2025 Edexcel IAL Biology (YBI11) 模擬試題連答案詳解

Red light of wavelength \(660\text{ nm}\) is absorbed by the inactive form of phytochrome, \(P_r\), converting it rapidly into the active form, \(P_{fr}\). The presence of \(P_{fr}\) triggers the physiological pathways that lead to seed germination. Far-red light of wavelength \(730\text{ nm}\) converts the active \(P_{fr}\) back into the inactive \(P_r\) form, which inhibits or fails to stimulate germination.

[1 mark] B is the only correct option.
- Reject A: Red light converts \(P_r\) to \(P_{fr}\), not vice versa.
- Reject C and D: Far-red light converts \(P_{fr}\) to \(P_r\), which does not stimulate germination.

Acetylcholinesterase is responsible for breaking down the neurotransmitter acetylcholine (ACh) in the synaptic cleft to stop the stimulus. If this enzyme is inhibited, acetylcholine remains bound to ligand-gated sodium channels on the postsynaptic membrane. This keeps the sodium channels open, allowing continuous influx of sodium ions and resulting in persistent depolarisation of the postsynaptic membrane.

[1 mark] B is the correct option.
- Reject A: The membrane cannot repolarise if acetylcholine is not broken down.
- Reject C: Hyperpolarisation does not occur because sodium channels remain open.
- Reject D: Action potentials are initially generated, and the membrane stays depolarised rather than failing to depolarise.

The cerebellum is the region of the brain responsible for the coordination of voluntary movement, posture, and balance. An fMRI scan measures brain activity by detecting changes in blood oxygenation and flow; increased oxygenation in the cerebellum indicates high metabolic activity in this region, which is associated with executing or planning physical movements.

[1 mark] A is the correct option.
- Reject B: Homeostatic regulation of body temperature is controlled by the hypothalamus.
- Reject C: Integration of visual images occurs in the visual cortex of the cerebrum.
- Reject D: Control of autonomic functions is located in the medulla oblongata.

For each turn of the Krebs cycle (which processes one molecule of acetyl CoA), there are three oxidation steps that reduce \(\text{NAD}^+\) to reduced NAD (NADH), one oxidation step that reduces FAD to reduced FAD (\(\text{FADH}_2\)), and one single substrate-level phosphorylation step that yields one molecule of ATP (or GTP, which is equivalent).

[1 mark] B is the correct option.
- Reject A, C, and D: These options provide incorrect stoichiometries for a single turn of the cycle.

During muscle contraction, calcium ions released from the sarcoplasmic reticulum bind to troponin. This binding causes a conformational change in troponin, which shifts the tropomyosin molecule away from the actin filament, exposing the myosin-binding sites. This allows the myosin head to bind to actin and form a cross-bridge.

[1 mark] B is the correct option.
- Reject A: Calcium binds to troponin, not tropomyosin.
- Reject C: ATP binds to the myosin head, not the actin head, to cause cross-bridge detachment.
- Reject D: The A-band remains constant in length while the I-band and H-zone shorten.

The correct sequence in a PCR cycle is: 1) Denaturation at approx. \(95^\circ\text{C}\) to break hydrogen bonds between complementary strands of DNA; 2) Annealing at approx. \(55^\circ\text{C}\) to allow primers to bind to the single-stranded DNA targets; 3) Extension at approx. \(72^\circ\text{C}\), which is the optimum temperature for the thermostable Taq DNA polymerase to synthesise the complementary strands.

[1 mark] C is the correct option.
- Reject A, B, and D because they mix up the correct temperatures with their biological purposes.

In the dark, rhodopsin remains intact (not bleached). High levels of cyclic GMP (cGMP) keep non-specific cation (sodium) channels in the outer segment membrane open, allowing sodium ions to flow continuously into the cell (known as the 'dark current'). This influx of positive ions keeps the rod cell membrane in a depolarised state, leading to the continuous release of the neurotransmitter glutamate.

[1 mark] C is the correct option.
- Reject A and D: Rhodopsin is only bleached in the presence of light.
- Reject B: In the dark, sodium channels are open and the cell is depolarised, not hyperpolarised.

High blood pressure stimulates baroreceptors, which send a higher frequency of sensory impulses to the cardiovascular control centre in the medulla oblongata. The centre responds by sending more nerve impulses along the parasympathetic (vagus) nerve to the sinoatrial node (SAN). This neurotransmitter action (acetylcholine) slows down the heart rate, reducing cardiac output and lowering blood pressure back to normal levels.

[1 mark] B is the correct option.
- Reject A: Sympathetic stimulation would increase heart rate, which would further increase blood pressure.
- Reject C: Decreased parasympathetic stimulation would increase heart rate.
- Reject D: Autonomic control of heart rate primarily targets the sinoatrial node (SAN), and sympathetic stimulation decreases rather than increases blood pressure control delay.

Red light converts inactive \(P_r\) (phytochrome red) to active \(P_{fr}\) (phytochrome far-red). The active \(P_{fr}\) moves into the nucleus and acts as a transcription factor, triggering the expression of genes involved in germination, such as those coding for amylase.

1. Red light converts inactive \(P_r\) to active \(P_{fr}\) (1 mark)
2. Active \(P_{fr}\) triggers the transcription of genes involved in germination (1 mark)

Oxygen acts as the terminal electron acceptor at the end of the electron transport chain, combining with electrons and protons to form water. Without oxygen, the electron transport chain cannot function, the proton gradient is not maintained, and ATP cannot be produced via oxidative phosphorylation.

1. Oxygen is the terminal electron acceptor / combines with electrons and protons to form water (1 mark)
2. Without oxygen, the electron transport chain stops, preventing the generation of the proton gradient / ATP synthesis via oxidative phosphorylation (1 mark)

When an action potential depolarises the presynaptic membrane, voltage-gated calcium channels open. Calcium ions (\(Ca^{2+}\)) diffuse into the presynaptic knob down their electrochemical gradient. This influx causes synaptic vesicles containing neurotransmitters (such as acetylcholine) to move towards and fuse with the presynaptic membrane, releasing their contents into the synaptic cleft by exocytosis.

1. Depolarisation opens voltage-gated channels, allowing calcium ions to diffuse into the presynaptic neurone (1 mark)
2. Calcium ions cause synaptic vesicles to fuse with the presynaptic membrane / release neurotransmitter by exocytosis (1 mark)

An increase in blood osmolarity is detected by osmoreceptors in the hypothalamus, which stimulates the pituitary gland to secrete antidiuretic hormone (ADH). ADH travels in the blood to the kidneys, where it binds to receptors on the collecting duct, increasing its permeability to water (via insertion of aquaporins). Consequently, more water is reabsorbed into the blood, resulting in a small volume of concentrated urine.

1. Osmoreceptors detect change, stimulating the pituitary gland to release more ADH (1 mark)
2. ADH increases permeability of the collecting duct to water, increasing water reabsorption (1 mark)

PCR involves cycling through high temperatures, including a step at approximately \(95^\circ\text{C}\) to denature double-stranded DNA. Standard DNA polymerases would denature and lose activity at this temperature. \(Taq\) polymerase is thermostable, meaning it remains functional after heating, enabling automated, repeated cycles of DNA amplification without manual replenishment of the enzyme.

1. PCR requires high temperatures (e.g. \(95^\circ\text{C}\)) to separate/denature DNA strands (1 mark)
2. \(Taq\) polymerase is thermostable / does not denature at high temperatures, allowing repeated amplification cycles (1 mark)

During muscle contraction, ATP binds to the myosin head, causing it to detach from the actin filament. The hydrolysis of ATP into ADP and inorganic phosphate (\(P_i\)) provides the energy to recock the myosin head into its high-energy state, ready to form a new cross-bridge further along the actin filament.

1. ATP binding causes the detachment of the myosin head from the actin filament (1 mark)
2. Hydrolysis of ATP provides the energy to recock/reset the myosin head (1 mark)

With repeated exposure to a harmless stimulus, calcium channels in the presynaptic membrane of the sensory neurone become less responsive and open less frequently. As a result, fewer calcium ions enter the presynaptic knob, which reduces the fusion of synaptic vesicles and the subsequent release of neurotransmitter into the synaptic cleft.

1. Calcium channels in the presynaptic membrane become less responsive / open less (1 mark)
2. Fewer calcium ions enter, leading to less neurotransmitter released by exocytosis (1 mark)

Upon water absorption, the embryo releases gibberellins, which diffuse to the aleurone layer. Gibberellins stimulate the transcription of the gene encoding amylase. The synthesized amylase then diffuses into the endosperm, where it hydrolyses starch reserves into soluble maltose and glucose, providing energy for embryo growth.

1. Gibberellins stimulate the transcription/synthesis of amylase in the aleurone layer (1 mark)
2. Amylase diffuses into the endosperm to hydrolyse starch into maltose/glucose (1 mark)

Red light is absorbed by Pr (phytochrome red), converting it into the active Pfr (phytochrome far-red) form. Short-day plants require a long, uninterrupted dark period to allow Pfr to slowly convert back into Pr. A brief flash of red light during the dark period rapidly regenerates Pfr. Because active Pfr inhibits flowering in short-day plants, its presence prevents the transcription of genes responsible for initiating flowering.

1. Red light converts Pr to Pfr (1 mark).
2. Pfr acts as a transcription inhibitor / represses flowering genes (1 mark).

When a harmless stimulus is repeated, the calcium ion channels on the presynaptic membrane become less responsive and do not open as readily. As a result, fewer calcium ions (\(\text{Ca}^{2+}\)) enter the presynaptic terminal. This leads to reduced exocytosis and less neurotransmitter release into the synaptic cleft. Consequently, there is insufficient neurotransmitter to bind to receptors on the postsynaptic membrane, failing to trigger an action potential.

1. Fewer calcium ions enter the presynaptic knob / calcium channels become less responsive (1 mark).
2. Less neurotransmitter is released by exocytosis, failing to trigger depolarisation in the postsynaptic membrane (1 mark).

Accumulation of lactate in the blood releases hydrogen ions (\(\text{H}^{+}\)), which lowers blood pH. This change is detected by chemoreceptors located in the carotid and aortic bodies, as well as the medulla oblongata. These chemoreceptors send nerve impulses to the ventilation centre in the medulla oblongata, which responds by sending more frequent impulses to the intercostal muscles and diaphragm to increase the rate and depth of breathing.

1. Lactate lowers blood pH / increases concentration of \(\text{H}^{+}\) ions (1 mark).
2. Detected by chemoreceptors which send impulses to the ventilation centre in the medulla to increase breathing rate (1 mark).

The ascending limb of the loop of Henle actively transports sodium (\(\text{Na}^{+}\)) and chloride (\(\text{Cl}^{-}\)) ions out of the tubule into the interstitial fluid of the renal medulla, while remaining impermeable to water. This creates a high solute concentration (low water potential) in the medulla. Because the descending limb is permeable to water but not to ions, water moves out of the descending limb by osmosis into the medulla, concentration-dependent, allowing for water reabsorption from the collecting duct.

1. Active transport of sodium / chloride ions out of the ascending limb into the interstitial fluid (1 mark).
2. Creates a low water potential in the medulla, so water moves out of the descending limb / collecting duct by osmosis (1 mark).

During the annealing step of PCR, the temperature is lowered to around \(55\text{ }^\circ\text{C}\). This allows short, single-stranded DNA primers to bind (anneal) via hydrogen bonding to complementary target sequences on the single-stranded DNA templates. Once bound, the primers provide a double-stranded attachment site for the Taq DNA polymerase enzyme to initiate the synthesis of the new DNA strands during the subsequent elongation step.

1. Allows primers to bind / hybridise to complementary sequences on single-stranded DNA (1 mark).
2. Provides a starting double-stranded section for DNA polymerase to bind / begin synthesis (1 mark).

Parkinson's disease is characterized by a loss of dopamine-producing neurons in the brain. Directly administering dopamine is ineffective because dopamine is a polar molecule that cannot cross the blood-brain barrier. In contrast, L-DOPA (levodopa) is a precursor that can cross this barrier via carrier proteins. Once in the brain, L-DOPA is decarboxylated by the enzyme dopa-decarboxylase to form dopamine, which replenishes the depleted neurotransmitter levels at synapses.

1. L-DOPA can cross the blood-brain barrier whereas dopamine cannot (1 mark).
2. L-DOPA is converted into dopamine in the brain to increase synaptic transmitter levels (1 mark).

ATP plays two critical roles in muscle contraction. First, the binding of a new ATP molecule to the myosin head causes it to detach from the actin binding site, breaking the cross-bridge. Second, the hydrolysis of this ATP into ADP and inorganic phosphate (\(\text{P}_i\)) by ATPase releases energy, which resets (re-cocks) the myosin head into its high-energy conformational state, ready to bind to another actin site further along the filament.

1. Binding of ATP to the myosin head causes it to detach from actin (1 mark).
2. Hydrolysis of ATP provides energy to reset / cock the myosin head (1 mark).

Transcription factors are proteins that regulate gene expression by binding to specific DNA sequences, such as promoter regions. To prevent transcription, a repressor transcription factor binds to the promoter or enhancer region of the target gene. This physically blocks RNA polymerase from binding to the DNA or prevents the formation of the transcription initiation complex. As a result, the gene cannot be transcribed into mRNA.

1. Transcription factor binds to the promoter / regulator region of the gene (1 mark).
2. This blocks RNA polymerase from binding, preventing the synthesis of mRNA (1 mark).

1. Phosphorylation of glucose to glucose phosphate activates the glucose molecule, making it more reactive for subsequent cleavage. 2. It adds a negative charge (phosphate groups) to the molecule, preventing it from passing through the cell membrane, thus trapping it inside the cell.

1 mark: For stating that phosphorylation activates glucose / makes it more reactive / lowers the activation energy. 1 mark: For stating that it adds a negative charge / polarises the glucose, preventing it from leaving the cell / diffusing across the cell membrane.

1. Exposure to far-red light causes the active form of phytochrome, Pfr, to be rapidly converted into the inactive form, Pr. This decreases the concentration of Pfr relative to Pr. 2. Since Pfr acts as a flowering inhibitor in short-day plants, the reduction of Pfr below a threshold level initiates flowering.

1 mark: Far-red light converts Pfr to Pr, reducing the levels of Pfr / increasing the proportion of Pr. 1 mark: Low levels of Pfr (or absence of Pfr) triggers / allows flowering in short-day plants.

1. During relaxation, calcium ions are actively transported back into the sarcoplasmic reticulum against their concentration gradient using ATP. 2. The decrease in sarcoplasmic calcium concentration causes calcium to detach from troponin, leading to tropomyosin moving back to block the myosin-binding sites on actin filaments, preventing cross-bridge formation.

1 mark: Calcium ions are actively transported / pumped back into the sarcoplasmic reticulum (using ATP). 1 mark: (This decrease in calcium concentration causes) tropomyosin to block the myosin-binding sites on actin, preventing cross-bridge formation.

1. PCR involves a high-temperature step (around 95°C) to denature double-stranded DNA into single strands. 2. Taq polymerase is thermostable (isolated from thermophilic bacteria) and does not denature at these high temperatures, whereas human DNA polymerase would denature and become inactive, requiring fresh enzyme to be added at every cycle.

1 mark: PCR requires a high temperature (around 90-95°C) to separate/denature DNA strands. 1 mark: Taq polymerase is thermostable / heat-tolerant and does not denature (whereas human DNA polymerase would denature / lose its tertiary structure).

1. Thermoreceptors in the hypothalamus monitor the core body temperature by detecting temperature changes in the blood flowing through the brain. 2. Thermoreceptors in the skin (peripheral thermoreceptors) monitor external environmental temperature changes, providing early warning signals to the hypothalamus before the core body temperature itself changes.

1 mark: Hypothalamus thermoreceptors detect core body temperature / blood temperature. 1 mark: Skin / peripheral thermoreceptors detect external / environmental temperature changes.

1. The myelin sheath acts as an electrical insulator, preventing ion flow across the axon membrane where it is present. 2. This forces depolarisation (and action potential generation) to occur only at the unmyelinated Nodes of Ranvier, allowing the action potential to 'jump' from one node to the next, which is much faster than continuous conduction.

1 mark: Myelin acts as an electrical insulator / prevents ion movement through the membrane (except at nodes). 1 mark: Depolarisation/action potentials can only occur at the Nodes of Ranvier, meaning the impulse jumps from node to node.

During recovery, lactate diffuses out of muscle fibres into the blood plasma and is transported to the liver. In hepatocytes, the enzyme lactate dehydrogenase converts lactate back to pyruvate by oxidizing it. Approximately 20% of this pyruvate enters the aerobic pathway (link reaction, Krebs cycle, and oxidative phosphorylation) to produce ATP. The remaining 80% of pyruvate is converted back into glucose and glycogen via gluconeogenesis, which is an anabolic process requiring the ATP generated by the oxidized portion. Repaying the oxygen debt (EPOC) involves elevated breathing rates post-exercise to supply the extra oxygen needed to: 1) restore muscle stores of ATP and creatine phosphate, 2) replenish oxygen bound to myoglobin and haemoglobin, and 3) meet the high oxygen demand of hepatocytes performing lactate conversion.

1. Lactate diffuses out of muscles and is transported via the blood to the liver (1 mark); 2. In liver hepatocytes, lactate is converted to pyruvate by lactate dehydrogenase (1 mark); 3. Around 20% of pyruvate is oxidized aerobically through the link reaction, Krebs cycle, and electron transport chain to produce ATP (1 mark); 4. Around 80% of pyruvate is converted back into glucose or glycogen via gluconeogenesis (1 mark); 5. Gluconeogenesis requires ATP produced by the aerobic oxidation of the remaining pyruvate (1 mark); 6. Post-exercise oxygen consumption (oxygen debt) provides extra oxygen to resynthesise creatine phosphate and ATP in muscle cells (1 mark); 7. Extra oxygen is also required to re-saturate myoglobin and haemoglobin with oxygen (1 mark).

When a shoot experiences unilateral light, IAA is translocated laterally from the illuminated side to the shaded side. This high concentration of IAA on the shaded side binds to receptors on the cell surface membrane, activating proton pumps (\(H^+\)-ATPases). These pumps actively transport hydrogen ions from the cytoplasm into the primary cell wall space, lowering the pH of the cell wall. The acidic environment activates expansins, which are specialized wall-loosening proteins. Expansins break the hydrogen bonds holding cellulose microfibrils and hemicelluloses together. This weakens and loosens the cell wall matrix. As the wall relaxes, water enters the vacuole by osmosis down a water potential gradient, generating turgor pressure. This internal turgor pressure stretches the weakened cell wall, permanently elongating the cells on the shaded side, causing the shoot to bend towards the light.

1. IAA is redistributed laterally to the shaded side of the shoot tip and diffuses down the shaded side (1 mark); 2. IAA stimulates proton pumps (\(H^+\)-ATPase) in the cell membrane (1 mark); 3. Hydrogen ions (\(H^+\)) are actively transported out of the cytoplasm and into the cell wall (1 mark); 4. The low pH / acidic environment in the cell wall activates expansin proteins (1 mark); 5. Expansins break hydrogen bonds between cellulose microfibrils and hemicelluloses, loosening the cell wall (1 mark); 6. Water enters the cell by osmosis down a water potential gradient (1 mark); 7. High turgor pressure stretches the weakened, flexible cell wall, causing cell elongation on the shaded side (1 mark).

When the siphon of Aplysia is repeatedly stimulated with a harmless stimulus, the sensory neurone is repeatedly depolarized. However, with repeated stimulation, the voltage-gated calcium ion channels on the presynaptic membrane of the sensory neurone become less sensitive and do not open as readily. This results in a significantly reduced influx of calcium ions (\(Ca^{2+}\)) into the presynaptic terminal. Because calcium ions are required to trigger vesicle transport, fewer synaptic vesicles undergo exocytosis to release neurotransmitters (specifically glutamate) into the synaptic cleft. Consequently, there is a lower concentration of neurotransmitter to diffuse across the cleft and bind to ligand-gated sodium channels on the postsynaptic membrane of the motor neurone. Fewer sodium channels open, meaning less sodium (\(Na^+\)) enters the motor neurone, leading to a smaller excitatory postsynaptic potential (EPSP) that fails to reach the threshold required to trigger an action potential. Without an action potential in the motor neurone, the effector muscle does not contract, and the gill withdrawal reflex is lost.

1. Repeated stimulation of the siphon sensory neurone causes presynaptic voltage-gated calcium (\(Ca^{2+}\)) channels to become less responsive (1 mark); 2. Fewer calcium ions (\(Ca^{2+}\)) enter the presynaptic neurone terminal (1 mark); 3. Fewer synaptic vesicles fuse with the presynaptic membrane (1 mark); 4. Less neurotransmitter (glutamate) is released by exocytosis into the synaptic cleft (1 mark); 5. Fewer neurotransmitter molecules bind to receptor-gated / ligand-gated sodium channels on the postsynaptic membrane (1 mark); 6. Fewer sodium ions (\(Na^+\)) enter the postsynaptic neurone, preventing depolarization from reaching the threshold (1 mark); 7. No action potential is generated in the motor neurone, so the effector muscle does not contract / gill withdrawal does not occur (1 mark).

The PCR process begins with denaturation, where the reaction mixture is heated to \(90-95^\circ\text{C}\) to break the hydrogen bonds holding the complementary base pairs of the double-stranded DNA template together, separating it into single strands. The temperature is then lowered to \(50-65^\circ\text{C}\) for the annealing step, which allows DNA primers to bind via complementary base pairing to the specific 3' ends of the target single-stranded DNA templates. Next, the mixture is heated to \(70-75^\circ\text{C}\) for the extension step, which is the optimum temperature for the thermostable DNA polymerase (typically Taq polymerase, isolated from Thermus aquaticus). Taq polymerase binds to the primed single strands and synthesizes a complementary strand by adding free deoxyribonucleoside triphosphates (dNTPs) to the 3' end of the primers. A thermostable DNA polymerase is absolutely essential because it is resistant to thermal denaturation at the high temperatures of \(90-95^\circ\text{C}\) required during denaturation. This eliminates the need to manually add fresh enzyme at the start of each new replication cycle, allowing the process to be fully automated and run through multiple exponential amplification cycles continuously.

1. Denaturation: Heat to \(90-95^\circ\text{C}\) to break hydrogen bonds between complementary strands, separating them into single strands (1 mark); 2. Annealing: Cool to \(50-65^\circ\text{C}\) to allow primers to bind / anneal to complementary target sequences (1 mark); 3. Extension: Heat to \(70-75^\circ\text{C}\), which is the optimum temperature for the DNA polymerase (1 mark); 4. DNA polymerase (Taq polymerase) synthesizes complementary DNA strands by adding free nucleotides (dNTPs) starting from the primers (1 mark); 5. The cycle of temperature changes is repeated multiple times to exponentially amplify the DNA sequence (1 mark); 6. Thermostable DNA polymerase (e.g., Taq polymerase) does not denature or lose its catalytic activity at \(95^\circ\text{C}\) (1 mark); 7. This allows the reaction to run continuously without the need to manually add new enzyme at each cycle, enabling automation (1 mark).

1. Recall the formula for calculating Respiratory Quotient: \(\text{RQ} = \frac{\text{volume of } CO_2 \text{ produced}}{\text{volume of } O_2 \text{ consumed}}\). 2. From the equation, 10 moles of \(CO_2\) are produced and 14 moles of \(O_2\) are consumed. 3. Substitute these values into the equation: \(\text{RQ} = \frac{10}{14} \approx 0.71428...\). 4. Rounding to 2 decimal places gives 0.71.

1. Award 1 mark for showing correct division of moles of \(CO_2\) by \(O_2\) in working: \(\frac{10}{14}\) or 0.714. 2. Award 2 marks for the correct final answer of 0.71. [Note: Do not award the second mark if incorrectly rounded, e.g., 0.7 or 0.72]

1. Calculate the reduction in length during contraction: 2.7 \(\mu\text{m}\) * 0.18 = 0.486 \(\mu\text{m}\). 2. Subtract this reduction from the relaxed length: 2.7 - 0.486 = 2.214 \(\mu\text{m}\). 3. Convert micrometres to nanometres: 2.214 * 1000 = 2214 \(\text{nm}\).

1. Award 1 mark for calculating the contracted length in micrometres (2.214 \(\mu\text{m}\)) OR for showing the correct conversion factor of 1000. 2. Award 2 marks for the correct final answer of 2214 (or 2214 \(\text{nm}\)).

1. Convert distance to metres: 82 cm = 0.82 m. 2. Convert time to seconds: 7.6 ms = 0.0076 s. 3. Calculate speed: 0.82 / 0.0076 = 107.895 \(\text{m s}^{-1}\). 4. Round to 3 significant figures: 108 \(\text{m s}^{-1}\).

1. Award 1 mark for correct conversion of units in working: 0.82 m and 0.0076 s, or showing the division 0.82 / 0.0076. 2. Award 2 marks for the correct final answer of 108 (or 108 \(\text{m s}^{-1}\)). [Note: Award maximum 1 mark if calculation is correct but rounded incorrectly.]

1. State the formula for PCR amplification: N = N0 * 2^n. 2. Substitute the values: N = (1.5 * 10^4) * 2^15 = 15,000 * 32,768 = 491,520,000. 3. Convert to scientific notation: 4.9152 * 10^8. 4. Round to 2 decimal places: 4.92 * 10^8 (or 4.92 x 10^8).

1. Award 1 mark for showing correct multiplication: 1.5 * 10^4 * 2^15 or 491,520,000. 2. Award 2 marks for correct final answer of 4.92 * 10^8 (or 4.92 x 10^8). [Note: Accept any standard formatting of the scientific notation. Reject answers not rounded to 2 decimal places, e.g., 4.9 * 10^8 or 4.915 * 10^8]