Cambridge IAL · Thinka-original Practice Paper

2023 Cambridge IAL Biology (9700) Practice Paper with Answers

Thinka Nov 2023 (V1) Cambridge International A Level-Style Mock — Biology (9700)

160 marks195 mins2023

An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V1) Cambridge International A Level Biology (9700) paper. Not affiliated with or reproduced from Cambridge.

Section AS Level Theory (Parallel Mock)

Answer all structured questions. Write your answers in the spaces provided. Show all working in calculations.

6 Question · 60 marks

Question 1 · structured

8 marks

Beta-glucosidase is an enzyme that catalyses the hydrolysis of cellobiose, a disaccharide, into glucose. An investigation was carried out to compare the activity of beta-glucosidase extracted from two different bacteria: Pseudomonas sp. (found in temperate soils) and Thermus sp. (found in hot springs).

(a) Explain what is meant by the term activation energy and describe how enzymes lower this energy barrier. [2]

(b) Predict how the optimum temperature of beta-glucosidase from Thermus sp. would compare to that from Pseudomonas sp., and explain this difference in terms of protein structure. [3]

(c) Gluconolactone is a competitive inhibitor of beta-glucosidase. Describe how a competitive inhibitor reduces the rate of an enzyme-catalysed reaction and explain how this inhibition can be overcome. [3]

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Worked solution

Part (a)
Activation energy is the minimum energy required to start a chemical reaction by raising substrate molecules to their transition state. Enzymes lower this by holding substrate molecules in the active site in a way that puts strain on chemical bonds, or by bringing reactants close together, aligning them in the correct orientation for reaction to occur. This provides an alternative pathway with lower activation energy.

Part (b)
Beta-glucosidase from Thermus sp. will have a higher optimum temperature than that from Pseudomonas sp. This is because the tertiary structure of the Thermus enzyme contains a greater density of stabilizing interactions, such as hydrogen bonds, ionic bonds, disulfide bridges, or hydrophobic interactions. These bonds require more thermal energy to disrupt, making the enzyme more thermostable and preventing denaturation of the active site at high temperatures.

Part (c)
A competitive inhibitor has a complementary shape to the enzyme's active site (similar to the substrate). It binds directly to the active site, physically blocking the substrate from entering and preventing the formation of enzyme-substrate (ES) complexes, thus lowering the rate. This inhibition can be overcome by increasing the substrate concentration. A high substrate-to-inhibitor ratio increases the probability of a substrate molecule colliding with and binding to an active site rather than an inhibitor molecule.

Marking scheme

Part (a) [2 marks max]
1. (definition) minimum energy required to start a reaction / to raise reactants to the transition state; [1]
2. (mechanism) active site holds substrates close together / strains bonds / provides alternative pathway with lower energy barrier; [1]

Part (b) [3 marks max]
1. (prediction) Thermus sp. has a higher optimum temperature than Pseudomonas sp.; [1]
2. (structural detail) Thermus enzyme has more hydrogen bonds / ionic bonds / disulfide bridges / hydrophobic interactions in its tertiary structure; [1]
3. (explanation) tertiary structure / active site is more stable / less easily denatured / bonds require more kinetic energy to break; [1]

Part (c) [3 marks max]
1. (structure) inhibitor has a similar shape to the substrate / complementary shape to the active site; [1]
2. (effect) binds to active site, blocking substrate / preventing enzyme-substrate (ES) complexes forming; [1]
3. (overcoming) can be overcome by increasing substrate concentration (increasing likelihood of substrate binding instead of inhibitor); [1]

Question 2 · structured

11 marks

(a) Active loading of sucrose is a key process in translocation in flowering plants.
Describe how sucrose is actively loaded into the companion cells and sieve tube elements at a source. [4]

(b) The accumulation of sucrose in the sieve tube elements leads to its movement via mass flow.
Explain how a hydrostatic pressure gradient is generated and maintained to allow mass flow between a source and a sink. [4]

(c) Sieve tube elements and xylem vessel elements are both highly specialized transport cells in plants.
State three structural differences between a sieve tube element and a mature xylem vessel element. [3]

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Worked solution

(a)
1. Hydrogen ions (\(\text{H}^+\)) are actively pumped out of the companion cell cytoplasm into the apoplast/cell wall using ATP.
2. This creates a high \(\text{H}^+\) concentration gradient in the cell wall relative to the companion cell cytoplasm.
3. \(\text{H}^+\) ions diffuse back down their electrochemical gradient into the companion cell through co-transporter proteins.
4. This movement is coupled to the transport of sucrose into the companion cell against its concentration gradient.
5. Sucrose then diffuses into the sieve tube element via plasmodesmata.

(b)
1. Loading of sucrose lowers the solute potential and water potential (\(\psi\)) in the sieve tube element at the source.
2. Water enters the sieve tube element from surrounding tissues (such as the xylem) by osmosis, down a water potential gradient.
3. This increases the hydrostatic pressure inside the sieve tube element at the source.
4. At the sink, sucrose is unloaded (for respiration, storage, or growth), which increases the water potential inside the sieve tube element.
5. Water leaves the sieve tube element at the sink by osmosis, which decreases the hydrostatic pressure.
6. This difference in hydrostatic pressure creates a gradient, driving the mass flow of assimilates from source to sink.

(c)
1. Sieve tube elements are living cells containing cytoplasm and a cell membrane, whereas mature xylem vessel elements are dead cells containing no cytoplasm, membrane, or organelles.
2. Sieve tube elements have perforated end walls called sieve plates, whereas xylem vessel elements have completely open end walls with no cross-walls.
3. Sieve tube elements have cellulose cell walls only, whereas xylem vessel elements have walls reinforced and thickened with impermeable lignin.

Marking scheme

(a) [Max 4 marks]
- (Active transport of) hydrogen ions / protons / \(\text{H}^+\) pumped out of companion cell [1]
- into cell wall / apoplast [1]
- using ATP / energy from ATP hydrolysis [1]
- proton / electrochemical gradient established (higher concentration in wall than cytoplasm) [1]
- \(\text{H}^+\) ions co-transport / symport with sucrose back into companion cell [1]
- through co-transporter / symport protein [1]
- sucrose diffuses into sieve tube element via plasmodesmata [1]

(b) [Max 4 marks]
- loading of sucrose lowers water potential (\(\psi\)) / solute potential inside sieve tube at source [1]
- water enters sieve tube by osmosis down water potential gradient [1]
- this increases hydrostatic pressure (at source) [1]
- unloading of sucrose at sink increases water potential [1]
- water leaves sieve tube at sink (by osmosis) [1]
- this lowers hydrostatic pressure at sink [1]
- mass flow occurs along the hydrostatic pressure gradient (from high to low pressure) [1]

(c) [Max 3 marks, allow any three differences]
- Sieve tube element is alive / has cytoplasm / has cell surface membrane / has some organelles AND xylem vessel element is dead / has no cytoplasm / no cell membrane / is a hollow tube [1]
- Sieve tube element has sieve plates / perforated end walls AND xylem vessel element has no end walls / open ends [1]
- Sieve tube element has cellulose walls only AND xylem vessel element has lignified walls / walls reinforced with lignin [1]
- Sieve tube element has companion cells / plasmodesmata AND xylem vessel element does not [1]

Question 3 · structured

12 marks

Endo-\(\beta\)-1,4-glucanase is an enzyme that catalyses the hydrolysis of \(\beta\)-1,4-glycosidic bonds in cellulose, producing glucose and shorter cellodextrin chains. (a) Explain how endo-\(\beta\)-1,4-glucanase increases the rate of cellulose hydrolysis with reference to activation energy and the lock-and-key or induced-fit hypothesis. [3] (b) A researcher investigated the effect of pH on the activity of endo-\(\beta\)-1,4-glucanase at \(40\ ^\circ\text{C}\). The rate of reaction was measured at different pH values, and the results showed that the optimum pH was 7.0, with activity decreasing sharply below pH 5.0 and above pH 8.5. Explain, in terms of protein structure, why the activity of the enzyme decreases at pH values far from the optimum. [4] (c) Cellobiose is a disaccharide produced during cellulose breakdown. It acts as a competitive inhibitor of endo-\(\beta\)-1,4-glucanase. Explain how cellobiose decreases the rate of cellulose breakdown and state how the effect of this inhibition can be overcome. [3] (d) When carrying out investigations into enzyme-catalysed reactions, it is important to measure the initial rate of reaction. Explain why the initial rate of reaction is used rather than the rate of reaction measured after 10 minutes. [2]

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Worked solution

Part (a): Enzymes are biological catalysts that speed up chemical reactions by offering an alternative reaction pathway with a lower activation energy. The substrate (cellulose) has a shape complementary to the active site of the enzyme. According to the lock-and-key model, the substrate fits exactly. According to the induced-fit model, the active site undergoes a conformational change to fit the substrate more tightly. This interaction forms an enzyme-substrate complex (ESC) and puts mechanical strain on the glycosidic bonds, making them easier to hydrolyse. Part (b): Changes in pH affect the concentration of hydrogen ions in solution. These ions interact with the charged R-groups of the amino acids making up the enzyme. This alters ionic and hydrogen bonding, which are crucial for maintaining the precise tertiary structure of the enzyme. As a result, the tertiary structure is disrupted and the enzyme denatures, causing the shape of the active site to change. The substrate can no longer fit, and fewer enzyme-substrate complexes can form. Part (c): Cellobiose, being a product of cellulose breakdown, has a similar three-dimensional structure to the substrate cellulose. It competes with cellulose for the active site of endo-\(\beta\)-1,4-glucanase. When cellobiose binds to the active site, it blocks cellulose from binding, reducing the frequency of successful collisions and ESC formation. This competitive inhibition can be overcome by significantly increasing the concentration of the substrate (cellulose), making it much more likely for a substrate molecule to bind to an active site than an inhibitor molecule. Part (d): At the very beginning of the reaction, the substrate concentration is at its maximum and is not a limiting factor. As the reaction proceeds, substrate is consumed, and its concentration decreases. By 10 minutes, the rate of reaction would slow down because there are fewer substrate molecules to collide with the enzyme, or because accumulation of product causes product inhibition. Measuring the initial rate ensures that the rate reflects only the independent variable being tested rather than substrate depletion.

Marking scheme

Part (a) [Max 3 marks]: 1. Enzyme provides an alternative pathway with a lower activation energy; 2. Substrate (cellulose) binds to the active site of the enzyme to form an enzyme-substrate complex / ESC; 3. Reference to complementary shapes (lock-and-key) OR conformational change of active site to fit substrate (induced-fit); 4. Binding puts strain on the glycosidic bond, lowering the energy needed to break it. Part (b) [Max 4 marks]: 1. pH changes alter the charge of R-groups on the constituent amino acids; 2. This disrupts ionic bonds and hydrogen bonds; 3. This leads to a loss of the specific tertiary / 3D structure of the enzyme (denaturation); 4. The active site changes shape; 5. The substrate is no longer complementary / cannot bind, resulting in fewer/no ESCs forming. Part (c) [Max 3 marks]: 1. Cellobiose has a similar shape/structure to the substrate (cellulose); 2. Cellobiose binds to / occupies the active site of the enzyme; 3. This prevents substrate molecules from entering the active site, reducing ESC formation; 4. The inhibition can be overcome by increasing the substrate (cellulose) concentration. Part (d) [Max 2 marks]: 1. At the start / initial rate, substrate concentration is not limiting / is at its maximum; 2. As the reaction proceeds, substrate is used up / concentration of substrate decreases; 3. (After 10 minutes) rate decreases because there are fewer collisions between enzyme and substrate / substrate concentration becomes limiting / accumulation of end-product.

Question 4 · structured

13 marks

Pectin methyl esterase (PME) is an enzyme used in the food industry to clarify fruit juices by breaking down pectin in plant cell walls. (a) Describe how the active site of a globular protein, such as PME, is formed and how it enables substrate specificity. [3] An investigation was carried out to compare the activity of free PME with PME immobilised in alginate beads at different temperatures. The results are shown below: Table 1.1: Temperature / °C | Rate of reaction of free PME / arbitrary units (a.u.) | Rate of reaction of immobilised PME / arbitrary units (a.u.) [20 °C | 1.2 a.u. | 0.8 a.u.], [30 °C | 2.8 a.u. | 1.9 a.u.], [40 °C | 4.5 a.u. | 3.1 a.u.], [50 °C | 3.8 a.u. | 3.8 a.u.], [60 °C | 0.5 a.u. | 3.5 a.u.], [70 °C | 0.0 a.u. | 2.1 a.u.]. (b) Describe and compare the effect of temperature on the rate of reaction of free PME and immobilised PME as shown in Table 1.1. [4] (c) With reference to the structure of enzymes, explain the differences in the rates of reaction of free and immobilised PME at temperatures of 60 °C and 70 °C. [3] (d) State three advantages of using immobilised enzymes rather than free enzymes in industrial food processing. [3]

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Worked solution

(a) PME is a globular protein with a complex tertiary structure formed by the folding of its polypeptide chain. The active site is a specific, three-dimensional cleft or pocket formed by the bringing together of specific amino acids from different parts of the primary sequence. The R-groups of these amino acids project into the active site, determining its shape and charge. This shape is complementary to the substrate, pectin, allowing the substrate to fit precisely and form an enzyme-substrate complex through hydrogen, ionic, or hydrophobic interactions. (b) For both forms, the rate of reaction increases with temperature up to an optimum and then decreases. However, the optimum temperature for free PME is lower (40 °C, 4.5 a.u.) than for immobilised PME (50 °C, 3.8 a.u.). Free PME has a higher rate than immobilised PME at lower temperatures (20 to 40 °C), for example, at 30 °C, free is 2.8 a.u. compared to 1.9 a.u. for immobilised. At 50 °C, the rates are identical at 3.8 a.u. At high temperatures (above 50 °C), immobilised PME has a significantly higher rate than free PME; at 70 °C, free PME is completely inactive (0.0 a.u.) due to denaturation, whereas immobilised PME still retains substantial activity (2.1 a.u.). (c) At high temperatures (60 °C and 70 °C), free PME molecules gain high kinetic energy, causing intense vibration. This breaks weak hydrogen and ionic bonds holding the tertiary structure, resulting in denaturation and loss of active site shape. For immobilised PME, the alginate matrix provides structural support and stability, restricting the movement of the enzyme's polypeptide chains. This prevents the enzyme from unfolding or denaturing easily, allowing the active sites to remain functional at higher temperatures. (d) Three industrial advantages of using immobilised enzymes are: 1. The enzyme can be easily recovered and reused, reducing production costs. 2. The final product is not contaminated with the enzyme, eliminating the need for complex downstream purification. 3. Immobilised enzymes are more stable at high temperatures and extreme pH values, allowing reactions to run under a wider range of industrial conditions.

Marking scheme

Part (a): [Max 3 marks] 1. (PME is a) protein with a tertiary structure / folded polypeptide chain; 2. Active site is formed by the bringing together of specific amino acids from different parts of the chain; 3. R-groups (side chains) of these amino acids project into the active site; 4. Active site shape is complementary to the substrate (pectin), allowing the formation of an enzyme-substrate complex. Part (b): [Max 4 marks] 1. Both increase rate as temperature increases up to an optimum and then decrease (above optimum); 2. Optimum temperature for free PME is lower (40 °C) than for immobilised PME (50 °C); 3. Free PME has a higher rate of reaction than immobilised PME at temperatures from 20 to 40 °C (or comparative data quote, e.g., at 30 °C, free is 2.8 a.u. vs immobilised 1.9 a.u.); 4. At 50 °C the rate is the same for both (3.8 a.u.); 5. Immobilised PME has a higher rate than free PME at temperatures above 50 °C / at 70 °C free is inactive (0.0 a.u.) while immobilised still functions (2.1 a.u.). Part (c): [Max 3 marks] 1. High temperature increases kinetic energy, causing molecules to vibrate; 2. In free PME, this vibration breaks hydrogen / ionic bonds that stabilise the tertiary structure, causing denaturation / loss of active site shape; 3. Alginate matrix / immobilisation physically supports / stabilises the enzyme's structure; 4. This restricts movement / prevents unfolding of the protein chain at high temperatures, keeping the active site functional. Part (d): [Max 3 marks] 1. Enzyme can be easily recovered/reused; 2. Product is uncontaminated by the enzyme / no purification step needed; 3. More stable at high temperatures / wider range of temperatures can be used; 4. More stable at extreme pH values; 5. Process can be run continuously.

Question 5 · structured

8 marks

Sucrose is synthesized in photosynthesising leaves (sources) and transported throughout the plant in the phloem. This process, known as translocation, requires the active loading of sucrose at the source.

(a) Describe the mechanism by which sucrose is actively loaded into the companion cells and subsequently enters the sieve tube elements at a source. [5]

(b) Explain how this loading process leads to the movement of phloem sap from the source to a sink (such as a developing root tuber). [3]

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Worked solution

(a) Active loading of sucrose occurs through the following steps:
1. Proton pumps in the cell surface membrane of the companion cell actively pump hydrogen ions (\(\text{H}^+\)) out of the cytoplasm and into the cell wall (apoplast) using energy from ATP hydrolysis.
2. This creates a high concentration gradient of \(\text{H}^+\)/protons in the apoplast relative to the inside of the companion cell.
3. The \(\text{H}^+\)/protons then diffuse back into the companion cell down their concentration gradient through co-transporter (symporter) proteins.
4. As the protons move through the co-transporter, they bring sucrose molecules into the companion cell against the sucrose concentration gradient.
5. Once inside the companion cell, the high concentration of sucrose allows it to diffuse into the adjacent sieve tube element through connecting cytoplasmic channels called plasmodesmata.

(b) The mass flow of phloem sap is driven by hydrostatic pressure gradients:
1. The accumulation of sucrose in the sieve tube element at the source lowers its water potential.
2. Water moves from the surrounding xylem vessels and tissue into the sieve tube element by osmosis, down a water potential gradient.
3. This intake of water generates a high hydrostatic pressure inside the sieve tube element at the source.
4. At the sink, sucrose is unloaded and used or converted, which increases water potential in the sieve tube, causing water to leave by osmosis and resulting in a low hydrostatic pressure.
5. The difference in hydrostatic pressure between the source (high pressure) and sink (low pressure) forces the phloem sap to flow along the sieve tube by mass flow.

Marking scheme

Part (a) [Max 5 marks]
- MP1: Hydrogen ions / \(\text{H}^+\)/ protons are actively pumped / transported out of the cytoplasm of the companion cell; [1]
- MP2: (Protons are pumped) into the cell wall / apoplast using ATP; [1]
- MP3: This establishes a hydrogen ion / \(\text{H}^+\)/ proton concentration / electrochemical gradient; [1]
- MP4: Hydrogen ions / protons diffuse back into the companion cell down their gradient through a co-transporter / carrier protein; [1]
- MP5: Sucrose is co-transported / symported into the companion cell against its concentration gradient; [1]
- MP6: Sucrose diffuses from the companion cell into the sieve tube element via plasmodesmata; [1]

Part (b) [Max 3 marks]
- MP7: The presence of sucrose in the sieve tube element lowers the water potential; [1]
- MP8: Water enters the sieve tube element from xylem / surrounding cells by osmosis (increasing the hydrostatic pressure at the source); [1]
- MP9: Unloading of sucrose at the sink lowers hydrostatic pressure at the sink (as water leaves by osmosis); [1]
- MP10: Sap moves from high to low hydrostatic pressure by mass flow down a hydrostatic pressure gradient; [1]

Question 6 · structured

8 marks

An investigation was carried out into the effect of an inhibitor, compound X, on the activity of the enzyme bromelain. Bromelain is a protease that catalyzes the hydrolysis of proteins.

(a) Describe how a competitive inhibitor differs in its action from a non-competitive inhibitor. [3]

(b) In the investigation, the rate of reaction was measured at a substrate concentration of \(20\text{ g dm}^{-3}\).
- Rate without inhibitor: \(1.80\text{ mg s}^{-1}\)
- Rate with compound X: \(0.45\text{ mg s}^{-1}\)

Calculate the percentage decrease in the rate of reaction when the inhibitor is added. Show your working. [2]

(c) Explain how the student could determine whether compound X is a competitive or non-competitive inhibitor by extending this investigation. [3]

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Worked solution

(a)
1. Competitive inhibitor has a similar/complementary shape to the substrate and binds to the active site of the enzyme.
2. Non-competitive inhibitor binds to an alternative site (allosteric site) other than the active site.
3. Binding of non-competitive inhibitor changes the tertiary structure/conformation of the active site so the substrate can no longer fit, whereas competitive inhibitor just blocks the substrate from entering the active site (without permanently altering it).
4. The effect of a competitive inhibitor can be overcome by increasing substrate concentration, whereas the effect of a non-competitive inhibitor cannot.

(b)
1. Calculate the difference in the rate of reaction:
\(1.80\text{ mg s}^{-1} - 0.45\text{ mg s}^{-1} = 1.35\text{ mg s}^{-1}\)
2. Express this difference as a percentage of the original rate (without inhibitor):
\(\frac{1.35}{1.80} \times 100 = 75\%\) (or \(75.0\%\))

(c)
1. Measure the rate of reaction at increasing / a range of higher substrate concentrations (whilst keeping the concentration of inhibitor X constant).
2. If compound X is a competitive inhibitor, the rate of reaction will increase as substrate concentration increases and will eventually reach the same maximum rate (\(V_{\text{max}}\)) as the reaction without the inhibitor.
3. If compound X is a non-competitive inhibitor, the rate of reaction will remain significantly lower and will never reach the original maximum rate (\(V_{\text{max}}\)) even at very high substrate concentrations.

Marking scheme

(a) Max 3 marks:
- MP1: Competitive inhibitor has a similar shape to the substrate and binds to the active site; [1]
- MP2: Non-competitive inhibitor binds to an allosteric site / site other than the active site; [1]
- MP3: Non-competitive inhibitor changes the shape / conformation of the active site (preventing ES complexes); [1]
- MP4: Competitive inhibition is overcome by high substrate concentration, but non-competitive is not; [1]

(b) Max 2 marks:
- MP1: Correct working shown (e.g., \(1.35 / 1.80 \times 100\) or equivalent step); [1]
- MP2: Correct final answer: 75 / 75.0 (ignore unit since % is in question, but accept 75%); [1]

(c) Max 3 marks:
- MP1: Suggests measuring rate of reaction over a range of higher / increasing substrate concentrations (with constant inhibitor concentration); [1]
- MP2: (For competitive) at high substrate concentration, the rate reaches/approaches the original maximum rate (\(V_{\text{max}}\)) / substrate outcompetes inhibitor; [1]
- MP3: (For non-competitive) the maximum rate (\(V_{\text{max}}\)) is never reached / remains lower than the uninhibited reaction; [1]

Section A Level Theory (Parallel Mock)

Answer all structured questions. Use appropriate genetic symbols and statistical statements where required.

10 Question · 100 marks

Question 1 · Structured

13 marks

Recombinant human proteins, such as alpha-galactosidase A used in the treatment of Fabry disease, can be produced using genetically engineered Chinese Hamster Ovary (CHO) cells.

(a) Explain why a eukaryotic promoter sequence, rather than a prokaryotic promoter sequence, must be included in the expression vector when transfecting CHO cells. [4]

(b) Explain why human therapeutic proteins produced by genetically engineered mammalian host cells, such as CHO cells, are often preferred over those produced by genetically engineered bacteria like Escherichia coli. [4]

(c) Before inserting the human gene into the expression vector, the gene is often amplified using the Polymerase Chain Reaction (PCR). Describe how PCR is used to amplify the specific coding sequence of the gene from a sample of cDNA, explaining the role of primers. [5]

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Worked solution

Part (a) [4 Marks]
1. The promoter acts as the binding site for RNA polymerase and transcription factors.
2. This is essential to initiate the transcription of the transgene into mRNA.
3. Eukaryotic RNA polymerases and transcription factors do not recognize or bind to prokaryotic promoter sequences.
4. Therefore, without a eukaryotic promoter, the recombinant human gene would not be expressed in CHO cells.

Part (b) [4 Marks]
1. Eukaryotic host cells like CHO cells possess the necessary organelles (rough endoplasmic reticulum and Golgi apparatus) for post-translational modifications, whereas prokaryotes like E. coli do not.
2. These modifications include glycosylation (addition of oligosaccharide chains) and correct disulfide bond formation.
3. Proper post-translational folding and modification are critical for the protein to adopt its correct 3D tertiary structure and be fully biologically active/functional.
4. Non-glycosylated or incorrectly folded proteins produced in bacteria are rapidly degraded in the human body or may trigger an immune response (immunogenicity) in the patient.

Part (c) [5 Marks]
1. Denaturation: The reaction mixture is heated to high temperature (typically \(90-95^\circ\text{C}\)) to break hydrogen bonds between complementary base pairs, separating the double-stranded cDNA template into single strands.
2. Annealing: The mixture is cooled to a lower temperature (typically \(50-65^\circ\text{C}\)) to allow the primers to bind/anneal to their complementary sequences on the single-stranded DNA templates.
3. Role of Primers: Primers are short, single-stranded DNA sequences designed to target the specific flanks of the gene. They provide a free 3'-OH group, which is required for Taq DNA polymerase to start synthesis.
4. Extension: The temperature is increased to \(70-75^\circ\text{C}\) (optimum for Taq polymerase). Taq polymerase adds free deoxyribonucleoside triphosphates (dNTPs) complementary to the template strand in the 5' to 3' direction.
5. The cycle of denaturation, annealing, and extension is repeated multiple times (typically 25-35 cycles) to achieve exponential amplification of the target sequence.

Marking scheme

Part (a) [Max 4 marks]:
1. Promoter is the binding site for RNA polymerase / transcription factors; [Accept: RNA polymerase binding site]
2. Required to initiate transcription / start mRNA synthesis; [Reject: translation]
3. Prokaryotic promoters have different consensus sequences (e.g., Pribnow box) that are not recognized by eukaryotic RNA polymerases / eukaryotic transcription factors; [Accept: different structural/sequence requirements]
4. Eukaryotes require specific transcription factors to recruit RNA polymerase, which only bind to eukaryotic promoters; [Accept: idea of host-specificity of transcription factors]
5. Without a eukaryotic promoter, the gene is silent / not expressed in CHO cells; [Accept: no transcription occurs]

Part (b) [Max 4 marks]:
1. Eukaryotic cells can perform post-translational modification(s) (which prokaryotes / E. coli cannot); [Accept: named modifications like glycosylation, phosphorylation, disulfide bond formation]
2. Correct folding of the polypeptide chain is achieved (in the RER/Golgi); [Accept: correct tertiary structure formed]
3. Essential for biological activity / function / efficacy of the therapeutic protein; [Accept: incorrect structure leads to inactive protein]
4. Reduces risk of immunogenicity / immune response in patients; [Accept: bacterial proteins may be recognized as foreign due to lack of human-like glycosylation]
5. CHO cells can easily secrete the protein into the culture medium, simplifying extraction/purification; [Accept: no cell lysis required, unlike in E. coli]

Part (c) [Max 5 marks]:
1. Heat to \(90 - 95^\circ\text{C}\) to denature double-stranded cDNA / break hydrogen bonds (to yield single strands); [Accept: \(92 - 98^\circ\text{C}\)]
2. Cool to \(50 - 65^\circ\text{C}\) to allow primers to anneal / bind; [Accept: \(55 - 60^\circ\text{C}\)]
3. Primers are short, single-stranded sequences of DNA complementary to the 3' ends of the target sequences; [Accept: prevent re-annealing of template strands]
4. Primers define the region to be amplified / provide a free 3'-OH group for DNA polymerase to bind; [Accept: starting point for replication]
5. Heat to \(70 - 75^\circ\text{C}\) for extension / synthesis of new strands by Taq DNA polymerase; [Accept: \(72^\circ\text{C}\)]
6. Taq polymerase uses free activated nucleotides / dNTPs to synthesize complementary strands; [Accept: thermostable DNA polymerase does not denature at high temperatures]
7. Cycle of heating, cooling, heating is repeated (to exponentially amplify DNA); [Accept: reference to thermocycler]

Question 2 · structured

14 marks

Fabry disease is a rare lysosomal storage disorder caused by a deficiency in the enzyme alpha-galactosidase A (\(\alpha\)-Gal A). To treat patients with Fabry disease, scientists have developed a method to produce recombinant human \(\alpha\)-Gal A (rh\(\alpha\)-Gal A) in the milk of transgenic rabbits.

(a) Describe how reverse transcriptase and the polymerase chain reaction (PCR) are used to obtain and amplify the human gene encoding \(\alpha\)-Gal A starting from mRNA extracted from human liver cells. [4]

(b) To ensure the recombinant protein is only produced in the milk of the rabbits, the gene must be cloned into an expression vector. Explain the roles of promoters, restriction endonucleases, and DNA ligase in constructing a recombinant plasmid that achieves this. [4]

(c) Green fluorescent protein (GFP) is often used as a marker gene instead of an antibiotic resistance gene to identify successfully transformed cells. Explain how GFP is used as a marker and why it is preferred over antibiotic resistance markers in this type of animal biotechnology. [3]

(d) Discuss three ethical or social concerns associated with using transgenic animals to produce therapeutic proteins. [3]

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Worked solution

(a) Reverse transcriptase uses the mRNA template extracted from liver cells to synthesize a single-stranded complementary DNA (cDNA) molecule using free nucleotides and a primer. This cDNA is then made double-stranded. PCR is then used to amplify this cDNA through repeated cycles of: (1) Denaturation: heating to 95°C to break hydrogen bonds and separate the DNA strands; (2) Annealing: cooling to 50–65°C to allow primers to bind to complementary sequences on the DNA template; (3) Extension: heating to 72°C to allow Taq polymerase (thermostable DNA polymerase) to synthesize complementary strands using free deoxynucleotide triphosphates (dNTPs).

(b) A mammary-gland-specific promoter is inserted upstream of the \(\alpha\)-Gal A gene. It acts as a binding site for RNA polymerase and transcription factors, ensuring that transcription occurs exclusively in the milk-secreting cells of the mammary glands. Restriction endonucleases cut both the plasmid vector and the cDNA at specific recognition sites, creating complementary sticky ends. DNA ligase then seals the sugar-phosphate backbone by forming phosphodiester bonds to join the cDNA and plasmid, producing the final recombinant plasmid.

(c) The GFP gene is incorporated into the plasmid alongside the target gene. When cells take up the plasmid, they express GFP and emit green fluorescence under ultraviolet (UV) light, allowing easy, non-destructive identification of transformed cells/embryos. It is preferred over antibiotic resistance markers because identifying GFP-expressing cells does not require killing non-transformed cells with antibiotics, and it eliminates the risk of environmental bacteria acquiring antibiotic resistance genes through horizontal gene transfer.

(d) Key concerns include: (1) Welfare issues for the transgenic rabbits, such as pain or physiological distress caused by the microinjection, embryo transfer, or the expression of the transgene; (2) Ethical or religious objections regarding genetic modification of animals and transferring human genes into animal genomes ('playing God'); (3) Environmental risk of transgenic rabbits escaping into the wild and breeding with wild populations, potentially disrupting local ecosystems; (4) High development and production costs, which could lead to inequitable access to life-saving treatments.

Marking scheme

**Part (a) [Max 4 marks]**
* Reverse transcriptase synthesizes cDNA from the mRNA template;
* Heat to 95 °C to denature DNA / separate strands;
* Cool to 50-65 °C to allow primers to anneal / bind;
* Heat to 72 °C to allow Taq polymerase / thermostable DNA polymerase to synthesize complementary strands / extend primers;
* Use of free nucleotides / dNTPs to build the new strands;
* Cycles are repeated exponentially to amplify the gene.

**Part (b) [Max 4 marks]**
* Promoter acts as a binding site for RNA polymerase (and transcription factors) to initiate transcription;
* Tissue-specific promoter (e.g. beta-casein promoter) ensures transcription occurs only in mammary gland cells;
* Restriction endonucleases cut the plasmid and gene at specific restriction sites / recognition sequences;
* Cutting produces compatible / complementary sticky ends (or blunt ends);
* DNA ligase joins the plasmid and cDNA by forming phosphodiester bonds (in the sugar-phosphate backbone).

**Part (c) [Max 3 marks]**
* GFP gene is co-expressed with the target gene;
* Transformed cells / embryos emit green light when exposed to UV light;
* Identification / screening is non-destructive / does not kill the cells;
* Antibiotic resistance markers require selection with antibiotics, which kills non-transformed cells;
* Eliminates risk of horizontal gene transfer / spreading antibiotic resistance genes to environmental bacteria.

**Part (d) [Max 3 marks]**
* Animal welfare / potential harm, distress, or physiological side effects to transgenic rabbits;
* Ethical / moral / religious objections to transferring human genes into animals / 'playing God';
* Ecological risk if transgenic rabbits escape and breed with wild rabbits / impact wild gene pools;
* High cost of research and production makes treatment expensive / inequitable access to healthcare.

Question 3 · Structured Question

16 marks

Drought is a major abiotic stress that limits crop productivity worldwide. Researchers isolated a gene, DREB1A, from a drought-tolerant wild grass species, Leymus chinensis. This gene encodes a transcription factor that activates several stress-response genes.

The DREB1A gene was amplified using the Polymerase Chain Reaction (PCR) before being cloned into a plasmid vector to transform barley (Hordeum vulgare) plants.

(a) Describe the role of primers in the PCR process and explain how the temperature changes in one PCR cycle allow the selective amplification of the DREB1A gene. [4]

(b) The amplified DREB1A gene is inserted into a plasmid vector. The plasmid contains a promoter sequence and a gene for green fluorescent protein (GFP) which acts as a marker.
(i) Explain why a promoter sequence must be transferred along with the DREB1A gene. [2]
(ii) Explain how restriction endonucleases and DNA ligase are used to insert the DREB1A gene into the plasmid to form a recombinant plasmid. [3]

(c) Agrobacterium tumefaciens is used to transfer the recombinant plasmid into barley plant cells.
(i) Describe how A. tumefaciens transfers the recombinant DNA into the barley cells. [2]
(ii) Explain how the GFP marker gene is used to identify the successfully transformed barley cells, and how these cells are regenerated into whole plants. [2]

(d) Discuss the potential ecological benefits and risks associated with the widespread cultivation of genetically modified (GM) drought-resistant barley. [3]

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Worked solution

Part (a)
Primers are short, single-stranded DNA sequences that are complementary to the molecular target regions at the \(3'\) ends of the DREB1A DNA template strands. They bind (anneal) to the template DNA, providing a free \(3'\)-OH group that is necessary for Taq polymerase to initiate replication. The PCR cycle involves three main temperature phases: first, the mixture is heated to \(94\text{ to }95\text{ }^\circ\text{C}\) to break hydrogen bonds, denaturing double-stranded template DNA into single strands. Second, the temperature is reduced to \(50\text{ to }65\text{ }^\circ\text{C}\) to allow primers to specifically anneal to complementary sequences on the single-stranded DNA templates. Finally, the mixture is heated to \(70\text{ to }75\text{ }^\circ\text{C}\), the optimum temperature for thermostable Taq polymerase to synthesize new complementary DNA strands starting from the primers.

Part (b)
(i) A promoter sequence acts as the binding site for RNA polymerase, which is essential for initiating transcription of the gene. Eukaryotic host enzymes cannot naturally recognize or transcribe a foreign gene without appropriate regulatory sequences; the promoter ensures that the DREB1A gene is actively and appropriately expressed in the barley cells to produce mRNA.
(ii) The same restriction endonuclease is used to cut both the DREB1A gene and the plasmid vector, ensuring they produce complementary sticky ends (single-stranded overhangs). The cut gene and plasmid are mixed together, allowing hydrogen bonds to form between complementary bases of the sticky ends. DNA ligase is then added to catalyze the formation of phosphodiester bonds, sealing the sugar-phosphate backbone and creating a stable recombinant plasmid.

Part (c)
(i) The recombinant plasmid is introduced into A. tumefaciens cells. When these bacteria are incubated with barley tissue, they naturally infect the host cells and transfer a specific segment of plasmid DNA (the T-DNA region containing the DREB1A and GFP genes) directly into the barley plant genome, where it is integrated.
(ii) To identify the transformed cells, the barley tissue is exposed to ultraviolet (UV) light. Cells expressing the GFP marker gene will fluoresce green and can be separated from untransformed cells. These selected cells are then grown on a nutrient agar medium containing specific plant growth regulators (auxins and cytokinins) to induce cell division and differentiation into shoots and roots, regenerating whole plants via tissue culture.

Part (d)
Benefits: Cultivating drought-resistant barley allows crop growth in arid regions, which improves food security and drastically reduces the volume of fresh water required for crop irrigation.
Risks: There is a potential risk of gene flow (via pollen) to wild relatives, creating highly invasive, drought-tolerant weeds. Additionally, widespread cultivation of GM monocultures may reduce agricultural biodiversity and disrupt local insect or soil ecosystems.

Marking scheme

Part (a) [Max 4 marks]:
1. Primers are short, single-stranded DNA sequences complementary to target DNA / provide a start point (free \(3'\)-OH group) for DNA polymerase. [1]
2. Heating to \(94-95\text{ }^\circ\text{C}\) denatures the double-stranded template DNA by breaking hydrogen bonds. [1]
3. Cooling to \(50-65\text{ }^\circ\text{C}\) allows primers to anneal / bind to complementary bases. [1]
4. Heating to \(70-75\text{ }^\circ\text{C}\) is the optimum temperature for Taq / thermostable DNA polymerase to synthesize complementary strands. [1]

Part (b) [Max 5 marks]:
(i) [Max 2 marks]:
5. Promoter is the binding site for RNA polymerase. [1]
6. Allows transcription / expression of the DREB1A gene / production of mRNA. [1]
(ii) [Max 3 marks]:
7. Same restriction endonuclease cuts both the DREB1A gene and plasmid. [1]
8. This produces complementary sticky ends / single-stranded overhangs. [1]
9. DNA ligase forms phosphodiester bonds to join the gene and plasmid / seals the sugar-phosphate backbone. [1]

Part (c) [Max 4 marks]:
(i) [Max 2 marks]:
10. A. tumefaciens naturally infects plant cells. [1]
11. The T-DNA (carrying the DREB1A / GFP gene) is integrated into the host barley genome / chromosome. [1]
(ii) [Max 2 marks]:
12. Transformed cells are identified by exposing them to UV light, causing them to fluoresce green. [1]
13. Selected cells are regenerated into whole plants using plant growth regulators / tissue culture / micropropagation. [1]

Part (d) [Max 3 marks]:
14. Benefit: reduces water/irrigation use OR allows crop survival in arid regions / increases crop yield in drought. [1]
15. Risk: gene flow / escape of gene to wild relatives via pollen transfer (creating superweeds). [1]
16. Risk: reduction in agricultural biodiversity / potential impact on non-target soil or insect species. [1]
Note: Award max 3 marks. Must include at least one benefit and one risk to achieve full marks.

Question 4 · structured

7 marks

Congenital nephrogenic diabetes insipidus (NDI) is a rare inherited disorder characterized by the inability of the kidneys to concentrate urine, despite normal or elevated plasma levels of antidiuretic hormone (ADH). NDI can be caused by a mutation in the *AVPR2* gene, which encodes the vasopressin V2 receptor (a G-protein coupled receptor) on the basolateral membrane of the collecting duct cells.

(a) Describe the sequence of cell signaling events that occurs in a healthy individual when ADH binds to the V2 receptor, leading to an increase in the permeability of the collecting duct cells to water. [4]

(b) Explain why a mutation in the *AVPR2* gene that prevents the G-protein from binding to the V2 receptor results in the production of a large volume of dilute urine. [3]

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Worked solution

### Part (a) - Cellular signaling pathway of ADH:
1. **Receptor activation**: ADH binds to the V2 receptor on the basolateral membrane, causing a conformational change that activates the associated G-protein.
2. **Enzyme activation**: The active G-protein activates the membrane-bound enzyme adenylyl cyclase.
3. **Second messenger production**: Active adenylyl cyclase converts ATP into cyclic AMP (cAMP).
4. **Protein kinase activation**: cAMP acts as a second messenger and activates protein kinase A (PKA).
5. **Vesicle translocation**: This triggers a signaling cascade that causes intracellular vesicles containing aquaporins (water channel proteins) to move towards and fuse with the apical (luminal) membrane, increasing its permeability to water.

### Part (b) - Effect of the mutation:
1. **Lack of signaling cascade**: When the G-protein cannot bind to the mutated V2 receptor, adenylyl cyclase is not activated. Consequently, cAMP is not produced and protein kinase A remains inactive.
2. **Lack of aquaporin insertion**: Vesicles containing aquaporins do not undergo exocytosis/fusion with the apical membrane, so water channels are absent from the luminal membrane.
3. **Reduced water reabsorption**: The apical membrane remains impermeable to water. Water in the collecting duct lumen cannot be reabsorbed by osmosis down the water potential gradient into the hypertonic medulla and blood capillaries.
4. **Excretion of dilute urine**: A large volume of water remains within the collecting duct, leading to the excretion of a high volume of highly dilute urine.

Marking scheme

**Part (a) [Max 4 marks]:**
* **M1**: ADH binds to the V2 receptor, leading to the activation of G-protein.
* **M2**: Activated G-protein activates adenylyl cyclase.
* **M3**: Adenylyl cyclase catalyzes the conversion of ATP to cyclic AMP (cAMP).
* **M4**: cAMP acts as a second messenger to activate protein kinase (A).
* **M5**: This causes vesicles containing aquaporins to move to and fuse with the apical / luminal membrane.
* *Accept*: increases the number/density of aquaporins in the apical membrane.

**Part (b) [Max 3 marks]:**
* **M6**: Mutation prevents G-protein activation, meaning no cAMP is produced / no vesicle translocation occurs.
* **M7**: Aquaporins are not inserted into the apical / luminal membrane.
* **M8**: The apical membrane / collecting duct remains impermeable to water.
* **M9**: Water cannot be reabsorbed by osmosis down the water potential gradient (into the tissue fluid / medulla / blood).
* **M10**: Water is not conserved, resulting in a large volume of dilute urine.

Question 5 · Structured Question

7 marks

Microarray technology is a powerful tool used in molecular biology to study the expression of thousands of genes simultaneously. A researcher used a microarray to compare the gene expression profiles of colorectal cancer cells and healthy colorectal cells. (a) Explain why single-stranded complementary DNA (cDNA), rather than genomic DNA, is used to hybridise with the probes on the microarray chip. [3] (b) Describe how the mRNA extracted from the cancer cells and healthy cells is processed, hybridised, and analysed to identify genes that are upregulated in the cancer cells. [4]

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Worked solution

Part (a): Complementary DNA (cDNA) is produced by reverse transcribing mature mRNA. This ensures that only the genes actively transcribed (expressed) in the cells at that specific time are represented, unlike genomic DNA which contains all genes regardless of whether they are being expressed. Additionally, cDNA contains only exons (no introns), representing the coding sequences. Being single-stranded is essential to allow the cDNA to complementary base-pair (hybridise) with the single-stranded oligonucleotide probes fixed on the microarray surface. Part (b): First, mRNA is extracted from both healthy and cancer cells. This mRNA is reverse transcribed into cDNA using reverse transcriptase and nucleotides tagged with distinct fluorescent dyes (for example, green fluorescent dye for healthy cells and red fluorescent dye for cancer cells). The two cDNA samples are then mixed and incubated with the microarray, allowing hybridisation to occur between the single-stranded cDNA molecules and their complementary probes on the chip. Unbound cDNA is washed away to ensure specificity. The microarray is then scanned with a laser scanner to detect fluorescence. Upregulated genes in cancer cells will show a higher ratio of the cancer-associated fluorescent dye (e.g., a dominant red signal), indicating higher levels of transcription of those specific genes in the colorectal cancer cells.

Marking scheme

Part (a) [max 3 marks]:
1. cDNA is synthesized from mRNA, representing only actively expressed/transcribed genes (whereas genomic DNA contains non-transcribed regions). [1]
2. cDNA lacks introns / contains only exons (whereas genomic DNA contains introns). [1]
3. Being single-stranded allows complementary base pairing / hybridisation / hydrogen bonding with the single-stranded probes on the microarray. [1]

Part (b) [max 4 marks]:
4. mRNA is reverse transcribed into cDNA using different fluorescent dyes / labels for cancer and healthy cells. [1]
5. The labelled cDNA samples are mixed and hybridised to the probes on the microarray. [1]
6. Unbound cDNA is washed off to prevent background / false signals. [1]
7. The microarray is scanned with a laser / fluorescence detector. [1]
8. A higher intensity of the cancer-associated dye (compared to healthy dye) indicates upregulation of that gene in cancer cells. [1]

Question 6 · Structured Question

11 marks

In a newly discovered plant species, Campanula novus, flower colour is determined by two genes, A/a and B/b, located on different chromosomes. A biochemical pathway for the synthesis of flower pigments is shown below: Precursor (white) --[Enzyme A]--> Intermediate (pink) --[Enzyme B]--> Product (purple). Gene A encodes enzyme A, which converts the white precursor into a pink intermediate. The recessive allele a produces a non-functional enzyme. Gene B encodes enzyme B, which converts the pink intermediate into a purple pigment. The recessive allele b produces a non-functional enzyme. (a) Explain how the interaction of genes A and B determines the purple, pink, and white flower phenotypes. [3] (b) Two plants, both heterozygous at both loci (AaBb), are crossed. State the expected phenotypes and their ratio in the offspring. [2] (c) A geneticist carried out the cross AaBb x AaBb and obtained 160 offspring with the following phenotypes: Purple: 85, Pink: 35, White: 40. Use the chi-squared (\chi^2) test to calculate the value of \chi^2 for these results. Show your working. \chi^2 = \sum \frac{(O - E)^2}{E}. [4] (d) The critical value for \chi^2 at p = 0.05 for the appropriate number of degrees of freedom is 5.99. Use this value and your calculated \chi^2 to explain the conclusion that can be drawn from this cross. [2]

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Worked solution

(a) For a plant to have purple flowers, both dominant alleles must be present (genotype A_B_) because enzyme A is required to make the pink intermediate and enzyme B is required to convert it to the purple product. For pink flowers, allele A must be present but with homozygous recessive bb (genotype A_bb) because enzyme A converts precursor to pink but lack of functional enzyme B prevents further conversion. For white flowers, homozygous recessive aa must be present (genotypes aaB_ or aabb) because no functional enzyme A is made, meaning no pink intermediate is produced regardless of gene B alleles. (b) The expected phenotypes and ratio are 9 Purple : 3 Pink : 4 White. (c) Total observed offspring = 160. Expected numbers (based on a 9:3:4 ratio): Expected Purple = 160 * (9/16) = 90. Expected Pink = 160 * (3/16) = 30. Expected White = 160 * (4/16) = 40. Calculation of (O - E)^2 / E: For Purple: (85 - 90)^2 / 90 = 25 / 90 = 0.278. For Pink: (35 - 30)^2 / 30 = 25 / 30 = 0.833. For White: (40 - 40)^2 / 40 = 0. Sum of these values gives \chi^2 = 0.278 + 0.833 + 0 = 1.11. (d) Since the calculated \chi^2 value of 1.11 is less than the critical value of 5.99, the difference between the observed and expected results is not statistically significant (p > 0.05). The difference is due to chance, and the genetic hypothesis of a 9:3:4 epistatic inheritance is supported.

Marking scheme

Part (a) [Max 3 marks]: 1. Allele A is required for active enzyme A to convert white precursor to pink intermediate OR genotype aa results in no enzyme A and leaves the flower white. 2. To obtain purple, dominant alleles for both genes must be present (genotype A_B_) to produce both active enzymes. 3. To obtain pink, dominant allele A must be present alongside homozygous recessive bb (genotype A_bb) so enzyme A is active but enzyme B is non-functional. Part (b) [Max 2 marks]: 1. Correct phenotypes: purple, pink, and white. 2. Correct expected ratio: 9 purple : 3 pink : 4 white (allow any order if correctly linked to phenotypes). Part (c) [Max 4 marks]: 1. Correct calculation of expected frequencies: Purple = 90, Pink = 30, and White = 40. 2. Correct calculation of (O - E)^2 values: Purple = 25, Pink = 25, White = 0. 3. Correct terms for (O - E)^2 / E: Purple = 0.28 (or 0.278), Pink = 0.83 (or 0.833), White = 0. 4. Correct final calculated \chi^2 value of 1.11 (accept 1.11 to 1.111). Part (d) [Max 2 marks]: 1. Correctly states that calculated \chi^2 (1.11) is less than the critical value (5.99) (consequential on part c). 2. Concludes that there is no significant difference (probability is greater than 0.05) / deviation is due to chance AND the hypothesis / epistatic ratio is supported.

Question 7 · structured

9 marks

Genetic technology is widely used in research, medicine, and agriculture.

(a) One important technique in genetic engineering is the use of marker genes to identify successfully transformed cells. Green fluorescent protein (GFP) is commonly used as a marker.

(i) Describe how GFP can be used to identify cells that have successfully taken up a recombinant plasmid. [2]

(ii) Explain the advantages of using GFP as a marker gene compared with using antibiotic resistance genes. [2]

(b) Once a gene is isolated or synthesized, polymerase chain reaction (PCR) is used to amplify the DNA.

(i) Calculate the theoretical number of double-stranded DNA molecules produced from a single starting double-stranded DNA template after 12 cycles of PCR. [1]

(ii) State two reasons why the actual yield of DNA molecules at the end of 30 cycles of PCR is usually much lower than the theoretical maximum. [2]

(c) Bioinformatics is an essential tool in genetic technology. Outline how bioinformatics can be used by scientists to analyze newly sequenced genes. [2]

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Worked solution

(a)(i) The GFP gene is cloned into the plasmid alongside the gene of interest under the control of a promoter. Transformed host cells express the GFP gene and produce green fluorescent protein. These cells can be identified easily because they fluoresce/glow green when exposed to ultraviolet (UV) light.

(a)(ii) GFP identification does not require killing untransformed cells (unlike antibiotic selection, which is destructive). It is also safer because there is no risk of horizontal gene transfer of antibiotic resistance genes to environmental or pathogenic bacteria. Additionally, it is faster and simpler as it only requires shining UV light instead of replica plating or long incubation on selective media.

(b)(i) Each cycle doubles the number of DNA molecules. After 12 cycles starting from 1 molecule, the theoretical yield is \(2^{12} = 4096\).

(b)(ii) The actual yield is lower because reactants (primers, nucleotides/dNTPs) eventually run out and become limiting. Also, Taq polymerase gradually denatures and loses its catalytic activity after repeated heating cycles to 95 °C.

(c) Bioinformatics databases (such as GenBank) allow scientists to run alignment searches (like BLAST) to compare the sequence of the new gene with known genes in other organisms to find homologous sequences. This helps predict the amino acid sequence of the encoded protein, model its 3D structure, and deduce its likely function or evolutionary origin.

Marking scheme

**Part (a)(i) [Max 2 marks]**
* MP1: GFP gene is incorporated into the plasmid/vector (along with the target gene/promoter). [1]
* MP2: Transformed cells express the GFP and glow/fluoresce (green) under ultraviolet (UV) light (allowing direct identification). [1]

**Part (a)(ii) [Max 2 marks]**
* MP1: No risk of transferring/spreading antibiotic resistance genes to pathogenic bacteria or the wider environment / no risk of horizontal gene transfer. [1]
* MP2: Non-destructive method / does not kill non-transformed cells (so replica plating is not needed). [1]
* MP3: Faster / easier to identify because it only requires shining UV light / no need to culture on antibiotic media. [1]

**Part (b)(i) [1 mark]**
* MP1: 4096 (or \(2^{12}\)) [1]

**Part (b)(ii) [Max 2 marks]**
* MP1: Primers or nucleotides (dNTPs) become depleted / run out / limit the reaction. [1]
* MP2: Taq polymerase denatures / loses activity over repeated heating cycles (to 95 °C). [1]
* MP3: DNA template strands re-anneal to each other rather than primers binding as DNA concentration increases. [1]

**Part (c) [Max 2 marks]**
* MP1: Compare the newly sequenced gene with sequences in databases (e.g. BLAST / GenBank) to find homologous/similar sequences in other species. [1]
* MP2: Predict the primary structure / amino acid sequence of the encoded polypeptide. [1]
* MP3: Predict/model 3D tertiary structure / active site of the protein. [1]
* MP4: Determine evolutionary relationships / phylogenetic analysis. [1]

Question 8 · Structured

6 marks

The dwarf phenotype in some varieties of pea plants (*Pisum sativum*) is caused by a mutation in a gene that encodes a DELLA protein. In normal wild-type plants, the binding of gibberellin (GA) to its receptor leads to the breakdown of DELLA proteins, promoting stem elongation.

(a) Explain how the breakdown of DELLA proteins by gibberellin leads to stem elongation in wild-type plants. [3]

(b) Suggest and explain how a mutation in the gene encoding the DELLA protein could result in a dwarf phenotype that does not respond to exogenous (external) applications of gibberellin. [3]

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Worked solution

**Part (a)**
In wild-type plants, DELLA proteins act as transcription inhibitors or repressors by binding to transcription factors (such as PIFs) and preventing them from binding to promoter regions of growth-regulating genes. When gibberellin (GA) is present, it binds to a soluble receptor protein (GID1). This GA-receptor complex then binds to the DELLA protein, targeting it for degradation via the ubiquitin-proteasome pathway. With the DELLA repressor protein degraded, the transcription factors are released and can bind to the promoter regions of target genes. This stimulates the transcription of genes involved in cell division and cell elongation (such as those encoding expansins, which loosen the cell wall), leading to stem elongation.

**Part (b)**
A mutation in the gene encoding the DELLA protein can lead to a change in the primary structure (amino acid sequence) of the DELLA protein, altering its tertiary structure (3D conformation). This conformational change may specifically occur in the domain that normally interacts with the GA-receptor complex. Consequently, the mutated DELLA protein can no longer bind to the GA-receptor complex and is not targeted for destruction by the proteasome. Since the mutant DELLA protein is resistant to degradation, it remains permanently bound to the transcription factors, continuing to repress the transcription of growth-related genes. Because the block on transcription cannot be removed, the plants exhibit a dwarf phenotype and fail to respond even when treated with high concentrations of exogenous gibberellin.

Marking scheme

**Part (a) [Max 3 marks]**
* **MP1** DELLA protein acts as a transcription repressor / inhibitor (of transcription factors / growth genes); [1]
* **MP2** GA binds to a receptor (protein) / GID1; [1]
* **MP3** GA-receptor complex binds to DELLA, leading to its breakdown / degradation (by proteasomes / ubiquitin pathway); [1]
* **MP4** Transcription factors / PIFs are released / can now bind to DNA / promoters; [1]
* **MP5** (This stimulates) transcription of genes for cell elongation / cell division / amylase / expansins; [1]

**Part (b) [Max 3 marks]**
* **MP6** Mutation changes the amino acid sequence / primary structure AND alters the tertiary structure / 3D shape of the DELLA protein; [1]
* **MP7** Mutated DELLA protein cannot bind to the GA-receptor complex / cannot be recognised for degradation; [1]
* **MP8** DELLA protein is not broken down / remains stable / accumulates; [1]
* **MP9** Transcription factors remain permanently repressed / growth genes are not transcribed (so no cell elongation / division occurs); [1]

Question 9 · Structured Question

10 marks

Researchers used the CRISPR-Cas9 system to knock out the OsSWEET14 gene in rice (Oryza sativa). This gene encodes a sugar transporter that is exploited by the bacterial pathogen Xanthomonas oryzae to obtain nutrients. Knocking out this gene makes the rice plants resistant to bacterial blight. (a) Describe how guide RNA (gRNA) directs the Cas9 enzyme to the target OsSWEET14 gene. [2] (b) The Cas9 enzyme creates a double-strand break in the DNA of the OsSWEET14 gene. During the cell's error-prone repair process, a single nucleotide is deleted, causing a frameshift mutation. Explain how a frameshift mutation results in a non-functional sugar transporter protein. [3] (c) To confirm that the gene editing was successful, researchers amplified the target region of the OsSWEET14 gene using the polymerase chain reaction (PCR). Describe the steps involved in using PCR to amplify this specific region of DNA. [3] (d) Suggest two advantages of using CRISPR-Cas9 to knock out a gene rather than using traditional genetic engineering methods that involve inserting an antibiotic resistance marker gene. [2]

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Worked solution

(a) Guide RNA (gRNA) contains a specific sequence of nucleotides that is complementary to the target sequence of the OsSWEET14 gene. It binds to the target DNA strand through complementary base pairing (hydrogen bonding). Because the Cas9 endonuclease is attached to the gRNA, this directs the enzyme to the exact site to make a double-stranded cut. (b) A single nucleotide deletion shifts the reading frame of codons from the point of mutation onwards (a frameshift). This changes the sequence of amino acids (primary structure) of the sugar transporter polypeptide. It may also introduce a premature stop codon, leading to a truncated, shorter polypeptide. This drastically alters the overall folding and tertiary structure of the protein, meaning it can no longer act as a functional sugar transporter in the cell membrane. (c) The three steps of PCR are: 1. Denaturation: The reaction mixture is heated to 90-95 \( ^\circ \)C to break the hydrogen bonds between complementary bases, separating the double-stranded DNA into single strands. 2. Annealing: The temperature is lowered to 50-65 \( ^\circ \)C, allowing the forward and reverse primers to bind to their complementary target sequences flanking the region of interest. 3. Extension: The temperature is raised to 72 \( ^\circ \)C to allow Taq DNA polymerase to synthesize new DNA strands by adding free DNA nucleotides (dNTPs) starting from the 3' end of the primers. (d) Two advantages are: 1. No foreign DNA, such as antibiotic resistance genes, remains in the crop, eliminating public health concerns and risks of environmental gene transfer. 2. CRISPR-Cas9 is extremely precise, resulting in far fewer off-target mutations compared to traditional random transgene insertion.

Marking scheme

(a) [Max 2 marks] 1. gRNA sequence is complementary to target DNA of OsSWEET14 gene; 2. gRNA binds/hybridizes to target DNA via complementary base pairing/hydrogen bonds; 3. Cas9 is bound to the gRNA, positioning the endonuclease at the specific cleavage site. (b) [Max 3 marks] 1. Deletion alters the triplet reading frame downstream of the mutation; 2. This changes the amino acid sequence/primary structure; 3. May introduce a premature stop codon, resulting in a truncated/shorter polypeptide; 4. Alters tertiary structure/3D conformation so transporter cannot bind/transport sugars. (c) [Max 3 marks] 1. Denaturation: Heat to 90-95 \( ^\circ \)C (accept 92-98 \( ^\circ \)C) to separate DNA strands; 2. Annealing: Cool to 50-65 \( ^\circ \)C to allow primers to bind/anneal to complementary sequences; 3. Extension: Heat to 70-75 \( ^\circ \)C (accept 72 \( ^\circ \)C) for Taq polymerase to synthesize complementary strands; 4. Correct mention of free DNA nucleotides/dNTPs used to build new strands. (d) [Max 2 marks] 1. No antibiotic resistance marker genes are left in the plant genome; 2. Reduces risk of horizontal gene transfer of resistance genes to environmental bacteria; 3. CRISPR-Cas9 is highly precise, leading to fewer off-target mutations/disrupted genes; 4. Faster/cheaper/more efficient than traditional vector cloning.

Question 10 · Structured Question

7 marks

Auxin (indol-3-acetic acid, IAA) is a plant hormone that plays a crucial role in regulating growth by promoting cell elongation in shoots.

(a) Explain how the presence of auxin in a shoot cell leads to the activation and increased activity of proton pumps in the cell surface membrane. [3]

(b) Describe how the transport of protons (\(\text{H}^+\)) into the cell wall results in the elongation of the plant cell. [4]

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Worked solution

(a) When auxin (IAA) binds to specific auxin receptors (such as ABP1) on or within target plant cells, it initiates a intracellular signaling cascade. This signal transduction pathway stimulates the transcription of genes encoding proton pump proteins (\(\text{H}^+\)-ATPase) in the nucleus, leading to increased translation of these proteins. Concurrently, auxin activates existing proton pumps already present in the cell surface membrane, which actively pump \(\text{H}^+\) ions out of the cytoplasm and into the cell wall using ATP as an energy source.

(b) The accumulation of protons in the cell wall lowers the extracellular pH (acidification). This acidic environment activates pH-dependent cell wall-loosening proteins called expansins. Expansins break the hydrogen bonds between cellulose microfibrils and matrix polysaccharides (like hemicellulose). As a result, the structural cross-links are disrupted, loosening the cell wall and increasing its elasticity/plasticity. The cell actively absorbs solutes (e.g., \(\text{K}^+\) ions), lowering its internal water potential, which drives water uptake via osmosis. The rising turgor pressure inside the vacuole pushes against the weakened, extensible cell wall, causing it to stretch and permanently expand.

Marking scheme

Part (a) [Max 3 marks]:
1. Auxin binds to an auxin receptor / receptor protein (e.g., ABP1);
2. This triggers a signal transduction pathway / intracellular cascade;
3. Activates genes / increases transcription / increases translation of proton (\(\text{H}^+\)-ATPase) pump proteins;
4. Activates existing proton pumps in the cell surface membrane;
5. Protons (\(\text{H}^+\)) are actively transported / pumped across the cell membrane into the cell wall using ATP;

Part (b) [Max 4 marks]:
6. Accumulation of protons lowers the pH / acidifies the cell wall;
7. Low pH activates expansins (wall-loosening proteins);
8. Expansins break hydrogen bonds between cellulose microfibrils and other cell wall components / hemicelluloses;
9. The cell wall becomes loose / plastic / extensible;
10. Potassium (\(\text{K}^+\)) / solutes enter the cell, lowering the internal water potential;
11. Water enters the cell / vacuole by osmosis down a water potential gradient;
12. Increased turgor pressure pushes against the weakened cell wall, causing cell expansion / stretching;

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