with a myeloma cell produces a hybridoma cell, which inherits both the ability to produce a single type of specific antibody and the ability to divide repeatedly in culture, ensuring a continuous source of monoclonal antibodies.

Marking scheme

[1 mark] A is the correct option.
- Reject B: Fusion does not aim to induce mutations to increase affinity.
- Reject C: Cancer cells are fused because of their division properties, not because they recognize the antigen.
- Reject D: Monoclonal antibodies are, by definition, identical and specific to a single epitope; hybridomas produce only one type of antibody, not several different types.

Question 30 · multiple-choice

1 marks

During the aerobic respiration of one molecule of glucose, how many molecules of carbon dioxide (\(\text{CO}_2\)) and reduced NAD are produced in total during the link reaction and the Krebs cycle combined?

A.4 \(\text{CO}_2\) and 6 reduced NAD
B.6 \(\text{CO}_2\) and 6 reduced NAD
C.6 \(\text{CO}_2\) and 8 reduced NAD
D.4 \(\text{CO}_2\) and 10 reduced NAD

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Worked solution

One molecule of glucose undergoes glycolysis to produce two molecules of pyruvate.

1. In the link reaction, each pyruvate is converted to acetyl-CoA, producing 1 \(\text{CO}_2\) and 1 reduced NAD. For two pyruvates, this yields 2 \(\text{CO}_2\) and 2 reduced NAD.
2. In the Krebs cycle, each acetyl-CoA yields 2 \(\text{CO}_2\) and 3 reduced NAD. For two acetyl-CoA molecules, this yields 4 \(\text{CO}_2\) and 6 reduced NAD.

Adding these together:
Total \(\text{CO}_2\) = 2 (link) + 4 (Krebs) = 6 molecules.
Total reduced NAD = 2 (link) + 6 (Krebs) = 8 molecules.

Marking scheme

[1 mark] C is the correct option.
- Reject A: This omits the products of the link reaction.
- Reject B: This incorrectly states that only 6 reduced NAD are produced in total.
- Reject D: This incorrectly states that only 4 \(\text{CO}_2\) are produced.

Question 31 · multiple-choice

1 marks

Red-green colour blindness is a sex-linked recessive condition caused by an allele carried on the X chromosome.

A woman with normal colour vision, whose father was red-green colour-blind, has children with a man who has normal colour vision.

What is the probability that their first child will be a red-green colour-blind son?

A.0.125
B.0.25
C.0.50
D.0.75

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Worked solution

Let \(X^B\) be the dominant allele for normal vision and \(X^b\) be the recessive allele for red-green colour blindness.

The woman's father was colour-blind (genotype \(X^b Y\)), meaning he must have passed his \(X^b\) chromosome to his daughter. Since she has normal vision, her genotype must be heterozygous: \(X^B X^b\).

The man has normal vision, so his genotype is \(X^B Y\).

Crossing these parents (\(X^B X^b \times X^B Y\)) yields the following offspring possibilities:
1. \(X^B X^B\) (normal-vision female)
2. \(X^B X^b\) (carrier female with normal vision)
3. \(X^B Y\) (normal-vision male)
4. \(X^b Y\) (colour-blind male)

The question asks for the probability that their first child will be a colour-blind son. Out of the 4 equally likely combinations, exactly 1 is a colour-blind son (\(X^b Y\)). Therefore, the probability is 1 in 4, or 0.25.

Marking scheme

[1 mark] B is the correct option.
- Reject A: 0.125 is incorrect.
- Reject C: 0.50 is the probability of a son being colour-blind given that the child is already known to be male, but the question asks for the overall probability of a first child being a colour-blind son.
- Reject D: 0.75 is incorrect.

Question 32 · multiple-choice

1 marks

Which statement correctly describes the movement of ions across the axon membrane during the depolarization and repolarization phases of an action potential?

A.During depolarization, sodium ions are actively pumped into the axon; during repolarization, potassium ions are actively pumped out of the axon.
B.During depolarization, sodium ions diffuse into the axon; during repolarization, potassium ions diffuse out of the axon.
C.During depolarization, potassium ions diffuse out of the axon; during repolarization, sodium ions diffuse into the axon.
D.During depolarization, sodium ions diffuse out of the axon; during repolarization, potassium ions diffuse into the axon.

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Worked solution

During depolarization, voltage-gated sodium channels open, allowing sodium ions (\(Na^+\)) to rapidly diffuse down their electrochemical gradient into the axon. During repolarization, voltage-gated sodium channels close and voltage-gated potassium channels open, allowing potassium ions (\(K^+\)) to diffuse down their electrochemical gradient out of the axon. Both phases involve passive diffusion through specific channel proteins, not active pumping.

Marking scheme

[1 mark] B is the correct option.
- Reject A: The movements during depolarization and repolarization are passive diffusion, not active pumping (the sodium-potassium pump is active but does not cause these phases directly).
- Reject C: This reverses the roles of potassium and sodium.
- Reject D: This incorrectly states that sodium diffuses out and potassium diffuses in.

Question 33 · multiple-choice

1 marks

A student isolates different structures and organelles from a healthy eukaryotic plant cell using cell fractionation. Four fractions (P, Q, R, and S) are obtained:

* **Fraction P** contains organelles with a double membrane that are the site of aerobic respiration.
* **Fraction Q** contains organelles bounded by a single membrane that are involved in the modification and packaging of proteins.
* **Fraction R** contains structures with no membrane that are the sites of translation.
* **Fraction S** contains organelles with a double membrane containing DNA and thylakoids.

Which organelles are present in fractions P, Q, R, and S?

A.P: Mitochondria; Q: Golgi apparatus; R: Ribosomes; S: Chloroplasts
B.P: Chloroplasts; Q: Lysosomes; R: Nucleoli; S: Mitochondria
C.P: Mitochondria; Q: Rough endoplasmic reticulum; R: Ribosomes; S: Nuclei
D.P: Nuclei; Q: Golgi apparatus; R: Microtubules; S: Chloroplasts

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Worked solution

* **Fraction P** has a double membrane and is the site of aerobic respiration, which corresponds to **mitochondria**.
* **Fraction Q** is bounded by a single membrane and modifies and packages proteins, which corresponds to the **Golgi apparatus**.
* **Fraction R** has no membrane and is the site of translation, which corresponds to **ribosomes**.
* **Fraction S** has a double membrane, DNA, and thylakoids, which corresponds to **chloroplasts**.

Marking scheme

Award 1 mark for the correct option (A).
- Reject other options because they mismatch organelle characteristics: nucleoli (option B) are sub-nuclear regions with no membrane; nuclei (option D) lack thylakoids; rough endoplasmic reticulum (option C) has a single membrane.

Question 34 · multiple-choice

1 marks

Which statements about glycogen, amylopectin, and cellulose are correct?

1. Glycogen and amylopectin both contain \(\alpha\)-1,4-glycosidic bonds and \(\alpha\)-1,6-glycosidic bonds.
2. Cellulose contains \(\beta\)-1,4-glycosidic bonds, where alternate glucose molecules are rotated by 180°.
3. Glycogen is more highly branched than amylopectin because it has a higher proportion of \(\alpha\)-1,6-glycosidic bonds.
4. Hydrogen bonds form cross-links between adjacent cellulose chains to form microfibrils.

A.1, 2, 3 and 4
B.1, 2 and 3 only
C.2 and 4 only
D.1 and 3 only

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Worked solution

All four statements are correct:
1. Both glycogen and amylopectin are branched molecules made of \(\alpha\)-glucose, with linear chains connected by \(\alpha\)-1,4-glycosidic bonds and branches formed by \(\alpha\)-1,6-glycosidic bonds.
2. Cellulose is a linear polymer of \(\beta\)-glucose with \(\beta\)-1,4-glycosidic bonds. To form these bonds, each successive glucose monomer must be rotated 180° relative to its neighbor.
3. Glycogen has branches occurring more frequently (every 8 to 12 glucose units) than in amylopectin (every 24 to 30 glucose units), meaning glycogen contains a higher density of \(\alpha\)-1,6-glycosidic bonds.
4. Parallel cellulose chains run side by side and are held together by numerous hydrogen bonds between hydroxyl groups, forming strong structural units called microfibrils.

Marking scheme

Award 1 mark for the correct option (A).
- Reject options B, C, and D as they omit correct statements.

Question 35 · multiple-choice

1 marks

An experiment was carried out to investigate the effect of an inhibitor on an enzyme-controlled reaction. The rate of reaction was measured at different substrate concentrations in the absence of the inhibitor, and in the presence of either inhibitor X or inhibitor Y.

The results are shown in the table below:

| Condition | \(V_{max}\) / arbitrary units | \(K_m\) / mmol dm\(^{-3}\) |
| :--- | :--- | :--- |
| Enzyme alone | 120 | 2.5 |
| Enzyme + Inhibitor X | 120 | 6.8 |
| Enzyme + Inhibitor Y | 45 | 2.5 |

What is the correct explanation of the mode of action of inhibitors X and Y?

A.Inhibitor X is a competitive inhibitor that binds reversibly to the active site; Inhibitor Y is a non-competitive inhibitor that binds to an allosteric site.
B.Inhibitor X is a non-competitive inhibitor that decreases the affinity of the enzyme for its substrate; Inhibitor Y is a competitive inhibitor that reduces the maximum velocity of the reaction.
C.Inhibitor X is a competitive inhibitor that increases the affinity of the enzyme for its substrate; Inhibitor Y is a non-competitive inhibitor that increases the rate of ES complex formation.
D.Inhibitor X is a non-competitive inhibitor that binds to the active site; Inhibitor Y is a competitive inhibitor that binds reversibly to an allosteric site.

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Worked solution

* **Inhibitor X** leaves the maximum velocity (\(V_{max}\)) unchanged at 120, but increases the Michaelis-Menten constant (\(K_m\)) from 2.5 to 6.8 mmol dm\(^{-3}\). This is characteristic of a **competitive inhibitor**, which binds reversibly to the active site. The inhibition can be overcome by high substrate concentrations, so \(V_{max}\) is unaffected, but more substrate is needed to reach half \(V_{max}\), meaning \(K_m\) increases.
* **Inhibitor Y** decreases \(V_{max}\) from 120 to 45 arbitrary units, but leaves \(K_m\) unchanged at 2.5 mmol dm\(^{-3}\). This is characteristic of a **non-competitive inhibitor**, which binds to an allosteric site. It reduces the concentration of active enzyme molecules, reducing \(V_{max}\), but does not alter the affinity of the unaffected active sites for the substrate, so \(K_m\) remains unchanged.

Marking scheme

Award 1 mark for the correct option (A).
- Reject B, C, and D because competitive inhibitors increase \(K_m\) without changing \(V_{max}\), and non-competitive inhibitors decrease \(V_{max}\) without changing \(K_m\).

Question 36 · multiple-choice

1 marks

Four transport mechanisms across a mammalian cell surface membrane are described below:

* **Mechanism 1**: Movement of a hydrophobic molecule directly through the phospholipid bilayer down its concentration gradient.
* **Mechanism 2**: Movement of ions through a specific membrane channel protein down their electrochemical gradient without the use of ATP.
* **Mechanism 3**: Movement of glucose against its concentration gradient using a membrane protein and energy from a sodium ion gradient established by the active transport of sodium ions.
* **Mechanism 4**: Movement of bulk secretory proteins out of the cell via vesicle fusion with the cell surface membrane.

Which row correctly identifies these transport mechanisms?

A.1: Simple diffusion, 2: Facilitated diffusion, 3: Secondary active transport, 4: Exocytosis
B.1: Simple diffusion, 2: Active transport, 3: Facilitated diffusion, 4: Endocytosis
C.1: Osmosis, 2: Facilitated diffusion, 3: Primary active transport, 4: Exocytosis
D.1: Facilitated diffusion, 2: Simple diffusion, 3: Secondary active transport, 4: Pinocytosis

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Worked solution

* **Mechanism 1** is **simple diffusion** because the hydrophobic molecule moves directly through the lipid bilayer down its concentration gradient without help.
* **Mechanism 2** is **facilitated diffusion** because ions move through a specific membrane channel protein down their gradient without requiring ATP.
* **Mechanism 3** is **secondary active transport** (co-transport) because glucose is transported against its gradient using energy from an ion gradient (established indirectly using ATP).
* **Mechanism 4** is **exocytosis** because bulk materials are released from vesicles fusing with the cell surface membrane.

Marking scheme

Award 1 mark for the correct option (A).
- Reject B because mechanism 3 is not simple facilitated diffusion, and bulk release is not endocytosis.
- Reject C because simple diffusion of hydrophobic molecules is not osmosis.
- Reject D because bulk export is not pinocytosis (which is the uptake of fluids).

Question 37 · multiple-choice

1 marks

The table below shows the solute potential (\(\psi_s\)) and pressure potential (\(\psi_p\)) for three adjacent cells (X, Y, and Z) in a plant root.

| Cell | Solute potential (\(\psi_s\)) / kPa | Pressure potential (\(\psi_p\)) / kPa |
| :--- | :---: | :---: |
| **X** | -700 | 200 |
| **Y** | -800 | 400 |
| **Z** | -900 | 300 |

Water potential (\(\psi\)) is calculated as \(\psi = \psi_s + \psi_p\).

In which direction will there be a net movement of water by osmosis between these cells?

A.from cell X to cell Y, and from cell Y to cell Z
B.from cell Y to cell X, and from cell Y to cell Z
C.from cell X to cell Y, and from cell Z to cell Y
D.from cell Y to cell X, and from cell Z to cell X

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Worked solution

First, calculate the water potential (\(\psi\)) for each cell using \(\psi = \psi_s + \psi_p\):
* \(\psi_X = -700 + 200 = -500\text{ kPa}\)
* \(\psi_Y = -800 + 400 = -400\text{ kPa}\)
* \(\psi_Z = -900 + 300 = -600\text{ kPa}\)

Water moves by osmosis down a water potential gradient, from a higher (less negative) water potential to a lower (more negative) water potential.
* Since \(\psi_Y\) (-400 kPa) is higher than \(\psi_X\) (-500 kPa), water moves from **cell Y to cell X**.
* Since \(\psi_Y\) (-400 kPa) is higher than \(\psi_Z\) (-600 kPa), water moves from **cell Y to cell Z**.

Therefore, there is a net movement of water from cell Y to cell X, and from cell Y to cell Z.

Marking scheme

Award 1 mark for the correct option (B).
- Reject A, C, and D because they suggest water moving from lower (more negative) water potential to higher (less negative) water potential.

Question 38 · multiple-choice

1 marks

During the cardiac cycle, pressure changes in the left atrium, left ventricle, and aorta cause the heart valves to open and close.

Which row correctly identifies the state of the valves at the point when the pressure in the left ventricle first rises above the pressure in the left atrium, and when the pressure in the left ventricle first falls below the pressure in the aorta?

A.Left ventricular pressure > left atrial pressure: AV valve closes, semilunar valve remains closed. Left ventricular pressure < aortic pressure: AV valve remains closed, semilunar valve closes.
B.Left ventricular pressure > left atrial pressure: AV valve opens, semilunar valve opens. Left ventricular pressure < aortic pressure: AV valve closes, semilunar valve opens.
C.Left ventricular pressure > left atrial pressure: AV valve closes, semilunar valve opens. Left ventricular pressure < aortic pressure: AV valve opens, semilunar valve closes.
D.Left ventricular pressure > left atrial pressure: AV valve remains open, semilunar valve closes. Left ventricular pressure < aortic pressure: AV valve remains open, semilunar valve remains open.

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Worked solution

1. When the pressure in the left ventricle first rises above the pressure in the left atrium: this forces the atrioventricular (bicuspid) valve to close to prevent backflow of blood into the left atrium. Since ventricular pressure has not yet exceeded the pressure in the aorta, the semilunar (aortic) valve remains closed.
2. When the pressure in the left ventricle first falls below the pressure in the aorta: this causes blood to start flowing backward towards the ventricle, which instantly catches the pockets of the semilunar (aortic) valve, forcing it to close. At this point, ventricular pressure is still higher than atrial pressure, so the atrioventricular valve remains closed (it only opens once ventricular pressure falls below atrial pressure).

Marking scheme

Award 1 mark for the correct option (A).
- Reject options B, C, and D because they mismatch the physiological pressure changes with the corresponding valve actions.

Question 39 · multiple-choice

1 marks

Which row correctly identifies the presence (+) or absence (-) of cartilage, smooth muscle, and elastic fibres in bronchioles and alveoli?

A.Bronchioles: Cartilage: absent (-), Smooth muscle: present (+), Elastic fibres: present (+). Alveoli: Cartilage: absent (-), Smooth muscle: absent (-), Elastic fibres: present (+).
B.Bronchioles: Cartilage: present (+), Smooth muscle: present (+), Elastic fibres: present (+). Alveoli: Cartilage: absent (-), Smooth muscle: present (+), Elastic fibres: present (+).
C.Bronchioles: Cartilage: absent (-), Smooth muscle: absent (-), Elastic fibres: present (+). Alveoli: Cartilage: absent (-), Smooth muscle: absent (-), Elastic fibres: absent (-).
D.Bronchioles: Cartilage: absent (-), Smooth muscle: present (+), Elastic fibres: absent (-). Alveoli: Cartilage: present (+), Smooth muscle: absent (-), Elastic fibres: present (+).

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Worked solution

* **Bronchioles** do not contain any cartilage (cartilage is absent: -), but they contain smooth muscle (+) to control the diameter of the airway, and elastic fibres (+) to allow expansion and recoil.
* **Alveoli** completely lack both cartilage (-) and smooth muscle (-). However, they contain a rich network of elastic fibres (+) that stretch during inhalation and recoil to push air out during exhalation.

Marking scheme

Award 1 mark for the correct option (A).
- Reject B because bronchioles lack cartilage, and alveoli lack smooth muscle.
- Reject C because alveoli contain elastic fibres.
- Reject D because bronchioles contain elastic fibres, and alveoli do not contain cartilage.

Question 40 · multiple-choice

1 marks

A newborn infant receives antibodies from its mother through breast milk. Later in life, the child receives a vaccine containing attenuated pathogens. Years later, the person is bitten by a venomous snake and is treated with an injection of pre-formed antitoxin antibodies (antivenom).

What types of immunity are acquired by the person in each of these three situations?

A.Breast milk: Natural passive; Vaccine: Artificial active; Antivenom: Artificial passive
B.Breast milk: Natural active; Vaccine: Artificial passive; Antivenom: Artificial active
C.Breast milk: Natural passive; Vaccine: Natural active; Antivenom: Artificial passive
D.Breast milk: Artificial passive; Vaccine: Artificial active; Antivenom: Natural passive

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Worked solution

1. **Breast milk**: Antibodies (IgA) are transferred naturally from mother to child. This is passive because the infant does not produce the antibodies themselves. Thus, it is **natural passive** immunity.
2. **Vaccines**: The introduction of an antigen (attenuated pathogen) medically stimulates the child's own immune response to produce antibodies and memory cells. Thus, it is **artificial active** immunity.
3. **Antivenom**: An injection of pre-formed antibodies (antitoxin) provides immediate protection but does not stimulate the recipient's immune system to make memory cells. Thus, it is **artificial passive** immunity.

Marking scheme

Award 1 mark for the correct option (A).
- Reject other options because they mismatch passive vs active or natural vs artificial categories.

Paper 22 - AS Structured Questions

Answer all structured questions in the spaces provided. Show all mathematical working and write in dark blue/black ink.

Part (a)(i) [2 marks]:
1. Chromatin condenses / coils / becomes visible as chromosomes; [1]
2. Chromosomes visible as two sister chromatids joined at the centromere; [1]

Part (a)(ii) [2 marks]:
1. Chromosomes line up along the equator / metaphase plate of spindle; [1]
2. Spindle fibres attach to centromere / kinetochore; [1]

Part (a)(iii) [2 marks]:
1. Centromere splits / divides; [1]
2. Sister chromatids separated / pulled to opposite poles (centromere leading); [1]

Part (b) [2 marks maximum]:
1. To ensure daughter cells are genetically identical / clones of parent cell; [1]
2. To prevent mutations / loss of genes / loss of function; [1]

Part (c) [2 marks maximum]:
1. Prevents the loss of essential / coding genetic information / genes during DNA replication; [1]
2. Consist of non-coding DNA that can be sacrificed as the chromosome ends shorten; [1]
(Accept: prevents chromosome ends from fusing / protects chromosome from degradation [1])

Paper 32 - Advanced Practical Skills

Carry out the experimental procedures as detailed. Record all data and draw biological structures using a sharp pencil.

2 Question · 40 marks

Question 1 · Practical Investigation

20 marks

Section instructions: Carry out the experimental procedures as detailed. Record all data and draw biological structures using a sharp pencil.

You are provided with:

E, a 1.0% amylase solution

S, a 1.0% starch solution

I, a 2.0% copper sulfate solution (non-competitive inhibitor)

W, distilled water

Iodine, iodine in potassium iodide solution

Part 1: Preparation of Inhibitor Concentrations

You are required to make a two-fold serial dilution of the 2.0% copper sulfate solution, I, to obtain the following concentrations: 2.0%, 1.0%, 0.5%, 0.25%, and 0.125%. The total volume of each solution must be 10 cm³.

(a) Complete Table 1.1 to show how you will prepare these five concentrations of inhibitor using solution I and distilled water W. [3 marks]

Table 1.1

Concentration of inhibitor / %
Volume of solution to be diluted / cm³
Volume of distilled water, W / cm³

2.0
10.0
0.0

1.0
...
...

0.5
...
...

0.25
...
...

0.125
...
...

Part 2: Investigation and Results

1. Label five clean test-tubes with the concentrations of inhibitor: 2.0%, 1.0%, 0.5%, 0.25%, and 0.125%.

2. Add 2 cm³ of the appropriate concentration of inhibitor to each labeled tube.

3. Add 2 cm³ of the amylase solution, E, to each tube. Swirl gently to mix and allow to stand for 3 minutes.

4. During this incubation, add one drop of iodine solution to each well of a clean spotting tile.

5. To the test-tube containing 2.0% inhibitor, add 2 cm³ of starch solution, S. Mix immediately and start a timer.

6. At 30-second intervals, remove a small drop of the mixture using a clean pipette and add it to a well of the spotting tile containing iodine solution.

7. Record the time taken (in seconds) for the blue-black color to first fail to appear. If starch is not completely hydrolyzed after 300 seconds, stop and record ">300".

8. Repeat steps 5 to 7 for each of the remaining inhibitor concentrations.

(b) Record your results in a single, appropriately designed table. [6 marks]

(c) Plot a line graph to show the rate of reaction (\(\text{s}^{-1}\)) on the y-axis against the concentration of inhibitor (%) on the x-axis. Calculate the rate as \(\frac{1}{\text{time taken (s)}}\). [4 marks]

(d) Explain the shape of your graph with reference to the mode of action of a non-competitive inhibitor like copper sulfate on amylase. [3 marks]

(e) Identify two sources of error in this experimental procedure that could affect the accuracy of your results, and suggest a specific improvement for each error. [4 marks]

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Worked solution

(a) Dilution Table Completion:

To perform a two-fold serial dilution of 2.0% inhibitor I to obtain 10 cm³ of each concentration:

2.0% inhibitor: 10.0 cm³ of 2.0% solution, 0.0 cm³ of W.

1.0% inhibitor: Transfer 5.0 cm³ from 2.0% tube, add 5.0 cm³ of W.

0.5% inhibitor: Transfer 5.0 cm³ from 1.0% tube, add 5.0 cm³ of W.

0.25% inhibitor: Transfer 5.0 cm³ from 0.5% tube, add 5.0 cm³ of W.

0.125% inhibitor: Transfer 5.0 cm³ from 0.25% tube, add 5.0 cm³ of W.

Completed table columns (Volume of solution / Volume of W):
- 2.0%: 10.0 cm³ / 0.0 cm³
- 1.0%: 5.0 cm³ of 2.0% / 5.0 cm³
- 0.5%: 5.0 cm³ of 1.0% / 5.0 cm³
- 0.25%: 5.0 cm³ of 0.5% / 5.0 cm³
- 0.125%: 5.0 cm³ of 0.25% / 5.0 cm³

(b) Sample Results Table:

Concentration of inhibitor, I / %
Time taken to reach end-point / s
Rate of reaction, \(1/t\) / \(\text{s}^{-1}\)

2.000
>300 (no reaction)
0.000

1.000
240
0.004

0.500
150
0.007

0.250
90
0.011

0.125
60
0.017

(c) Graph Construction:

- X-axis: Concentration of inhibitor / % (scale from 0 to 2.0% with linear intervals).
- Y-axis: Rate of reaction / \(\text{s}^{-1}\) (scale from 0 to 0.020 \(\text{s}^{-1}\)).
- Plotted Points: Plotted with small 'x' marks. Points show a decrease in rate as concentration of inhibitor increases.
- Line: Points connected by straight ruled lines or a line of best fit.

(d) Explanation:

- Copper sulfate acts as a non-competitive inhibitor because copper ions (\(\text{Cu}^{2+}\)) bind to an allosteric site of the amylase enzyme molecule.
- This binding disrupts ionic/hydrogen bonds, altering the tertiary structure of the protein and therefore the specific shape of the active site.
- The substrate (starch) can no longer fit into the active site, preventing the formation of enzyme-substrate complexes, reducing the rate of starch hydrolysis even at high substrate concentrations.

(e) Errors and Improvements:

- Error 1: The 30-second interval is too large, leading to low precision of the end-point. Improvement: Test the mixture at shorter, more frequent intervals, such as every 10 or 15 seconds.
- Error 2: Fluctuations in room temperature affect enzyme kinetic energy during the experiment. Improvement: Incubate all tubes and carry out the reactions within a thermostatically controlled water bath set to 35°C.

Marking scheme

(a) Dilution table [3 marks maximum]:

- 1 mark for correct volumes of solution and W for all concentrations (5.0 cm³ of previous concentration and 5.0 cm³ of distilled water for all serial dilutions except the starting 2.0% which has 10 cm³ and 0 cm³).
- 1 mark for showing a consistent two-fold dilution pattern (transferring 5.0 cm³ of the preceding solution to the next tube).
- 1 mark for writing all volumes to one decimal place consistently (e.g., 5.0, 10.0, 0.0).

(b) Results table [6 marks maximum]:

- 1 mark for a single table with clear headers including appropriate units (Concentration of inhibitor / %, Time / s, and Rate / \(\text{s}^{-1}\)).
- 1 mark for having no units in the body of the table (only in the column headers).
- 1 mark for recording all five inhibitor concentrations in a logical descending or ascending order.
- 1 mark for recording times as whole numbers (multiples of 30, corresponding to the sampling interval).
- 1 mark for showing a correct trend where the time to end-point increases (or no reaction occurs) as the inhibitor concentration increases.
- 1 mark for calculating rates correctly to an appropriate number of significant figures / decimal places (e.g., 3 decimal places).

(c) Graph plotting [4 marks maximum]:

- 1 mark for correct axes labels with units (x-axis: Concentration of inhibitor / %, y-axis: Rate of reaction / \(\text{s}^{-1}\)).
- 1 mark for a linear scale where the plotted points cover more than half of the grid in both dimensions.
- 1 mark for plotting all points accurately with thin, precise lines/points (e.g., small 'x' or encircled dot within 1 mm of correct coordinate).
- 1 mark for connecting points with a sharp, ruled straight line from point to point, or a smooth line of best fit with no feathering.

(d) Explanation [3 marks maximum]:

- 1 mark for identifying copper sulfate / copper ions as non-competitive inhibitors.
- 1 mark for explaining that the inhibitor binds to an allosteric site (a site other than the active site) of the amylase enzyme.
- 1 mark for stating that this alters the tertiary structure / active site shape, preventing substrate binding / enzyme-substrate complex (ESC) formation.

(e) Errors and Improvements [4 marks maximum]:

- 1 mark for identifying a valid error (e.g., 30-second sampling interval is too wide / subjective end-point determination / lack of temperature control).
- 1 mark for a matching realistic improvement (e.g., sample every 10 or 15 seconds / use a colorimeter to measure absorbance / use a thermostatically controlled water bath).
- Max 2 marks for errors, Max 2 marks for corresponding improvements.

Question 2 · Practical Investigation

20 marks

Section instructions: Carry out the experimental procedures as detailed. Record all data and draw biological structures using a sharp pencil.

You are provided with a high-resolution photomicrograph of a transverse section through a dicotyledonous stem, labeled Figure 2.1 (representing a specimen under a light microscope).

[ Figure 2.1: Photomicrograph of a Dicotyledonous Stem TS - Magnification \(\times 120\) ]

Outer boundary: Epidermis & Collenchyma

Cortex parenchyma cells

Ring of Collateral Vascular Bundles

(Xylem vessels on inside, Phloem on outside)

Scale Bar: 10 mm on image = 83.3 \(\mu\)m actual

(a) Draw a large, low-power plan diagram of a sector of the stem shown in Figure 2.1. Your diagram should include at least two vascular bundles to show the tissue organization. Do not draw any individual cells. Label the epidermis and the phloem on your diagram. [6 marks]

(b) Focus on the xylem tissue within one vascular bundle in Figure 2.1. Draw a high-power, detailed diagram of three adjacent xylem vessel elements. Use double lines to represent the cell walls. Label the lumen of one vessel and its cell wall. [5 marks]

(c) You are required to calculate the actual maximum width of one vascular bundle shown in Figure 2.1.

- The magnification of the photomicrograph is \(\times 120\).

- Use a ruler to measure the maximum width of one vascular bundle on Figure 2.1. Give your measurement in millimeters (mm).

- Assume your measurement of the vascular bundle on the diagram is 48 mm.

Calculate the actual width of this vascular bundle in micrometers (\(\mu\text{m}\)). Show your working clearly. [5 marks]

(d) Prepare a table to compare the observable structural features of the xylem vessels with those of the phloem sieve tube elements, based on Figure 2.1 and your biological knowledge. [4 marks]

Show answer & marking scheme

Worked solution

(a) Low-Power Plan Diagram:

- The diagram should depict a wedge-shaped sector of the stem.
- Outermost layer: Epidermis drawn as a single, thin, continuous layer of uniform width.
- Cortex: A layer of parenchyma tissue between the epidermis and the vascular bundles.
- Vascular bundles: Wedge-shaped or oval bundles arranged in a ring. Within each bundle, a distinct line or boundary should separate the outer phloem from the inner xylem.
- Pith/Medulla: Central region of the stem.
- No individual cells should be drawn in any tissue layer.
- Clear label lines pointing to the outermost cell layer (epidermis) and the tissue located outer/above the xylem within the vascular bundle (phloem).

(b) High-Power Drawing:

- Drawing should contain exactly three adjacent cells.
- The cell walls must be drawn as double lines to represent their thickness, with a clear middle lamella boundary if applicable, and no gaps between adjacent cells (shared walls).
- Cell shape should be angular / polygonal (typical of xylem vessels in cross-section).
- Lumens must be completely empty (no nuclei, cytoplasm, or organelles).
- Labels: Cell wall (pointing to the double boundary) and lumen (pointing to the empty center of a cell).

(c) Calculation:

- Measured image width of vascular bundle = 48 mm.
- Convert mm to micrometers (\(\mu\text{m}\)):
\(48\text{ mm} \times 1000 = 48000\text{ }\mu\text{m}\).
- Magnification = \(\times 120\).
- Using the formula:
\(\text{Actual size} = \frac{\text{Image size}}{\text{Magnification}}\)
\(\text{Actual size} = \frac{48000\text{ }\mu\text{m}}{120} = 400\text{ }\mu\text{m}\).
- Therefore, the actual width of the vascular bundle is 400 \(\mu\text{m}\).

(d) Comparison Table:

Feature
Xylem Vessels
Phloem Sieve Tube Elements

Thickness of cell walls
Thick, lignified secondary cell walls
Thin, non-lignified primary cellulose walls

Cell contents / lumen
Completely hollow lumen, no cytoplasm or organelles (dead cells)
Peripheral cytoplasm, no nucleus or vacuole (living cells)

Relative cell diameter
Larger/wider diameter
Smaller/narrower diameter

End walls
Completely absent / perforated to form continuous tubes
Sieve plates present with sieve pores

Marking scheme

(a) Low-power plan diagram [6 marks maximum]:

- 1 mark for clean, continuous, sharp lines drawn with a sharp pencil (no feathering, no shading).
- 1 mark for drawing a sector of the stem encompassing at least two vascular bundles.
- 1 mark for drawing tissues in correct proportions (e.g., width of cortex proportional to the size of the vascular bundle).
- 1 mark for representing vascular bundles as distinct wedge-shaped or oval areas with a clear boundary line separating xylem and phloem.
- 1 mark for ensuring NO individual cells are drawn anywhere in the entire plan diagram.
- 1 mark for correct labels of 'epidermis' (outermost layer) and 'phloem' (outer region of vascular bundle) with straight, uncrossed label lines.

(b) High-power cell drawing [5 marks maximum]:

- 1 mark for drawing exactly three adjacent xylem vessel cells with no gaps between them.
- 1 mark for drawing all cell walls as clear double lines to represent thickness.
- 1 mark for showing correct polygonal/angular shapes with lumens that are completely empty of any contents.
- 1 mark for drawing cells to a large scale (at least 4 cm across for each cell).
- 1 mark for correct, clear label lines pointing to 'lumen' and 'cell wall'.

(c) Actual size calculation [5 marks maximum]:

- 1 mark for stating the correct relationship: \(\text{Actual size} = \frac{\text{Image size}}{\text{Magnification}}\).
- 1 mark for converting the measured image size from millimeters to micrometers (e.g., \(48\text{ mm} \times 1000 = 48000\text{ }\mu\text{m}\)).
- 1 mark for correct substitution of values: \(\frac{48000}{120}\).
- 1 mark for the correct final numerical answer of 400 (or equivalent based on measurement if slightly different, e.g., 48 mm gives exactly 400).
- 1 mark for expressing the final answer with correct units (\(\mu\text{m}\)) and to appropriate significant figures (2 or 3 significant figures, e.g., 400 or 400.0 is acceptable, reject incorrect units like mm or cm).

(d) Comparison table [4 marks maximum]:

- 1 mark for drawing a structured table with appropriate column/row headers ('Feature', 'Xylem Vessels', 'Phloem Sieve Tube Elements').
- 1 mark for a correct comparison of cell wall thickness (thick/lignified vs thin/cellulose).
- 1 mark for a correct comparison of cell contents or lumen (empty lumen/no cytoplasm vs peripheral cytoplasm present/sieve plates).
- 1 mark for a correct comparison of cell size/diameter (larger/wider lumen vs smaller/narrower lumen).

Paper 42 - A Level Structured Questions

Answer all 10 long-form structured questions. Ensure biological explanations are detailed and precise.

Part (a) [Max 3 marks]:
1. ATP hydrolysed to release energy / provides energy (for the conversion). [1]
2. Reduced NADP provides hydrogen / electrons / acts as a reducing agent. [1]
3. To reduce glycerate 3-phosphate (GP) to triose phosphate (TP). [1]

Part (b)(i) [Max 2 marks]:
1. Ribulose bisphosphate (RuBP) / 5C compound combines with carbon dioxide / \(CO_2\). [1]
2. To produce two molecules of glycerate 3-phosphate (GP) / 3C compound (via an unstable 6C intermediate). [1]

Part (b)(ii) [Max 3 marks]:
1. Oxygen acts as a competitive inhibitor / competes with \(CO_2\) for the active site of Rubisco. [1]
2. Less RuBP reacts with carbon dioxide / less GP is produced. [1]
3. Less triose phosphate (TP) is formed to regenerate RuBP / less TP to make glucose / starch / organic molecules. [1]
4. Energy (ATP) and reducing power (reduced NADP) are wasted / used inefficiently, lowering crop biomass. [1]

Part (c) [Max 2 marks]:
1. Grana / stacked thylakoids provide a large surface area for the attachment of photosystems / light-harvesting complexes / photosynthetic pigments. [1]
2. Large surface area holds many electron carriers / electron transport chains / ATP synthase molecules. [1]
3. Thylakoid membranes are impermeable to protons, allowing a proton gradient to be established between the thylakoid lumen and the stroma. [1]

Paper 52 - Planning, Analysis and Evaluation

Design, plan, and statistically evaluate experimental investigations based on the provided prompts.

2 Question · 30 marks

Question 1 · Design & Analysis

15 marks

A student wants to investigate the effect of different concentrations of sodium malonate on the rate of oxygen consumption by mitochondria isolated from germinating mung beans (*Vigna radiata*). Succinate is used as the respiratory substrate, and a dissolved oxygen probe connected to a data logger is used to measure the rate of oxygen consumption. Sodium malonate is a metabolic inhibitor of the enzyme succinate dehydrogenase in the Krebs cycle.

(a) Describe a method the student could use to carry out this investigation. Your description should include:
- how the independent variable is varied
- how key confounding variables are controlled
- how the mitochondria are isolated and kept active
- how the rate of oxygen consumption is measured
- how safety and reliability are ensured. [10]

(b) Outline how the student could extend this investigation to determine whether sodium malonate acts as a competitive or a non-competitive inhibitor of succinate dehydrogenase. [3]

(c) Predict and explain the effect of adding an excess concentration of succinate on the rate of oxygen consumption in the presence of sodium malonate, assuming malonate is a competitive inhibitor. [2]

Show answer & marking scheme

Worked solution

(a) Method for the investigation:
1. Independent variable: Prepare at least five different concentrations of sodium malonate (e.g., 0.0, 0.2, 0.4, 0.6, 0.8, and 1.0 mol dm^{-3}) by performing simple dilution of a 1.0 mol dm^{-3} stock solution with distilled water.
2. Mitochondrial isolation: Homogenize germinating mung beans in ice-cold, isotonic phosphate buffer (pH 7.4). The cold temperature prevents enzyme denaturation, the buffer maintains pH, and the isotonic medium prevents the osmotic lysis or shrinkage of mitochondria. Filter the homogenate through muslin to remove intact cells and cell walls. Centrifuge the filtrate at a low speed (e.g., 1000g) for 5 minutes, discard the pellet (nuclei and heavy debris), then centrifuge the supernatant at a high speed (e.g., 10000g) for 15 minutes to pellet the mitochondria. Resuspend this mitochondrial pellet in fresh ice-cold isotonic buffer and store on ice.
3. Control of variables: Keep temperature constant (e.g., 25\u00b0C) using a thermostatically controlled water bath. Keep the volume and concentration of the respiratory substrate (succinate) constant in each test (e.g., 1.0 cm^{3} of 0.1 mol dm^{-3} succinate).
4. Measurement: Set up a sealed reaction vessel (to prevent atmospheric oxygen from dissolving in the mixture). Add standard volumes of mitochondrial suspension, buffer, succinate, and the selected concentration of sodium malonate. Insert the dissolved oxygen probe connected to a data logger. Record the oxygen concentration (%) at regular intervals (e.g., every 10 seconds) for 5 minutes. Calculate the initial rate of oxygen consumption from the linear gradient of the oxygen concentration against time graph.
5. Reliability & Safety: Repeat the measurement at least three times for each concentration of sodium malonate, identify any anomalous results, and calculate a mean rate. Wear eye protection, gloves, and a lab coat as sodium malonate is an irritant, and keep electrical equipment (data loggers/probes) away from water baths.

(b) Extension to determine inhibition type:
1. Measure the rate of oxygen consumption at a range of different succinate (substrate) concentrations (e.g., 0.02 to 0.2 mol dm^{-3}).
2. Carry out this process twice: once in the absence of sodium malonate, and once in the presence of a fixed, constant concentration of sodium malonate.
3. Plot a graph of the rate of oxygen consumption against succinate concentration for both treatments. If the maximum rate (Vmax) is the same in both treatments (achieved at higher succinate concentrations), the inhibition is competitive. If the maximum rate (Vmax) is significantly lower in the presence of the inhibitor, the inhibition is non-competitive.

(c) Prediction and explanation:
- Prediction: The rate of oxygen consumption will increase and eventually return to the maximum uninhibited rate (Vmax).
- Explanation: Since sodium malonate is a competitive inhibitor, it binds reversibly to the active site of succinate dehydrogenase. An excess concentration of succinate greatly increases the probability of a succinate molecule binding to the active site rather than a malonate molecule, successfully outcompeting the inhibitor.

Marking scheme

(a) [Max 10 marks]
- 1. Dilution: Describes simple or serial dilution of stock sodium malonate to produce at least 5 different concentrations (must state example concentrations or volumes).
- 2. Homogenisation: Homogenise mung beans in ice-cold, isotonic, and buffered solution.
- 3. Isolation Role: Explains that ice-cold temperature prevents denaturation / isotonicity prevents osmotic lysis of mitochondria / buffer maintains optimum pH for enzymes.
- 4. Centrifugation: Centrifuge homogenate at low speed first to remove cell debris/nuclei, then supernatant at high speed to pellet mitochondria.
- 5. Standardisation (Temp): Keep temperature constant using a water bath.
- 6. Standardisation (Substrate): Keep the concentration and volume of succinate constant.
- 7. Sealed Vessel: Use a sealed reaction chamber to prevent oxygen entering from the air.
- 8. Detection: Use a dissolved oxygen probe and data logger to record oxygen concentration.
- 9. Rate Calculation: Find the rate by calculating the initial gradient of the oxygen concentration vs. time curve.
- 10. Reliability: Repeat each concentration at least 3 times and calculate the mean / identify anomalies.
- 11. Safety: Goggles / gloves / lab coat because sodium malonate is an irritant / safety precaution with electrical equipment near water.

(b) [Max 3 marks]
- 1. Measure the rate of oxygen consumption across a range of different succinate (substrate) concentrations.
- 2. Repeat this in the presence of a constant concentration of sodium malonate and in its absence (control).
- 3. Compare the maximum rate (Vmax) reached: if Vmax is unchanged at high substrate concentrations, it is competitive; if Vmax is reduced, it is non-competitive.

(c) [Max 2 marks]
- 1. (Prediction) The rate of oxygen consumption increases / returns to maximum / Vmax is restored.
- 2. (Explanation) Succinate and malonate compete for the same active site on succinate dehydrogenase; high substrate concentration increases the frequency of substrate-active site collisions, outcompeting the inhibitor.

Question 2 · Design & Analysis

15 marks

An ecologist investigated the hypothesis that wind exposure affects the mean stomatal density on the lower epidermis of *Plantago maritima* leaves. Leaf samples were collected from two populations: Site A (a sheltered inland valley) and Site B (an exposed coastal cliff).

At each site, the ecologist collected 15 leaves from 15 different plants of *P. maritima*.

The stomatal density (number of stomata per \( \text{mm}^2 \)) was determined for each leaf using clear nail varnish impressions viewed under a light microscope. The results are summarized below:
- Site A (sheltered): Mean = \( 124.6 \text{ stomata mm}^{-2} \), Standard Deviation (\( s_1 \)) = \( 12.3 \), \( n_1 = 15 \)
- Site B (exposed): Mean = \( 141.2 \text{ stomata mm}^{-2} \), Standard Deviation (\( s_2 \)) = \( 16.8 \), \( n_2 = 15 \)

(a) Describe a method the ecologist could use to ensure that the selection of leaves at each site was random and representative. [3]

(b) State a null hypothesis for this investigation. [1]

(c) Calculate the value of the Student's t-test 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. [3]

(d) Calculate the degrees of freedom for this test. Using the critical values table below, state whether the difference in means is significant at the \( p < 0.05 \) probability level, and explain how you reached your conclusion. [5]

Critical values table for Student's t-test at \( p = 0.05 \):
- \( \text{df} = 26 \): 2.056
- \( \text{df} = 28 \): 2.048
- \( \text{df} = 30 \): 2.042

(e) Suggest one physiological reason why a higher stomatal density might be advantageous to *P. maritima* in an exposed coastal environment. [3]

Show answer & marking scheme

Worked solution

(a) Representative and random sampling method:
1. Establish a sampling grid at each site using two long tape measures laid out at right angles to each other (e.g., 20 m x 20 m grid).
2. Generate random coordinates using a computer random number generator or calculator to remove investigator bias. Avoid 'throwing' quadrats.
3. To ensure representativeness, standardise the leaf selected from each plant located at the coordinates (e.g., always select the third fully expanded leaf from the base of the plant of a similar size/age/height).

(b) Null hypothesis:
There is no significant difference between the mean stomatal density of *Plantago maritima* leaves at Site A (sheltered) and Site B (exposed).

(c) Calculations:
1. Difference in means: \( |\bar{x}_1 - \bar{x}_2| = |124.6 - 141.2| = 16.6 \)
2. Variance terms:
\( s_1^2 = 12.3^2 = 151.29 \) and \( \frac{s_1^2}{n_1} = \frac{151.29}{15} = 10.086 \)
\( s_2^2 = 16.8^2 = 282.24 \) and \( \frac{s_2^2}{n_2} = \frac{282.24}{15} = 18.816 \)
3. Sum of variances: \( 10.086 + 18.816 = 28.902 \)
4. Square root of sum: \( \sqrt{28.902} \approx 5.376 \)
5. Calculate t-value: \( t = \frac{16.6}{5.376} = 3.0878 \approx 3.09 \)

(d) Degrees of freedom & Significance:
1. Degrees of freedom: \( \text{df} = (n_1 - 1) + (n_2 - 1) = (15 - 1) + (15 - 1) = 28 \).
2. Critical value: The critical value at \( \text{df} = 28 \) for \( p = 0.05 \) is 2.048.
3. Comparison: The calculated t-value (3.09) is greater than the critical value (2.048).
4. Significance: Because the calculated t-value exceeds the critical value, the difference in mean stomatal density between the two populations is statistically significant.
5. Conclusion: There is a less than 5% probability (\( p < 0.05 \)) that this difference is due to chance alone. The null hypothesis is rejected.

(e) Physiological advantage of higher stomatal density:
In an exposed coastal environment, wind speeds are high, which rapidly removes the boundary layer of water vapour from around the leaves, potentially causing high transpiration rates but also maintaining a steep diffusion gradient for carbon dioxide. A higher stomatal density allows the plant to rapidly take in carbon dioxide for photosynthesis during brief periods when conditions are favorable (e.g., when the air is humid or when the plant has sufficient water to keep stomata fully open). Rapid photosynthesis allows the plant to synthesize organic compounds quickly to withstand stress, and the stomata can be shut tightly during dry, highly windy periods to prevent excessive water loss.

Marking scheme

(a) [Max 3 marks]
- 1. Method of randomisation: Use of tape measures at right angles to construct a grid and generating random coordinates using a random number generator.
- 2. Location: Select the plant closest to the generated coordinates.
- 3. Standardisation: Select leaves of the same age / position on the plant / leaf size to ensure the samples are representative.

(b) [1 mark]
- 1. States that there is no significant difference between the mean stomatal density of *Plantago maritima* at Site A and Site B / sheltered and exposed sites.

(c) [Max 3 marks]
- 1. Calculates the numerator correctly as 16.6.
- 2. Calculates the denominator correctly as 5.38 (or \( \sqrt{28.90} \)).
- 3. Final calculated t-value of 3.09 (accept 3.08 to 3.10 based on rounding steps, but must be to 2 decimal places as requested).

(d) [Max 5 marks]
- 1. States degrees of freedom = 28.
- 2. Correctly identifies the critical value of 2.048.
- 3. States that the calculated t-value is greater than the critical value.
- 4. Concludes that the difference is statistically significant (or that the probability of the difference occurring by chance is less than 0.05 / 5%).
- 5. States that the null hypothesis is rejected.

(e) [Max 3 marks]
- 1. Wind removes the boundary layer of water vapour, promoting high diffusion rates.
- 2. A higher stomatal density maximizes carbon dioxide uptake for photosynthesis when stomata are open / when water is available.
- 3. Enables high transpiration pull to transport essential mineral ions from nutrient-poor / saline coastal soils.
- 4. Facilitates rapid gas exchange during brief periods of stomatal opening, allowing quick closing to prevent dehydration in high winds.

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