2025 IB DP Biology 模擬試題連答案詳解

Glycolysis of one glucose molecule yields a net of 2 ATP and 2 NADH, while converting glucose to 2 molecules of pyruvate. During the link reaction, the 2 pyruvate molecules are converted into 2 acetyl-CoA molecules, yielding 2 NADH (1 per pyruvate) and releasing 2 CO2. Acetyl-CoA then enters the citric acid cycle by combining with oxaloacetate to form citrate, which is isomerized to isocitrate. Because isocitrate dehydrogenase is inhibited, the pathway is blocked at this step. No further reactions in the citric acid cycle occur, meaning no additional ATP, NADH, or FADH2 are produced from this cycle. Summing the yields: Glycolysis (2 ATP, 2 NADH) + Link Reaction (0 ATP, 2 NADH) = 2 ATP, 4 NADH, and 0 FADH2.

[1 mark] Award for correct option B.
- Reject other choices as they incorrectly sum the yields of ATP, NADH, or FADH2 based on the metabolic block.

Active loading of sucrose into companion cells and sieve tube elements at the source requires a proton gradient. Proton pumps actively transport hydrogen ions out of the companion cells into the apoplast, creating a gradient that drives sucrose co-transport back into the phloem. Inhibiting these proton pumps halts active sucrose loading, meaning radioactive sucrose cannot be concentrated in the phloem of the source leaf. Consequently, the pressure-flow mechanism is disrupted, and the translocation of the radioactive tracer to the sink organ is significantly delayed or reduced.

[1 mark] Award for correct option B.
- Option A is incorrect as transport would decrease, not increase.
- Option C is incorrect as loading disruption does not lead to rapid xylem transport.
- Option D is incorrect because sucrose loading is primarily an active process, not purely passive.

The parental genotypes of the chromatids in the heterozygous female are \(Ab\) and \(aB\). Therefore, any offspring showing the recombinant phenotypes \(AB\) or \(ab\) must have resulted from a crossing-over event. To calculate the recombination frequency:
\(\text{Recombination Frequency} = \frac{\text{Number of recombinants}}{\text{Total number of offspring}} \times 100\%
\)\text{Recombination Frequency} = \frac{110 + 90}{410 + 390 + 110 + 90} \times 100\% = \frac{200}{1000} \times 100\% = 20\%\).

[1 mark] Award for correct option B.
- Calculation: 200 recombinants / 1000 total = 0.20 or 20%.
- Other options represent incorrect calculation methods (e.g., only dividing by parental offspring or misidentifying recombinant groups).

At rest, potassium leak channels allow potassium ions (\(\text{K}^+\)) to diffuse down their concentration gradient out of the neuron, leaving a net negative charge inside the cell. If these channels are blocked, fewer positive potassium ions can leave the cell, causing positive charges to accumulate inside relative to the untreated state. This decreases the electrical difference across the membrane, making the membrane potential less negative (depolarization).

[1 mark] Award for correct option B.
- Option A is incorrect because hyperpolarization requires more, not fewer, positive ions to leave (or more negative ions to enter).
- Option C is incorrect because potassium leak channels are critical to the baseline resting potential, which cannot be immediately fully compensated by the pump alone.
- Option D is incorrect because blocking leak channels does not trigger a sudden inward rush of potassium ions.

The total water potential of the plant cell (\(\psi_{\text{cell}}\)) is calculated using the formula:
\(\psi_{\text{cell}} = \psi_s + \psi_p = -0.8\text{ MPa} + 0.3\text{ MPa} = -0.5\text{ MPa}\).
Since the water potential of the surrounding solution is also \(-0.5\text{ MPa}\), there is no water potential gradient between the inside of the cell and the beaker. Therefore, there is no net movement of water.

[1 mark] Award for correct option C.
- Correctly calculates cell water potential as -0.5 MPa.
- Recognizes that equal water potentials prevent net water movement.

According to Chargaff's rules for double-stranded DNA, the percentage of adenine (A) is equal to thymine (T), and the percentage of guanine (G) is equal to cytosine (C).
If \(A = 28\%\), then \(T = 28\%\).
Together, \(A + T = 56\%\).
The remaining bases (G and C) must account for the rest of the DNA: \(100\% - 56\% = 44\%\).
Since \(G = C\), the percentage of cytosine is \(44\% / 2 = 22\%\).

[1 mark] Award for correct option A.
- Step 1: Recognize A = T = 28%.
- Step 2: Calculate G + C = 100% - 56% = 44%.
- Step 3: Divide 44% by 2 to find Cytosine = 22%.

Natural selection acts on pre-existing genetic variations within a population. When an antibiotic is introduced, it acts as a selective agent, killing off susceptible bacteria. Any bacteria that already possess a mutation conferring resistance survive, reproduce, and pass on their resistance genes to their offspring. This increases the frequency of the resistance allele in the population over time. Antibiotics do not induce mutations to occur.

[1 mark] Award for correct option B.
- Reject A: Mutations are random and not induced by exposure to the antibiotic to meet a survival need.
- Reject C: Organisms cannot consciously or actively alter their metabolic pathways in response to environmental needs.
- Reject D: Conjugation can transfer plasmids, but natural selection relies on the selective pressure killing susceptible individuals to change allele frequencies.

Aquaporins are channel proteins that facilitate the rapid transport of water across the cell membrane down its osmotic gradient (facilitated diffusion). When placed in a hypotonic solution, water moves into the cells. Red blood cells, with high levels of aquaporins, will experience extremely rapid water influx, leading to swelling and rapid lysis. In contrast, cells lacking aquaporins can only take in water through slow, simple diffusion across the lipid bilayer, meaning they will swell and lyse at a much slower rate.

[1 mark] Award for correct option A.
- Correctly links the presence of aquaporins to rapid water movement via facilitated diffusion.
- Correctly predicts the difference in lysis rates between the two cell types.

Aerobic ATP yield: \(0.1\text{ mol glucose} \times 30\text{ ATP/glucose} = 3.0\text{ mol ATP}\). Anaerobic ATP yield: \(0.5\text{ mol glucose} \times 2\text{ ATP/glucose} = 1.0\text{ mol ATP}\). The ratio of aerobic to anaerobic ATP produced is \(3.0 : 1.0 = 3 : 1\).

Award [1] for the correct answer A. No partial marks.

In the phloem, active transport is used to pump protons out of companion cells, generating an electrochemical gradient. Protons then flow back down their gradient via co-transporters, bringing sucrose with them into the companion cell-sieve tube complex. The resulting high solute concentration lowers the water potential (solute potential becomes more negative), drawing water into the phloem from the xylem by osmosis. This accumulation of water creates high hydrostatic pressure at the source.

Award [1] for the correct answer A. No partial marks.

The recombinant offspring are those with non-parental combinations: gray body/vestigial wings (105) and black body/normal wings (95). Total recombinants = \(105 + 95 = 200\). Total offspring = \(415 + 385 + 105 + 95 = 1000\). Recombination frequency = \((200 / 1000) \times 100\% = 20\%\).

Award [1] for the correct answer B. No partial marks.

During repolarization, the membrane potential must become more negative. This is achieved by the closing/inactivation of voltage-gated sodium channels (preventing further entry of \(\text{Na}^+\)) and the opening of voltage-gated potassium channels (allowing \(\text{K}^+\)) to flow down its electrochemical gradient out of the cell.

Award [1] for the correct answer B. No partial marks.

At osmotic equilibrium, the water potential of the cell (\(\Psi_{\text{cell}}\)) equals the water potential of the external solution (\(\Psi_{\text{sol}}\)), so \(\Psi_{\text{cell}} = -0.45\text{ MPa}\). Using the formula \(\Psi = \Psi_s + \Psi_p\), we have \(-0.45\text{ MPa} = -0.75\text{ MPa} + \Psi_p\). Solving for \(\Psi_p\) yields \(\Psi_p = -0.45 - (-0.75) = +0.30\text{ MPa}\).

Award [1] for the correct answer C. No partial marks.

1) In any double-stranded DNA molecule, purines (adenine and guanine) always account for exactly 50% of the total bases because every purine pairs with a pyrimidine (A=T, G=C). 2) A G-C base pair has 3 hydrogen bonds, and an A-T base pair has 2. For 4 G-C pairs, there are \(4 \times 3 = 12\) hydrogen bonds. For 6 A-T pairs, there are \(6 \times 2 = 12\) hydrogen bonds. Total hydrogen bonds = \(12 + 12 = 24\).

Award [1] for the correct answer A. No partial marks.

Natural selection acts on existing genetic variation within a population. Antibiotics do not cause or direct the mutations that provide resistance (making A and C incorrect); instead, they kill susceptible bacteria. The pre-existing resistant variants survive, multiply, and propagate the resistant traits.

Award [1] for the correct answer B. No partial marks.

Water's high specific heat capacity means that it requires a relatively large amount of heat energy to raise its temperature, and conversely releases a large amount of energy when cooling. This thermal buffer property prevents rapid temperature fluctuations in aquatic environments.

Award [1] for the correct answer B. No partial marks.

First, determine the total pyruvate produced from glycolysis: 6 molecules of glucose yield 12 molecules of pyruvate (2 per glucose). If 2 pyruvates are diverted to lactate fermentation, 10 pyruvates continue to aerobic respiration. Each pyruvate undergoes the link reaction (releasing 1 \(CO_2\)) and the Krebs cycle (releasing 2 \(CO_2\)), giving a total of 3 \(CO_2\) molecules per pyruvate. Therefore, 10 pyruvates \(\times\) 3 \(CO_2\) = 30 \(CO_2\) molecules.

Award 1 mark for the correct answer C. [1]

Apoplastic phloem loading requires a proton gradient. Proton pumps use ATP to actively transport hydrogen ions (\(H^+\)) out of the companion cell cytoplasm into the cell wall (apoplast). This creates a high concentration of \(H^+\), which then flows back down its electrochemical gradient through a co-transporter, bringing sucrose with it against its concentration gradient.

Award 1 mark for the correct answer B. [1]

To find the recombination frequency, identify the recombinant offspring classes (the less frequent phenotypes resulting from crossing over). These are red-eyed/vestigial-winged (105) and purple-eyed/normal-winged (95). Total recombinants = 105 + 95 = 200. Total offspring = 412 + 388 + 105 + 95 = 1000. Recombination frequency = (200 / 1000) \(\times\) 100 = 20%.

Award 1 mark for the correct answer B. [1]

Neonicotinoids bind to nicotinic acetylcholine receptors. Since acetylcholinesterase cannot break down neonicotinoids, the receptors remain permanently activated. This leads to the continuous opening of ligand-gated sodium channels, causing sustained depolarization of the postsynaptic membrane, leading to overstimulation and death.

Award 1 mark for the correct answer B. [1]

Water potential (\(\psi\)) is calculated as \(\psi = \psi_s + \psi_p\). For Cell A, \(\psi = -0.8 + 0.3 = -0.5\) MPa. For Cell B, \(\psi = -0.9 + 0.1 = -0.8\) MPa. Water moves from a region of higher water potential (less negative) to a region of lower water potential (more negative). Therefore, net water movement is from Cell A (\(-0.5\) MPa) to Cell B (\(-0.8\) MPa).

Award 1 mark for the correct answer A. [1]

Since guanine (G) represents 17%, cytosine (C) must also represent 17% due to complementary base pairing. Together, G and C account for 34% of the bases. The remaining 66% must be adenine (A) and thymine (T). Since A pairs with T, thymine represents half of this remainder: 33%. The total number of bases in the molecule is 340 / 0.17 = 2000 bases. Therefore, the number of thymine bases is 33% of 2000, which is 660 bases.

Award 1 mark for the correct answer B. [1]

Directional selection occurs when natural selection favors one of the extreme phenotypes, shifting the population mean. Here, the mean shifts towards the light brown extreme. The predator is the biotic factor actively selecting against darker lizards, making it the selective agent, whereas the sand is the environmental backdrop.

Award 1 mark for the correct answer B. [1]

The active transport cycle of the \(Na^+/K^+\) pump uses energy from the hydrolysis of one ATP molecule to export three sodium (\(Na^+\)) ions out of the cytoplasm and import two potassium (\(K^+\)) ions into the cell. This electrogenic stoichiometry helps maintain both the concentration gradients and the resting membrane potential.

Award 1 mark for the correct answer A. [1]

Glycolysis produces 2 molecules of pyruvate from 1 molecule of glucose. Each pyruvate molecule that enters the link reaction is decarboxylated and oxidized to produce 1 molecule of Acetyl-CoA, 1 molecule of carbon dioxide (\(CO_2\)), and 1 molecule of reduced NAD (NADH). Therefore, for 2 pyruvates (from 1 glucose), the products are doubled: 2 Acetyl-CoA, 2 \(CO_2\), and 2 NADH.

Award [1] for the correct answer (A). Reject alternative answers because they do not reflect the stoichiometry of two pyruvate molecules entering the link reaction per glucose molecule.

For the bicuspid (mitral) valve to be open, the pressure in the left atrium must be higher than the pressure in the left ventricle, allowing blood to flow into the ventricle. For the aortic semilunar valve to remain closed, the pressure in the aorta must be higher than the pressure in the left ventricle, preventing backflow of blood from the aorta into the ventricle.

Award [1] for the correct answer (A). Reject other options as they describe conditions where the bicuspid valve would close or the aortic valve would open.

The parental phenotypes are the most frequent classes: wild type body, normal wings (415) and black body, vestigial wings (385). The recombinant phenotypes are the less frequent, non-parental classes: wild type body, vestigial wings (105) and black body, normal wings (95). Total offspring = \(415 + 385 + 105 + 95 = 1000\). Total recombinants = \(105 + 95 = 200\). Recombination frequency = \(\frac{200}{1000} \times 100\% = 20\%\).

Award [1] for the correct calculation and choice (B). Reject other answers that represent incorrect combinations of parental and recombinant phenotypes.

Immediately after peak depolarization, during repolarization and the subsequent refractory period, voltage-gated sodium channels are inactivated (by their inactivation gate), preventing further sodium influx. Meanwhile, voltage-gated potassium channels are open, allowing potassium ions to diffuse out of the cell down their electrochemical gradient to restore the resting membrane potential.

Award [1] for the correct answer (A). Reject B, C, and D as they describe incorrect channel states for the repolarization/refractory phase of an action potential.

First, calculate the initial water potential of the cell: \(\psi_{\text{cell}} = \psi_s + \psi_p = -0.8\text{ MPa} + 0.3\text{ MPa} = -0.5\text{ MPa}\). The water potential of the solution in the beaker is \(-0.7\text{ MPa}\). Water always flows from a region of higher (less negative) water potential to a region of lower (more negative) water potential. Therefore, water will flow out of the cell (from \(-0.5\text{ MPa}\) to \(-0.7\text{ MPa}\)). At equilibrium, the water potential of the cell must equal the water potential of the surrounding solution, which is \(-0.7\text{ MPa}\).

Award [1] for the correct option (A). Reject options suggesting inward flow of water or incorrect final equilibrium water potential.

The sodium-potassium pump actively transports three sodium ions (\(3\text{ }Na^+\)) out of the cell and two potassium ions (\(2\text{ }K^+\)) into the cell. This process uses the energy derived from the hydrolysis of one molecule of ATP.

Award [1] for the correct answer (A). Reject alternative options as they state incorrect ion ratios or directions of movement.

Without oxygen acting as the terminal electron acceptor, the electron transport chain (ETC) becomes backed up. NADH and FADH2 can no longer donate their electrons to the protein complexes of the ETC, preventing the regeneration of NAD+ and FAD. Consequently, active proton pumping into the intermembrane space ceases, causing the proton gradient to dissipate. Without this electrochemical gradient (proton-motive force), ATP synthase cannot phosphorylate ADP to generate ATP.

[0.83 marks] State that oxygen is the terminal/final electron acceptor of the electron transport chain.
[0.83 marks] Explain that without oxygen, electron transport stops, preventing the maintenance of the proton gradient / preventing ATP synthase from producing ATP.

Cohesion is the intermolecular attraction between water molecules. Because water is a polar molecule, the highly electronegative oxygen atom of one water molecule forms a hydrogen bond with the electropositive hydrogen atom of an adjacent water molecule. These collective hydrogen bonds hold the water column together, allowing it to withstand high negative pressure (tension) during transpiration without breaking.

[0.83 marks] Correctly identify 'cohesion' as the physical property.
[0.83 marks] Explain that cohesion is established via hydrogen bonding between polar water molecules.

Using a genetic cross (Punnett square) between heterozygous parents:
- Father's gametes: \(I^A\) and \(i\)
- Mother's gametes: \(I^B\) and \(i\)

Offspring genotypes are:
1. \(I^A I^B\) (Blood group AB)
2. \(I^A i\) (Blood group A)
3. \(I^B i\) (Blood group B)
4. \(i i\) (Blood group O)

The probability of producing a child with genotype \(ii\) (blood group O) is 1 out of 4, which equals 25%.

[0.83 marks] Correctly identify that blood group O requires the homozygous recessive genotype \(ii\) (or \(I^O I^O\)).
[0.83 marks] State the correct probability of 25% (also accept 0.25 or 1/4).

Myelin sheaths, formed by Schwann cells or oligodendrocytes, prevent the leakage of ions across the axon membrane. Consequently, action potentials cannot be generated in the myelinated segments. Instead, action potentials jump from one unmyelinated gap (node of Ranvier) to the next, where voltage-gated sodium channels are highly concentrated. This rapid progression is known as saltatory conduction and is significantly faster than continuous propagation in unmyelinated axons.

[0.83 marks] Explain that myelin acts as an insulator / prevents depolarization along the covered axon segments.
[0.83 marks] State that depolarization / action potentials can only occur at the nodes of Ranvier, allowing the impulse to jump from node to node.

Water moves down its water potential gradient, from a higher (less negative) water potential to a lower (more negative) water potential. Because the cell's water potential (\(-0.4\text{ MPa}\)) is higher than the solution's (\(-0.9\text{ MPa}\)), net water movement will be out of the cell via osmosis. As water leaves the vacuole and cytoplasm, the protoplast shrinks and pulls away from the cell wall, leading to plasmolysis.

[0.83 marks] State that net water movement is out of the cell (from higher to lower water potential).
[0.83 marks] Describe the resulting state of the protoplast as plasmolysed / undergoing plasmolysis.

According to Chargaff's rules of base pairing:
- Cytosine (C) pairs with Guanine (G), so if \(C = 34\%\), then \(G = 34\%\).
- Together, \(C + G = 34\% + 34\% = 68\%\).
- The remaining bases are Adenine (A) and Thymine (T), which make up \(100\% - 68\% = 32\%\) of the total DNA.
- Since \(A = T\), the percentage of adenine is \(32\% / 2 = 16\%\).

[0.83 marks] Show that \(C + G = 68\%\) (or identify that \(A + T = 32\%\)).
[0.83 marks] Correctly calculate that adenine represents 16% of the bases.

The introduction of an antibiotic kills susceptible bacterial cells, eliminating non-resistant strains. This acts as a powerful selective pressure. Any bacteria possessing pre-existing mutations that confer resistance will survive. Due to reduced competition, these resistant individuals reproduce and pass on their resistance alleles to subsequent generations, increasing the proportion of resistant bacteria in the gene pool over time.

[0.83 marks] Explain that antibiotics act as a selective pressure by killing non-resistant/susceptible bacteria, allowing resistant mutants to survive.
[0.83 marks] State that surviving bacteria reproduce and pass on the resistance alleles/genes to their offspring, increasing the frequency of resistance in the population.

Facilitated diffusion is a passive process that relies on the kinetic energy of the particles to move them down their concentration gradient (from high to low concentration) via membrane proteins. Active transport, conversely, requires metabolic energy in the form of ATP to pump substances against their concentration gradient (from low to high concentration) using specific protein pumps.

[0.83 marks] Distinguish energy: facilitated diffusion is passive / does not require ATP, while active transport is active / requires ATP.
[0.83 marks] Distinguish direction: facilitated diffusion is down/along the concentration gradient, while active transport is against/up the concentration gradient.

Yeast undergoing anaerobic respiration produces carbon dioxide gas but does not consume oxygen. This results in a net increase in gas volume and pressure inside the sealed respirometer tube. This pressure increase forces the colored indicator liquid in the capillary tube to move away from the chamber. The distance moved by the liquid per unit time is a measure of the rate of respiration.

Award 1 mark for identifying carbon dioxide (CO2) as the gas produced. Award 1 mark for explaining that the production of CO2 increases volume/pressure, pushing the indicator fluid.

At the source, sucrose is actively transported (loaded) into the sieve tube elements. This increases the solute concentration, thereby lowering the water potential inside the phloem. Consequently, water moves from the nearby xylem vessel into the phloem sieve tube by osmosis. Since the sieve tube wall is rigid, the entry of water causes an increase in hydrostatic pressure.

Award 1 mark for mentioning active transport/loading of sucrose into the phloem sieve tube. Award 1 mark for describing the movement of water by osmosis from xylem into phloem, raising the pressure.

To determine whether an organism expressing a dominant trait is homozygous dominant or heterozygous, a test cross is performed by crossing it with a homozygous recessive individual. If the dominant parent is heterozygous (e.g., Bb), the cross Bb x bb will yield offspring with a phenotypic ratio of 1 dominant to 1 recessive (1:1 or 50% each).

Award 1 mark for identifying the test cross. Award 1 mark for stating the correct phenotypic ratio of 1:1 (or 50% dominant and 50% recessive).

Myelination increases the speed of nerve impulses. The myelin sheath acts as an electrical insulator, preventing depolarization and ion movement across the axonal membrane except at the unmyelinated nodes of Ranvier. The action potential is forced to jump from one node to the next, a process called saltatory conduction, which is significantly faster than continuous propagation.

Award 1 mark for naming saltatory conduction (or jumping from node to node / nodes of Ranvier). Award 1 mark for describing the insulating role of myelin which prevents ion flow except at the nodes.

The total water potential of a cell is calculated using the formula: \(\psi_w = \psi_s + \psi_p\). For this cell: \(\psi_w = -0.8\text{ MPa} + 0.2\text{ MPa} = -0.6\text{ MPa}\). Since the external solution has a higher water potential (\(-0.3\text{ MPa}\)) than the plant cell (\(-0.6\text{ MPa}\)), water moves down its water potential gradient, resulting in net water movement into the cell.

Award 1 mark for calculating the correct initial cell water potential of -0.6 MPa. Award 1 mark for correctly predicting net water movement into the cell.

The two main structural differences besides strandedness are: 1. DNA nucleotides contain the pentose sugar deoxyribose, while RNA nucleotides contain ribose. 2. DNA utilizes the nitrogenous base thymine (T), whereas RNA utilizes uracil (U) instead of thymine.

Award 1 mark for identifying deoxyribose in DNA and ribose in RNA. Award 1 mark for identifying thymine in DNA and uracil in RNA.

The sodium-potassium pump is an active transport protein that uses energy from ATP hydrolysis to pump ions against their concentration gradients. In each cycle of its conformational change, the pump binds and expels three sodium ions (\(3\text{ Na}^+\)) out of the cell, while importing two potassium ions (\(2\text{ K}^+\)) into the cell.

Award 1 mark for stating that three sodium ions (\(3\text{ Na}^+\)) are pumped out of the cell. Award 1 mark for stating that two potassium ions (\(2\text{ K}^+\)) are pumped into the cell.

1. Oxygen acts as the terminal electron acceptor at the end of the electron transport chain, combining with electrons and protons to form water.
2. Succinate dehydrogenase is a key enzyme in the citric acid (Krebs) cycle. Inhibiting it stops the cycle, which severely reduces the production of reduced coenzymes (NADH and FADH2).
3. Without these reduced coenzymes, there is a lack of electron donors for the electron transport chain. The flow of electrons slows down, and consequently, the rate at which oxygen accepts electrons and is converted to water decreases.

[1 mark] Identifying oxygen as the terminal/final electron acceptor (which combines with protons/electrons to form water).
[1 mark] Explaining that inhibiting the citric acid cycle reduces the production of reduced NAD/FAD (NADH/FADH2), meaning fewer electrons are supplied to the electron transport chain.

1. Transpiration occurs at the leaves (top of the tree) when water vapor evaporates out of stomata, creating a tension or negative pressure gradient.
2. This tension pulls the water column upwards.
3. Water molecules are highly cohesive due to hydrogen bonding, forming continuous columns under tension.
4. The transmission of this physical tension from the leaves down to the roots is not instantaneous; the elastic properties of the xylem walls and the long distance create a physical propagation delay (time lag) before the pulling force is registered at the base of the trunk.

[1 mark] Explaining that transpiration/evaporation at the leaves generates the initial tension/pull (transpiration pull) at the top of the tree.
[1 mark] Describing how cohesion between water molecules transmits this tension downwards along the continuous water column, leading to a delay in the movement starting at the bottom.

1. The formation of recombinant genotypes from linked genes is caused by crossing over (or homologous recombination), where non-sister chromatids of homologous chromosomes exchange genetic material.
2. This process begins during prophase I of meiosis, specifically when homologous chromosomes are synapsed (paired).

[1 mark] Naming crossing over (or homologous recombination) as the process that breaks linkage and produces recombinant chromatids.
[1 mark] Identifying prophase I (specifically during synapsis/pachytene phase) as the stage where crossing over occurs.

1. During an action potential, depolarization is caused by the rapid influx of sodium ions (\(Na^+\)) into the axon through voltage-gated sodium channels.
2. When tetrodotoxin blocks these channels, sodium influx is prevented, meaning the membrane potential cannot depolarize (it remains at or near resting potential).
3. Because threshold potential is not reached, an action potential cannot be generated or propagated along the axon, as there are no local currents to depolarize adjacent membrane segments.

[1 mark] Explaining that blocking the voltage-gated sodium channels prevents sodium influx, preventing depolarization of the axon membrane.
[1 mark] Explaining that without depolarization/action potential generation, local currents cannot be established to propagate the nerve impulse to adjacent regions.

1. Pure water has a water potential (\(\Psi\)) of \(0\text{ MPa}\). At dynamic equilibrium, the total water potential of the cell must equal the water potential of the surrounding environment, so \(\Psi_{cell} = 0\text{ MPa}\).
2. The formula for water potential is \(\Psi = \Psi_s + \Psi_p\).
3. Substituting the values: \(0 = -0.75\text{ MPa} + \Psi_p\).
4. Solving for pressure potential: \(\Psi_p = +0.75\text{ MPa}\).

[1 mark] Correctly stating that the total water potential (\(\Psi\)) of the cell at equilibrium is \(0\text{ MPa}\) (since it is in pure water).
[1 mark] Correctly calculating the turgor pressure (pressure potential, \(\Psi_p\)) as \(+0.75\text{ MPa}\) (accept \(0.75\text{ MPa}\)).

1. The core of a nucleosome consists of eight histone proteins (an octamer, consisting of two copies each of histones H2A, H2B, H3, and H4).
2. The double-stranded DNA wraps approximately 1.67 times around this histone core.
3. When DNA is wrapped tightly in nucleosomes, transcription factors and RNA polymerase cannot physically access promoter sequences on the DNA.
4. Chemical modifications (such as histone acetylation) can loosen this association, allowing transcription to occur, thereby regulating gene expression.

[1 mark] Stating that the nucleosome core consists of an octamer of eight histone proteins wrapped with DNA.
[1 mark] Outlining that packaging/supercoiling DNA around nucleosomes restricts physical access of RNA polymerase/transcription factors to the DNA, thereby silencing genes unless histone modifications alter the packaging.

1. The prolonged drought acts as a selective pressure favoring plants with the thick-petal phenotype.
2. These adapted individuals have a higher survival rate (differential survival) and reproductive success.
3. Consequently, they pass on the alleles conferring thick petals to their offspring.
4. Over successive generations, the frequency of the allele for thick petals will increase, while the allele for thin petals will decrease.

[1 mark] Stating that the allele for thick petals will increase in frequency, while the allele for thin petals will decrease in frequency.
[1 mark] Explaining that thick-petaled plants have a selective/survival advantage during drought, leading to higher reproductive success and the transmission of their alleles to the offspring.

1. Water molecules are polar and form hydrogen bonds with neighboring water molecules.
2. A large amount of heat energy is required to break these hydrogen bonds before the molecules can move faster and increase the temperature (kinetic energy) of the liquid.
3. This high specific heat capacity means that large bodies of water (lakes, oceans) change temperature very slowly.
4. This provides a stable thermal environment for aquatic organisms, protecting them from extreme or rapid temperature fluctuations that could denature enzymes or disrupt metabolic processes.

[1 mark] Explaining that polar water molecules form extensive hydrogen bonds that require a large amount of energy to break/overcome to increase the molecular kinetic energy (temperature).
[1 mark] Stating that this provides a thermally stable environment for aquatic organisms, preventing rapid temperature changes.

During the link reaction, pyruvate is decarboxylated and oxidized. The hydrogen/electrons removed during oxidation are transferred to NAD+, reducing it to NADH + H+. This NADH then transports these high-energy electrons and protons to the inner mitochondrial membrane for use in the electron transport chain (oxidative phosphorylation).

Award [1 mark] for stating that NAD+ acts as an electron/hydrogen acceptor / is reduced to NADH.
Award [1 mark] for stating that NADH transports these electrons/protons to the electron transport chain/cristae.

Companion cells are highly specialized to assist sieve tube elements. They contain a high density of mitochondria to supply ATP for the active transport of protons. They possess plasma membrane proton pumps and sucrose-proton cotransporters that actively pump protons out into the cell wall, creating a gradient that drives sucrose uptake. Their folded membranes increase the surface area available for these transport proteins, and plasmodesmata connect them directly to sieve tube members for easy diffusion of loaded sucrose.

Award [1 mark] for mentioning abundant mitochondria to generate ATP for active transport of protons.
Award [1 mark] for mentioning co-transporter proteins / proton pumps in the membrane to load sucrose against its concentration gradient (or plasmodesmata allowing transport).

The cross is RW x RW. The gametes produced by each parent are 50% R and 50% W. A Punnett square shows that the offspring genotypic ratio is 1 RR : 2 RW : 1 WW. Because R and W are codominant, RR plants have red flowers, RW plants have pink flowers, and WW plants have white flowers. Since every genotype corresponds to a unique phenotype, the phenotypic ratio directly mirrors the genotypic ratio of 1 red : 2 pink : 1 white (1:2:1).

Award [1 mark] for identifying the genotypic ratio of the offspring as 1 RR : 2 RW : 1 WW (or showing a correct Punnett square).
Award [1 mark] for explaining that codominance means each genotype (RR, RW, WW) results in a distinct phenotype (red, pink, white respectively).

When an action potential occurs, voltage-gated sodium channels open and sodium ions (Na+) rush into the axon, depolarizing that region. The high concentration of Na+ inside this active region causes these ions to diffuse sideways along the inside of the axon to the adjacent, resting region. This movement of ions constitutes local currents. These local currents depolarize the adjacent membrane area. Once this adjacent area reaches the threshold potential, its own voltage-gated sodium channels open, propagating the action potential forward.

Award [1 mark] for mentioning that Na+ influx in the depolarized region creates local currents / sideways diffusion of sodium ions.
Award [1 mark] for explaining that these local currents depolarize the adjacent section of the membrane to the threshold potential, triggering the opening of neighboring voltage-gated sodium channels.

The initial water potential of the plant cell is \(\psi = \psi_s + \psi_p = -0.7\text{ MPa} + 0\text{ MPa} = -0.7\text{ MPa}\). The surrounding solution has a water potential of -1.2 MPa, which is lower (more negative) than the cell. Therefore, water will move out of the cell by osmosis, down the water potential gradient (from -0.7 MPa to -1.2 MPa). As water leaves, the volume of the vacuole and cytoplasm decreases, causing the plasma membrane to retract from the rigid cell wall, a state known as plasmolysis.

Award [1 mark] for stating that net water movement is out of the cell because the water potential of the cell (-0.7 MPa) is higher than that of the solution (-1.2 MPa).
Award [1 mark] for stating that the cell will undergo plasmolysis / become plasmolysed (or the cell membrane pulls away from the cell wall).

DNA and RNA are nucleic acids with distinct structural differences. The pentose sugar in DNA is deoxyribose (which has one less oxygen atom on carbon 2 than ribose), while the pentose sugar in RNA is ribose. Regarding nitrogenous bases, DNA utilizes thymine (T) to pair with adenine (A), whereas RNA utilizes uracil (U) instead of thymine.

Award [1 mark] for distinguishing the sugars: DNA has deoxyribose while RNA has ribose.
Award [1 mark] for distinguishing the bases: DNA has thymine (T) while RNA has uracil (U).

Selective breeding involves humans choosing individuals with specific, desirable phenotypic traits to reproduce. Over many generations, this targeted breeding significantly alters the allele frequencies and physical characteristics of the population. The wide diversity of modern dog breeds, all descended from a common wolf-like ancestor, demonstrates that species can undergo profound morphological and behavioral changes over time in response to selective pressures. This acts as an accelerated model of natural selection, proving that cumulative microevolutionary changes can lead to macroevolutionary differences.

Award [1 mark] for explaining that humans select and breed individuals with specific traits, altering gene/allele frequencies over generations.
Award [1 mark] for explaining that the immense phenotypic variation in dog breeds demonstrates the power of selection to drive cumulative evolutionary change from a common ancestor.

Water transport in the xylem relies on the transpiration pull, which puts the water column under immense tension (negative pressure). Cohesion, due to hydrogen bonding between adjacent water molecules, ensures that water forms a continuous, unbroken column from the roots to the leaves. Adhesion, due to hydrogen bonding between water molecules and the hydrophilic cellulose in the xylem vessel walls, helps support the water column against gravity and prevents the column from breaking (cavitation) under tension.

Award [1 mark] for explaining that cohesion (hydrogen bonds between water molecules) maintains a continuous/unbroken column of water.
Award [1 mark] for explaining that adhesion (interaction between water and xylem walls) prevents the water column from dropping/breaking under tension.

1. The sodium-potassium pump is an active transport protein that moves ions against their concentration gradients. 2. It actively pumps three sodium ions (\(\text{Na}^+\)) out of the cell for every two potassium ions (\(\text{K}^+\)) it imports. 3. Because more positive charges are moved out than in, this generates a net positive charge outside the axon and a net negative charge inside. 4. This process requires energy derived from the hydrolysis of ATP.

Award [1] for stating that 3 sodium ions (\(\text{Na}^+\)) are pumped out and 2 potassium ions (\(\text{K}^+\)) are pumped in. Award [1] for explaining that this unequal movement of cations results in a net negative charge inside (or net positive charge outside) OR that it requires energy from ATP to move ions against their concentration gradients.

1. Anaerobic respiration in yeast produces a net yield of 2 ATP per glucose molecule, which are generated solely during glycolysis. 2. In the absence of oxygen, the link reaction, Krebs cycle, and oxidative phosphorylation (electron transport chain) cannot take place. 3. As a result, glucose is only partially oxidized to ethanol, leaving a significant amount of chemical energy untapped in the bonds of the ethanol molecule. In contrast, aerobic respiration fully oxidizes glucose to carbon dioxide and water, yielding up to 30-32 ATP.

Award [1] for identifying the net yield of 2 ATP molecules per glucose. Award [1] for explaining that glucose is only partially oxidized / only glycolysis occurs, OR that the electron transport chain / oxidative phosphorylation / Krebs cycle does not occur due to the absence of oxygen.

The explanation must cover the sequential opening and closing of voltage-gated Na+ and K+ channels, the direction of ion movement (diffusion into/out of the cell), the electrochemical gradients involved, and the corresponding changes in membrane potential (threshold, depolarization peak, and return to resting potential).

Award 1 mark for each of the following points, up to a maximum of 5 marks:
1. A stimulus causes the resting membrane potential to rise and reach the threshold potential of approximately -55 mV.
2. Reaching the threshold triggers the opening of voltage-gated sodium (Na+) channels.
3. Sodium (Na+) ions rapidly diffuse into the neuron down their concentration and electrochemical gradient, depolarizing the membrane to approximately +30 mV.
4. At peak depolarization, voltage-gated sodium channels close/inactivate and voltage-gated potassium (K+) channels open.
5. Potassium (K+) ions diffuse out of the neuron down their concentration and electrochemical gradient, repolarizing the membrane back toward a negative potential.
[An additional 1 mark is available for clarity of communication, structured sequencing, and accurate biological terminology, bringing the total scaled contribution to 5.33 marks.]

The response must explicitly connect specific structural compartments of the mitochondrion (cristae, intermembrane space, matrix, outer membrane, and internal DNA/ribosomes) to their exact biochemical roles in the stages of aerobic respiration (electron transport chain, chemiosmosis, Link reaction, and Krebs cycle).

Award 1 mark for each of the following points, up to a maximum of 5 marks:
1. The inner membrane is folded into cristae, which increases the surface area for electron transport chain carriers and ATP synthase enzymes.
2. The intermembrane space is small/narrow, which allows for the rapid accumulation of a high concentration of protons (H+) to generate a steep gradient for chemiosmosis.
3. The fluid matrix contains decarboxylase and dehydrogenase enzymes necessary to catalyze the Link reaction and Krebs cycle.
4. The outer mitochondrial membrane contains specialized transport proteins (porins) to allow pyruvate to enter from the cytoplasm.
5. The presence of circular DNA and 70S ribosomes allows the mitochondrion to independently synthesize its own respiratory enzymes and electron transport proteins.
[An additional 1 mark is available for clarity of communication, structured sequencing, and accurate biological terminology, bringing the total scaled contribution to 5.33 marks.]

The explanation must detail the key meiotic phases where variation is generated. This includes prophase I (homologous pairing, crossing over, recombinants) and metaphase I (random orientation, independent assortment), followed by a brief mention of how this results in genetically diverse haploid gametes.

Award 1 mark for each of the following points, up to a maximum of 5 marks:
1. During prophase I, homologous chromosomes undergo synapsis to pair up and form bivalents.
2. Crossing over occurs between non-sister chromatids at points called chiasmata, where genetic material is exchanged.
3. Crossing over results in recombinant chromatids, creating new combinations of maternal and paternal alleles.
4. During metaphase I, homologous chromosome pairs align randomly at the equator of the spindle (random orientation).
5. This random orientation leads to the independent assortment of maternal and paternal chromosomes into gametes during anaphase I.
6. Meiosis involves two divisions to produce four haploid gametes, each containing a unique combination of alleles.
[An additional 1 mark is available for clarity of communication, structured sequencing, and accurate biological terminology, bringing the total scaled contribution to 5.33 marks.]