Biology 9700/41 — October/November 2024

Background Concept

An action potential is a brief, all-or-nothing depolarisation of the membrane that propagates along an axon. Its standard features, read from a graph of membrane potential (mV) against time (ms), are:

• Resting potential – the membrane sits at around –40 mV in these unusual (rat/snail) neurones (typical mammalian value is –70 mV).
• Depolarisation – voltage-gated Na⁺ channels open; Na⁺ rushes in down its electrochemical gradient, and the membrane potential reverses toward Eₙₐ (~+60 mV). The peak reached depends on how many Na⁺ channels open and how long they stay open.
• Repolarisation – voltage-gated Na⁺ channels inactivate and voltage-gated K⁺ channels open; K⁺ leaves the cell, returning the membrane toward resting potential.
• Hyperpolarisation (undershoot) – K⁺ channels close slowly, so K⁺ efflux continues briefly and the membrane potential dips briefly below resting potential before the Na⁺/K⁺ pump restores ionic gradients.
• Refractory period – the window during and immediately after an action potential when another cannot easily be initiated (absolute refractory: Na⁺ channels are inactivated; relative refractory: K⁺ channels still open, requiring a larger stimulus). The longer each action potential takes, the longer the refractory period, and the lower the maximum firing frequency of the neurone.

Different species tune these features to their lifestyle. Vertebrate (rat) myelinated axons prioritise very fast conduction, with brief, sharply-peaked action potentials. Many invertebrate (snail) neurones have broader action potentials – sometimes with different ion channel complements – that conduct more slowly but may be metabolically cheaper.

Understanding the Question

Fig. 1.2 (rat) and Fig. 1.3 (snail) plot membrane potential against time for a single action potential in each. You must contrast them – i.e. make point-by-point comparisons of measurable features (peak height, duration, hyperpolarisation, refractory period). The command word 'contrast' (not 'describe') means every point must explicitly say which is greater/longer or which differs from which.

Approach

Read each trace carefully and pull out the features on which they differ:

• Peak height (look at the y-value at the top of the spike).
• Total duration (look at the time from the start of the upstroke to the return to resting potential).
• Size of the undershoot (how far below resting the trace dips after the spike).
• Refractory period (a logical deduction from the duration).
  Aim to give three crisp, contrast-style points – each phrased as 'snail is X compared with rat' or vice versa.

Step-by-Step Reasoning

1. Peak depolarisation – Fig. 1.2 (rat) peaks at about +25 mV; Fig. 1.3 (snail) peaks at about +45 mV. The snail therefore undergoes greater depolarisation / a higher peak / reaches a more positive membrane potential.
2. Duration – the rat spike rises and falls within roughly 5–6 ms; the snail spike takes about 20 ms from start to return to resting. The snail action potential is therefore slower / longer in duration / takes longer to return to resting potential.
3. Hyperpolarisation – both traces dip below the resting line after the spike, but the snail's undershoot is clearly more pronounced and longer. The snail therefore has a longer (and deeper) hyperpolarisation.
4. Refractory period – because the action potential itself lasts longer and the hyperpolarisation is more prolonged, the refractory period is longer in the snail. A longer refractory period means the maximum rate at which successive action potentials can be fired is lower in the snail than in the rat.

A candidate scoring all three marks typically gives any three of the four points above; the marking scheme accepts any valid contrast.

Key Takeaways

• 'Contrast' requires a comparative statement: always state BOTH which is greater and which is smaller.
• The features that vary between action potentials are: peak depolarisation, rate of rise/fall, duration, hyperpolarisation depth/duration, and refractory period.
• A longer action potential ⇒ longer refractory period ⇒ lower maximum firing frequency.
• Read peak values and times DIRECTLY from the graph axes – do not approximate.

Common Mistakes

• Describing one trace only, without comparing it to the other – this loses the 'contrast' mark.
• Saying the snail is 'bigger' without specifying what is bigger (peak? duration?).
• Confusing hyperpolarisation with depolarisation – hyperpolarisation is the dip BELOW resting, not the spike above.
• Quoting values that are not on the graph (e.g. 'snail peaks at +60 mV' – the trace clearly tops out near +45 mV).
• Writing only that the snail is 'slower' without giving a measurable reason – 'longer duration' or 'longer time to return to resting' is the credit-worthy phrasing.

Things to Be Careful About

• Use the exact command word's demands: contrast → compare two; describe → state features; explain → give reasons.
• Refractory period is not directly drawn on the trace; you must infer it from the duration/hyperpolarisation and you may need to say so explicitly.
• Quote the units shown on the axes (mV on the y-axis, ms on the x-axis).
• If you can only confidently spot two differences, do not invent a third – better to give two crisp, accurate contrasts than three vague ones.

Background Concept

Stomata are small pores, mainly on the underside of leaves, bounded by two guard cells. Each guard cell has a thicker inner wall (facing the pore) and a thinner outer wall. When the guard cells are turgid they bow apart and the pore opens; when they are flaccid the pore closes.

The stomatal pore is the main route by which two gases enter and leave the leaf:

• CO₂ diffuses in for photosynthesis.
• Water vapour diffuses out as transpiration.

These two requirements pull in opposite directions: photosynthesis needs CO₂ (so the pore should be open), but minimising water loss needs the pore to be closed. Plants therefore open stomata when light is available for photosynthesis and close them when it is dark and photosynthesis cannot occur anyway.

Understanding the Question

The question asks why the daily rhythm matters, not how it works. The command word is "explain", so each point must give a reason, not just describe what happens. You need to cover both the day (opening) and the night (closing) and link each to a consequence for the plant.

Approach

For each phase of the cycle, name what happens to the stomata and state the benefit (or avoided harm) to the plant:

• Day → CO₂ uptake → photosynthesis.
• Night → stomata closed → reduced water loss.

Step-by-Step Reasoning

• During the day light provides the energy for the light-dependent reactions, but the Calvin cycle needs a continuous supply of CO₂. With stomata open, CO₂ from the atmosphere diffuses through the pore into the air spaces of the spongy mesophyll and dissolves in the wet cell walls, where rubisco can fix it.
• At night there is no light, so the light-dependent reactions stop and the Calvin cycle has no ATP or reduced NADP coming from them. Taking in CO₂ would be pointless — and would also waste water. Closing the stomata reduces the diffusion gradient for water vapour from the leaf interior to the drier atmosphere, lowering transpiration and conserving water.
• A predictable rhythm lets the plant anticipate dawn and dusk and adjust the pore before conditions change, rather than reacting slowly after them.

Key Takeaways

• Stomatal opening is a trade-off between gaining CO₂ and losing water.
• Day: stomata open → CO₂ in → photosynthesis.
• Night: stomata close → less transpiration → water conserved.

Common Mistakes

• Saying stomata open "to let oxygen in" — oxygen is a waste product of photosynthesis, not a requirement.
• Saying stomata close "to stop photosynthesis" — photosynthesis stops because of lack of light, not because of stomatal closure.
• Only mentioning the day and ignoring the night phase.

Things to Be Careful About

• "Explain" requires the reason, not just the observation. "Stomata open in the day" alone is insufficient without a link to photosynthesis.
• Use the precise term transpiration for water loss, not "sweating" or "evaporation" alone.

Background Concept

Stomatal opening is an osmotic event. The guard cell is the osmotic engine: anything that lowers its water potential causes water to enter by osmosis, and the resulting turgor pressure pushes the pore open. Anything that raises its water potential lets water leave and the pore closes.

The water potential of a plant cell is governed mainly by the concentration of dissolved solutes inside it. The key solutes in guard cells are K⁺ and Cl⁻ ions (and the organic anion malate²⁻). The cell raises its solute concentration by pumping ions in.

Understanding the Question

The stem describes Fig. 2.1, in which the percentage of open stomata in Arabidopsis thaliana rises during each day and falls at night. The question asks you to describe the sequence of changes in the guard cells that makes the pore open. The mark scheme lists 8 separate points and accepts any 6 of them, so you should write a connected, ordered mechanism — not a random list.

The command word is "describe … the sequence of changes", so the answer must be a chain: ion movement → ion movement → water movement → turgor → wall mechanics.

Approach

Work from the trigger (the proton pump) through to the pore opening:

1. Active transport of H⁺ out.
2. Electrochemical consequence (membrane hyperpolarised).
3. K⁺ enters through opened channels (facilitated diffusion).
4. Cl⁻ enters to balance charge.
5. Solute accumulation lowers Ψ.
6. Water enters by osmosis.
7. Cell swells / becomes turgid.
8. Asymmetric wall thickness → cells curve → pore opens.

Step-by-Step Reasoning

• In the light, a blue-light receptor activates an H⁺-ATPase in the guard-cell plasma membrane. This pump hydrolyses ATP to pump H⁺ out of the cell against its concentration gradient. The cell therefore loses positive charge from its interior, becoming more negative (hyperpolarised).
• The more-negative interior opens voltage-gated K⁺ channels. K⁺ ions now diffuse into the cell, down both the chemical and electrical gradient — this is facilitated diffusion, not active transport.
• Anions, mainly Cl⁻ (and malate²⁻ produced from starch breakdown), also accumulate inside the cell. They balance some of the positive charge from K⁺ and contribute to the lowered water potential.
• The combined K⁺, Cl⁻ and malate²⁻ raise the internal solute concentration and so lower (make more negative) the water potential of the guard cell relative to the surrounding epidermal cells.
• Water therefore moves into the guard cell by osmosis, down the water potential gradient, across the partially permeable plasma membrane.
• The guard cell swells and its turgor pressure rises.
• The cell wall of a guard cell is not uniform: the wall adjacent to the pore (the inner wall) is thicker and less elastic than the wall on the outside. As turgor rises, the thin outer wall expands more than the thick inner wall, so the cell bows apart from its neighbour and a pore forms between the two guard cells.

Key Takeaways

• The opening of stomata is driven by active transport of H⁺ out of the guard cell.
• Ion accumulation (K⁺, Cl⁻) lowers water potential so water enters by osmosis.
• The asymmetric wall structure of the guard cell converts a uniform swelling into a directional opening of the pore.

Common Mistakes

• Saying "K⁺ is actively transported into the cell" — K⁺ entry is by facilitated diffusion through an opened channel; the only active step is the proton pump.
• Saying "water enters by active transport" or "water is pumped in" — water only moves by osmosis.
• Forgetting to mention Cl⁻ and only mentioning K⁺; the mark scheme explicitly credits Cl⁻ entry.
• Saying "the cell expands and opens the pore" without linking it to wall-thickness asymmetry.
• Writing the events out of order so it is not a "sequence".

Things to Be Careful About

• Use the term osmosis (water movement across a partially permeable membrane down a water-potential gradient), not just "diffusion of water".
• Use water potential (Ψ) when describing the driving force for water entry.
• The H⁺ pump is the only ATP-dependent step in this sequence — make sure you mark it clearly as such.
• Facilitated diffusion must be used for K⁺ entry, not just "diffusion".

Background Concept

Many biological rhythms are circadian — they have an approximately 24-hour cycle that persists even when the obvious environmental cue (light/dark) is removed. Such free-running rhythms are generated by an internal circadian clock, which is encoded by genes whose products form feedback loops with a period of roughly 24 hours. Environmental cues (zeitgebers), especially light, then entrain the clock so that its period matches the real day and its amplitude is maintained.

In Arabidopsis, several clock genes (e.g. TOC1, CCA1, LHY) interact in a transcriptional–translational feedback loop that drives many outputs, including the rhythm of stomatal opening.

Understanding the Question

The question gives you two graphs to compare:

• Fig. 2.1: normal 14 h light / 10 h dark cycle, and the rhythm is clear and large.
• Fig. 2.2: continuous darkness from 14 h onwards, and you must describe what you see there in relation to Fig. 2.1.

You are asked specifically about the role of genes and the role of the environment. That wording signals the classic biology-of-variation split: internal (genetic) control versus external (environmental) influence.

Approach

Two things to look for in Fig. 2.2:

1. Does the rhythm persist? If yes, an internal mechanism (genes) must be generating it.
2. Does the amplitude (size of the peaks) change compared with Fig. 2.1? If yes, the environment (light) must normally boost or sustain that amplitude.

Step-by-Step Reasoning

• In Fig. 2.2, even though no light has been given since hour 14, the percentage of open stomata still rises and falls in roughly 24-hour cycles. The cycle therefore does not require the light/dark signal to exist at all — it is generated internally. This is the classic signature of a circadian clock, and is taken as evidence that genes control the rhythm.
• However, the peaks in Fig. 2.2 are lower than those in Fig. 2.1, and they damp (become progressively smaller) over the days. In Fig. 2.1, the peaks are repeatedly restored to ~80–90% each day by the light period. So light (an environmental factor) is needed to maintain the full amplitude of stomatal opening.
• Together, the two graphs illustrate the standard genetic–environmental interaction: genes supply the underlying rhythm; the environment (light) sets its strength and entrains it to the real day.

Key Takeaways

• A rhythm that persists without its usual environmental cue is evidence of an internal, gene-based circadian clock.
• The environment (light) modulates the amplitude of that rhythm rather than its existence.
• Many plant behaviours, including stomatal opening, show this dual control.

Common Mistakes

• Saying "genes control everything" or "environment controls everything" — the question is asking you to separate the two contributions.
• Stating only one of the two points (rhythm persists OR peaks are smaller). The mark scheme awards one mark for each.
• Confusing the period (still ~24 h) with the amplitude (reduced) and saying "the cycle gets longer/shorter in the dark".

Things to Be Careful About

• "With reference to Fig. 2.1" means you must explicitly compare with Fig. 2.1, not describe Fig. 2.2 in isolation.
• The credit-bearing statement for genes is that the rhythm continues in darkness, not merely that stomata open at all.
• Use the term circadian rhythm (or describe a 24-hour internal clock) when explaining the genetic component.

Background Concept

Aerobic respiration occurs in four successive stages, each in a different location:

• Glycolysis (G) — in the cytoplasm. Glucose (6C) is broken down into two molecules of pyruvate (3C), with a net gain of ATP and reduced NAD.
• Link reaction (LR) — in the matrix of the mitochondria. Pyruvate is decarboxylated (one $\text{CO}_2$ released) and dehydrogenated, and the remaining 2C acetyl group is joined to coenzyme A to form acetyl CoA. Reduced NAD is also produced.
• Krebs cycle (KC) — in the matrix. Acetyl CoA is decarboxylated twice (releasing two $\text{CO}_2$) and dehydrogenated several times, producing reduced NAD, reduced FAD and ATP by substrate-level phosphorylation.
• Oxidative phosphorylation (OP) — on the inner mitochondrial membrane. Reduced NAD and reduced FAD are reoxidised, releasing electrons that pass along the electron transport chain; the energy released is used to pump protons and generate ATP via chemiosmosis. No new $\text{CO}_2$ is released and no covalent bonding of a coenzyme to a new intermediate occurs.

The events in the question are tested across these four stages.

Understanding the Question

The stem has already introduced the three events (catabolism, coenzymes being reduced/oxidised, and $\text{CO}_2$ release) plus a fourth one: covalent bonding of a coenzyme to a respiratory intermediate. We must place a tick or a cross in each of the four stage columns for each of the four events, so the table has sixteen cells to fill.

Approach

Go through each event in turn and ask in which stage(s) does this happen? and then fill the rest with crosses. The trick is to think about each stage's chemistry, not to guess.

Step-by-Step Reasoning

1. Catabolism — defined in the stem as the breakdown of the substrate into smaller molecules. Glucose is broken down in glycolysis; pyruvate is broken down further in the link reaction; acetyl CoA's 2C group is fully oxidised to $\text{CO}_2$ across the Krebs cycle. In oxidative phosphorylation, the substrate is already $\text{CO}_2$ and water — no further breakdown of an organic substrate occurs. So: G ✓, LR ✓, KC ✓, OP ✗.

2. Coenzyme is reduced or oxidised — In glycolysis NAD is reduced. In the link reaction NAD is reduced. In the Krebs cycle NAD and FAD are reduced. In oxidative phosphorylation the reduced coenzymes are oxidised (releasing electrons to the electron transport chain). So every column has a tick: G ✓, LR ✓, KC ✓, OP ✓.

3. A coenzyme forms a covalent bond with a respiratory intermediate — This is the classic step of the link reaction: coenzyme A binds via a thioester bond to the acetyl (2C) group to form acetyl CoA. In the Krebs cycle the coenzymes are reduced but they do not become covalently attached to a carbon intermediate in the cycle itself. In glycolysis and OP, no coenzyme–intermediate covalent bond is formed. So: G ✗, LR ✓, KC ✗, OP ✗.

4. Carbon dioxide is released — $\text{CO}_2$ is released by decarboxylation in the link reaction (one $\text{CO}_2$ per pyruvate) and twice per turn of the Krebs cycle (two $\text{CO}_2$). Glycolysis and oxidative phosphorylation do not release $\text{CO}_2$. So: G ✗, LR ✓, KC ✓, OP ✗.

Key Takeaways

• Catabolism and oxidation/reduction of coenzymes occur in every stage except that OP is not catabolic in the substrate-breakdown sense.
• The link reaction is unique: it is the only stage in which a coenzyme forms a covalent bond with a respiratory intermediate (acetyl CoA).
• $\text{CO}_2$ is released only in the link reaction and the Krebs cycle.

Common Mistakes

• Putting a cross in OP for "coenzyme is reduced or oxidised" — oxidation of the reduced coenzymes happens on the electron transport chain, so OP does involve redox of the coenzymes.
• Ticking KC for "covalent bond with intermediate" — the Krebs cycle does involve coenzymes but they are reduced while free in solution, not by forming a covalent bond with an intermediate in the cycle.
• Ticking G for $\text{CO}_2$ release — pyruvate is the end-product of glycolysis, so no carbon is lost as $\text{CO}_2$ in this stage.

Things to Be Careful About

• "Catabolism" in the question means the breakdown of the substrate into smaller molecules; oxidative phosphorylation does not break the substrate down further, even though the original glucose has now become $\text{CO}_2$ and water.
• A coenzyme is "reduced or oxidised" in every stage — this phrasing covers both directions of the redox.

Background Concept

Both Mimosa pudica (the 'sensitive plant') and the Venus fly trap (Dionaea muscipula) respond rapidly to touch, but their cellular mechanisms differ in direction and in the ions involved. Both use electrical signalling — an action potential travels through the leaf — but the downstream ionic and osmotic events that produce the mechanical movement work in OPPOSITE directions.

In M. pudica (Fig. 4.3), a touch triggers an action potential, and protons ($\text{H}^+$) are pumped INTO the extensor cells at the base of each pinna. This makes the cell interior more positive, so $\text{Cl}^-$ and $\text{K}^+$ ions move OUT down their electrochemical gradients. Water follows by osmosis, the extensor cells become flaccid, and the pinnae close because the flexor cells on the opposite side remain turgid.

In the Venus fly trap, the response requires TWO stimulations of sensory hairs (or two hairs), which generates an action potential. Protons then LEAVE the hinge cells (and enter the cell walls, where acidic pH activates expansin and loosens the wall). $\text{Ca}^{2+}$ ions move INTO the hinge cells, water follows, and the cells become turgid — they swell rapidly, snapping the two lobes of the trap shut.

Understanding the Question

The question provides Fig. 4.3 (the M. pudica mechanism) and asks the candidate to describe how the Venus fly trap mechanism is different. The mark scheme lists five contrasting features and awards any three. The command word is 'describe', so the answer needs the key Venus fly trap features stated clearly, with the contrast to M. pudica made obvious.

Approach

Work through the Venus fly trap mechanism in the same order as Fig. 4.3, identifying the points of contrast. Three marks are available, so the answer needs three clean contrasting points — adding the 'two hairs' point at the start is a strong, distinctive feature worth a mark.

Step-by-Step Reasoning

1. Trigger requirement: In Venus fly traps, two sensory hairs must be touched (or the same hair twice within ~20 s) before the action potential fires. In M. pudica, a single touch is enough.
2. Proton movement: In M. pudica, protons are pumped INTO the extensor cells. In Venus fly traps, protons LEAVE the hinge cells (or move into the cell wall, lowering wall pH and activating expansins).
3. Ion movement: In M. pudica, $\text{Cl}^-$ and $\text{K}^+$ move OUT. In Venus fly traps, $\text{Ca}^{2+}$ ions move INTO the hinge cells.
4. Water movement: In M. pudica, water moves OUT of the extensor cells by osmosis. In Venus fly traps, water moves INTO the hinge cells.
5. Resulting cell state: In M. pudica, the extensor cells become flaccid. In Venus fly traps, the hinge cells become turgid — they swell, increasing in volume, which physically forces the two lobes together.

Any three of these five points earn the three marks available.

Key Takeaways

• Both touch-sensitive plants use action potentials but the direction of the ionic and osmotic response is opposite.
• M. pudica closes by loss of turgor in the extensor cells (cells become flaccid).
• The Venus fly trap snaps shut by gain of turgor in the hinge cells (cells become turgid, driven by $\text{Ca}^{2+}$ influx).
• The Venus fly trap requires two hair stimulations as a 'memory' filter to prevent wasteful closure from a single accidental touch (e.g. raindrop).

Common Mistakes

• Describing both mechanisms in the same direction (e.g. claiming both involve water leaving the cells). The directions are opposite.
• Stating 'ions move in' without specifying which ions — $\text{Ca}^{2+}$ in the Venus fly trap; $\text{Cl}^-$ and $\text{K}^+$ in M. pudica.
• Confusing 'flaccid' and 'turgid' — reversing these reverses the entire mechanism.
• Omitting the requirement to touch two sensory hairs in the Venus fly trap.
• Writing about M. pudica instead of describing the Venus fly trap (the question asks for the differences, so the answer must be the contrasting features).

Things to Be Careful About

• Each contrasting point should be a separate, clearly stated sentence; the mark scheme separates the points with ';'.
• Use 'turgid' (not 'bigger' or 'expanded') — 'turgid' is the precise wording the mark scheme requires.
• Use correct ion notation: $\text{Ca}^{2+}$, $\text{K}^{+}$, $\text{Cl}^{-}$, $\text{H}^{+}$ (or 'proton').
• Only three marks are available, so three contrasting points are enough — a longer answer wastes time but should still earn full marks if each point is correct.

Background Concept

The rate of photosynthesis is set by the slowest of its requirements: light intensity, carbon dioxide concentration, and (at high light intensities) temperature. Any change that reduces the leaf's exposure to light or its uptake of CO₂ lowers the rate.

The M. pudica leaf is a compound leaf made of many pinnae arranged along a central stalk. When the pinnae fold together (Fig. 4.2), they cover each other, hiding the upper surface of each pinna from sunlight and pressing the lower-surface stomata against one another so they can no longer communicate with the atmosphere.

Understanding the Question

The question states that photosynthesis decreases by 40% when the pinnae close and asks the candidate to explain why. The 'why' is mechanistic — the candidate must identify the two factors (light and CO₂) that are reduced when the leaf closes, and connect each to the relevant limiting factor of photosynthesis.

The command word is 'explain', so each point needs the reason, not just the observation.

Approach

Think about what the closed leaf is hiding: less surface area is exposed, so less light falls on chlorophyll, and fewer stomata are open to the air. Both effects reduce the rate of photosynthesis.

Step-by-Step Reasoning

1. Surface area and light: When the pinnae fold, the leaflets overlap. The total surface area exposed to direct sunlight is reduced, so less light reaches the chlorophyll in the chloroplasts (or, more precisely, the photosystems in the thylakoid membranes). With less light absorbed, the light-dependent stage produces less ATP and less reduced NADP, limiting the rate at which the Calvin cycle can fix CO₂.
2. Stomata and CO₂: The pinnae close with their lower surfaces pressed against each other. The stomata on these surfaces are no longer exposed to the atmosphere, so the diffusion gradient for CO₂ from air into the leaf is reduced. Less CO₂ reaches the mesophyll, and the carbon-fixation step of the Calvin cycle slows down.

Any two of the four creditable points (less surface area, less light absorbed, fewer stomata exposed, less CO₂ entering) earn the two marks.

Key Takeaways

• Leaf closure reduces the leaf's exposure to BOTH light AND CO₂ simultaneously.
• The two main limiting factors of photosynthesis (light intensity and CO₂ concentration) are both reduced when a leaf closes — this is why the rate drops substantially (40% in this case).
• The explanation must connect the physical change (closure) to the specific photosynthetic factor it affects (light or CO₂).

Common Mistakes

• Stating 'less photosynthesis because the leaf is closed' — this is circular and scores nothing. The candidate must say WHICH factor is reduced (light or CO₂) and WHY (less surface area or fewer stomata).
• Saying 'less oxygen' or 'less water' instead of less CO₂. The gas taken in for photosynthesis is CO₂; O₂ is released.
• Saying 'less chlorophyll' — the leaf does not lose chlorophyll when it closes; what is lost is light reaching the chlorophyll.
• Giving only one factor (e.g. light alone) — the question expects two, and the two-mark allocation makes this clear.

Things to Be Careful About

• The command word is 'explain', so the answer needs the cause AND the consequence (e.g. 'fewer stomata exposed → less CO₂ enters → slower Calvin cycle').
• Two marks are available — two clear points, each linking the closure to a specific limiting factor, are sufficient.
• 'Surface area' alone is acceptable, but the more precise statement is 'less surface area exposed to light' or 'less light absorbed'.

Background Concept

Both cyclic and non-cyclic photophosphorylation occur in the thylakoid membranes of the grana during the light-dependent stage of photosynthesis. In both, light energy is captured by chlorophyll and used to drive electrons through an electron transport chain (ETC), pumping protons ($\text{H}^+$) across the thylakoid membrane. The resulting proton gradient is then used by ATP synthase to produce ATP (chemiosmosis).

The two processes differ in:

• the photosystems involved (PSI and/or PSII);
• whether water is split (photolysis), producing $\text{O}_2$ and protons;
• whether reduced NADP is produced as well as ATP;
• and the fate of the electrons after they pass down the ETC.

In non-cyclic photophosphorylation, light energy is absorbed by PSII (and then PSI). Electrons leave PSII, pass down the ETC between PSII and PSI (via plastoquinone, the cytochrome complex, and plastocyanin), and replace the electrons lost from PSI. The electrons excited from PSI are passed to NADP⁺, reducing it to reduced NADP. The lost electrons from PSII are replaced by electrons from the photolysis of water, which also releases $\text{O}_2$.

In cyclic photophosphorylation, only PSI is involved. Excited electrons leave PSI, pass down the (shorter) ETC, and return to PSI. No photolysis, no $\text{O}_2$, no reduced NADP — only ATP is made.

Understanding the Question

The question asks the candidate to outline the similarities AND differences between the two processes. The command word 'outline' means the answer should be a structured summary that covers the main points in a logical order. The 6 marks suggest that around 3–4 similarities and 2–3 differences are expected (the mark scheme lists 4 similarities and 4 differences, awarding any 6).

Approach

Group the answer into two clear sections — Similarities and Differences. The similarities are: photoactivation, electron transport, chemiosmosis, ATP production. The differences are: photosystems used, photolysis/$\text{O}_2$, reduced NADP, and the fate of the electrons (returned to the same photosystem in cyclic, passed to NADP in non-cyclic).

Step-by-Step Reasoning

Similarities

1. Photoactivation: In both processes, light energy excites electrons in chlorophyll to a higher energy level (photoactivation).
2. Electron transport chain: In both, the energised electrons move along/down an electron transport chain in the thylakoid membrane, releasing energy that is used to pump protons across the membrane.
3. Chemiosmosis: In both, the resulting proton gradient across the thylakoid membrane drives protons through ATP synthase, producing ATP.
4. ATP production: Both processes produce ATP.

Differences
5. Photosystems: Non-cyclic uses BOTH PSI and PSII. Cyclic uses ONLY PSI.
6. Photolysis / $\text{O}_2$: Photolysis of water (and hence $\text{O}_2$ release) occurs ONLY in non-cyclic photophosphorylation. (Cyclic does not split water.)
7. Reduced NADP: Reduced NADP is produced ONLY in non-cyclic photophosphorylation. (Cyclic produces ATP only.)
8. Fate of electrons: In cyclic, the electrons return to the SAME photosystem (PSI). In non-cyclic, the electrons from PSI are passed to NADP (and the electrons from PSII are replaced by electrons from water photolysis).

Any six of these eight points earn the six marks available.

Key Takeaways

• Both processes convert light energy into chemical energy (ATP) by chemiosmosis — that is the central similarity.
• The presence of photolysis, $\text{O}_2$ release, and reduced NADP uniquely identifies non-cyclic photophosphorylation.
• Cyclic photophosphorylation is a 'top-up' mechanism for ATP only, with no reduced NADP produced — useful when the Calvin cycle needs more ATP than reduced NADP.
• Both processes occur in the thylakoid membranes of the grana.

Common Mistakes

• Saying 'electrons are recycled in both' — they are only recycled in cyclic; in non-cyclic they end up on NADP.
• Saying 'photophosphorylation produces glucose' — it produces ATP (and reduced NADP in non-cyclic). Glucose is made in the Calvin cycle.
• Confusing which photosystem is involved — cyclic uses only PSI; non-cyclic uses PSII (first) then PSI.
• Omitting chemiosmosis — this is the mechanism of ATP production in both processes and is a required point.
• Stating that 'water is split in cyclic photophosphorylation' — it is not; photolysis is unique to non-cyclic.
• Vague phrasing: 'ATP is made by the ETC' — it is made by ATP SYNTHASE using the proton gradient generated by the ETC; this distinction is important.

Things to Be Careful About

• The mark scheme separates similarities and differences. Grouping the answer this way in the solution helps the examiner award marks.
• 'Chemiosmosis' is a specific term — the candidate must use it (not 'diffusion of protons' or 'ATP production by the ETC').
• 'Photolysis' is the term for the light-dependent splitting of water — say it, not 'splitting of water' alone.
• The 'ORA' (or reverse argument) on the mark scheme means the reverse statement is also acceptable (e.g. 'cyclic does not produce reduced NADP' is equivalent to 'non-cyclic produces reduced NADP').
• Six marks are available, so six clear points are needed. The candidate should aim to cover at least three similarities and three differences for balance.

Background Concept

Every sexually reproducing species has a characteristic chromosome number. In each cell of a diploid (2n) organism, the chromosomes are present in homologous pairs — one of each pair inherited from the mother, the other from the father. The two chromosomes of a homologous pair carry genes at the same loci but may have different alleles.

A haploid (n) cell, in contrast, contains only one of each type of chromosome. There is no homologous partner for any chromosome in the nucleus. Haploid cells are produced by meiosis (gametes in animals; spores and gamete-producing cells in plants) and by mitosis of a haploid parent cell.

Understanding the Question

The stem tells us that the thallus and the reproductive structures of Lunularia cruciata are haploid, and asks for a definition. The mark scheme expects two independent ideas: the chromosome count (n, half the diploid number) and the absence of homologous pairs. The wording "explain what is meant by" requires a definition, not just one word.

Approach

Reach for the standard CIE definition: haploid = one set of chromosomes, no homologous pairs. The two mark-scheme ideas are (1) one set / half the diploid number, and (2) chromosomes not in homologous pairs. State both clearly and add a third permitted point — that the chromosomes are all different from each other — to be safe.

Step-by-Step Reasoning

• In a diploid cell of L. cruciata (e.g. the cells of the sporophyte, which is the brief diploid phase), chromosomes would be in pairs.
• The thallus, gemma cup, gemmae, antheridia and egg-producing structures are all part of the gametophyte generation, which is haploid.
• So each cell contains one of each chromosome type. They cannot be paired because no second copy is present.
• Each remaining chromosome is therefore unique in size, shape and the gene loci it carries — the second creditable point.

Key Takeaways

• Haploid = n, one set, no homologous pairs.
• Diploid = 2n, two sets, homologous pairs present.
• Gametes (in animals) and spores / gamete-producing bodies (in plants with alternation of generations) are haploid.

Common Mistakes

• Writing "half the number of chromosomes" without mentioning homologous pairs — this is incomplete.
• Confusing haploid with having "one chromosome" — haploid refers to one set (and the set contains many different chromosomes).
• Saying "no chromosomes" instead of "one set of chromosomes".

Things to Be Careful About

Use the precise term "homologous" (not "similar" or "matched"). The CIE mark scheme rejects paraphrases that lose the genetic meaning.

Background Concept

Lunularia cruciata is a thalloid liverwort — a primitive land plant that shows alternation of generations. The visible, flattened green body (thallus) is the gametophyte, which is haploid. It produces:

• Gemma cups (A) on the upper surface, each containing small discs of tissue called gemmae (C). Gemmae are clones: when raindrops splash them out, they grow into new gametophytes that are genetically identical to the parent. This is asexual reproduction, achieved entirely by mitosis because the thallus is already haploid.
• Antheridia (B), the male reproductive structures, which produce motile, biflagellate sperm. Sperm in a liverwort are also haploid and are produced by mitosis of cells inside the antheridium.
• Archegonia (not labelled here) on a neighbouring thallus, which produce eggs. Sperm swim through a film of water to reach them.
• When sperm and egg fuse, the diploid zygote (D) is formed. The zygote develops into the sporophyte, which remains attached to the gametophyte and produces spores by meiosis. The haploid spores then germinate into new gametophytes.

Understanding the Question

The stem describes:

• A: a gemma cup containing discs C that grow into genetically identical new plants → asexual reproduction by mitosis.
• B: a structure containing sperm → sexual reproduction; sperm made by mitosis because the parent is haploid.
• D: a structure formed by fusion of egg and sperm → sexual reproduction; develops to produce spores (made by meiosis) that grow into genetically different plants.

We must place each of A, B, C, D into three categories: "sexual reproduction", "produced by mitosis", and "site of meiosis". Each letter may be used once, more than once or not at all.

Approach

Decide, for each labelled structure:

1. Is it part of the sexual cycle? (involves gamete fusion or its product)
2. Is it produced by mitosis? (formed by mitotic division of a haploid cell)
3. Is it the site of meiosis? (where reduction division happens)

Then put the appropriate letters in each box.

Step-by-Step Reasoning

• A — gemma cup: part of the asexual cycle. The cup itself is a multicellular outgrowth of the haploid thallus, so it is built by mitosis. It is NOT part of sexual reproduction. It is NOT the site of meiosis. → asexual, mitosis.
• B — antheridium (contains sperm): the sperm inside are made by mitosis of haploid cells, and the structure is involved in sexual reproduction. It is not the site of meiosis. → sexual, mitosis.
• C — gemmae: multicellular discs of haploid tissue produced by mitosis of the thallus. Asexual only. Not the site of meiosis. → asexual, mitosis.
• D — zygote: formed by fertilisation (egg + sperm) — the defining event of sexual reproduction. The zygote develops into the sporophyte, and within the sporophyte capsule, spore mother cells undergo meiosis to form haploid spores. So D is associated with sexual reproduction, and is the structure from which the site of meiosis develops. The mark scheme credits D as the site of meiosis because the meiotic divisions take place in the sporophyte that grows from D. → sexual, site of meiosis.

Categories:

• Sexual reproduction: B (produces the gametes) and D (the fertilised cell).
• Mitosis: any two of A, B, C, D. The clearest answer is A and C — the gemma cup and the gemmae are both products of mitotic divisions of the haploid thallus and are the entire asexual pathway. (B is also a credible alternative because sperm are made by mitosis; D is a borderline case as the embryo then divides by mitosis.)
• Site of meiosis: D — the sporophyte that develops from the zygote produces spores by meiosis.

Key Takeaways

• Asexual reproduction in plants is mediated by mitosis and produces clones; sexual reproduction involves gamete fusion (fertilisation) and produces genetically variable offspring.
• In a plant with alternation of generations, the gametophyte is haploid and produces gametes by mitosis; the sporophyte is diploid and produces spores by meiosis.
• Asexual structures (gemmae) and sexual structures (antheridia/archegonia) can coexist on the same thallus.

Common Mistakes

• Putting C in "sexual reproduction" — gemmae are clones, not gametes; the new plants from C are genetically identical to the parent.
• Putting A or C as "the site of meiosis" — neither is diploid, so meiosis cannot occur there.
• Putting D in "produced by mitosis" — D (the zygote) is produced by fertilisation, not mitosis. (Once the zygote begins to divide, those later divisions are mitotic, but the zygote itself was not produced by mitosis.)
• Naming only one letter for "produced by mitosis" when two are required.

Things to Be Careful About

The CIE mark scheme explicitly allows any two of A, B, C, D for "mitosis" because each of these structures does involve mitosis somewhere in its formation. Choose the two you can defend most strongly — for a liverwort, A and C are the most obvious products of mitotic asexual reproduction.

Background Concept

In a sexually reproducing animal, body (somatic) cells are diploid (2n) — each chromosome is present as a homologous pair, one from each parent. Gametes (sperm and eggs) are haploid (n). When two gametes fuse at fertilisation, the zygote once again has 2n chromosomes — one set from the sperm, one from the egg.

The only way to produce haploid gametes from a diploid parent is to halve the chromosome number during their formation. That halving is meiosis — a reduction division in which one round of DNA replication is followed by two divisions, so that each daughter cell ends up with half the original chromosome number.

Meiosis also shuffles the genetic material: in prophase I, homologous chromosomes pair up and crossing over occurs between non-sister chromatids, exchanging segments of DNA. At metaphase I, homologous pairs line up independently of other pairs, so when they separate in anaphase I, each gamete receives a different mix of maternal and paternal chromosomes (independent assortment). Together these mechanisms generate huge genetic variation among the gametes.

Understanding the Question

The stem introduces horses as diploid, sexually reproducing animals whose gametes fuse to form genetically different offspring, and asks us to explain why a reduction division is needed in gamete production. The expected marks cover three threads:

1. Maintenance of the chromosome number across generations.
2. Generation of genetic variation in gametes/offspring.
3. The consequence of NOT having reduction division (chromosome number would double each generation, causing problems).

Approach

State the central reason first: fertilisation doubles the chromosome number, so gamete production must halve it. Then add the bonus reason: halving is achieved by meiosis, which simultaneously creates genetic variation. Mention the consequence of failing to halve to round out the answer.

Step-by-Step Reasoning

• Horses have a diploid chromosome number of 2n = 64. Every somatic cell of a horse has 32 homologous pairs.
• If gametes were produced by mitosis, they would also contain 32 pairs (64 chromosomes), and fertilisation would give the zygote 4n = 128 chromosomes. After many generations, the chromosome number would keep doubling — a runaway increase called polyploidy.
• Polyploidy in animals is usually lethal or causes serious developmental problems (e.g. Down syndrome in humans is a partial trisomy from a single chromosome pair failing to separate). So a strict halving at gamete formation is essential to keep the chromosome number constant generation after generation.
• Meiosis performs this halving AND, in the same process, generates variation: crossing over in prophase I recombines maternal and paternal alleles on each chromosome; independent assortment at metaphase I mixes whole maternal and paternal chromosomes into different gametes. The gametes are therefore genetically non-identical, and so are the offspring produced when they fuse.
• The mark scheme credits four points: (1) maintains chromosome / diploid number; (2) from parents to offspring / across generations; (3) gives genetic variation in gametes/offspring/horses; (4) extra sets of chromosomes cause (named) problems. Any three are needed for three marks.

Key Takeaways

• Meiosis is essential because fertilisation doubles the chromosome number; without meiosis, the chromosome number would double every generation.
• Meiosis produces haploid gametes so fertilisation restores the diploid state — keeping the chromosome number constant across generations.
• A second, equally important, role of meiosis is to generate genetic variation through crossing over and independent assortment.
• In animals, polyploidy is almost always lethal, so chromosome number must be tightly controlled.

Common Mistakes

• Saying only "meiosis produces four cells" without explaining why halving is needed.
• Saying "meiosis creates variation" without linking it back to chromosome number.
• Giving the consequence of NOT having reduction division as "offspring will have extra chromosomes" without naming a specific problem (e.g. genetic disorders, lethality, abnormal development).
• Confusing mitosis with meiosis, or stating that gametes are produced by mitosis.

Things to Be Careful About

The CIE mark scheme rewards why a reduction division is needed, not just what it does. Always tie the answer back to fertilisation restoring the diploid number. For the third mark, name at least one source of variation (crossing over, independent assortment, or random fusion of gametes).

Background Concept

The first step in urine formation is ultrafiltration in the renal corpuscle. The renal corpuscle consists of a tuft of capillaries called the glomerulus sitting inside the cup-shaped Bowman's capsule. Blood enters the glomerulus via the afferent arteriole and leaves via the efferent arteriole (which is narrower). This narrowing creates a high hydrostatic pressure inside the glomerular capillaries — high enough to push fluid and small solutes out through the capillary walls into the Bowman's capsule. The filtrate is called the glomerular filtrate.

The filtration barrier has three layers, each contributing to selectivity:

• Fenestrations (pores) in the endothelium of the glomerular capillary — small gaps (~70–100 nm) that allow most plasma constituents through but block blood cells.
• Basement membrane — a thin layer of collagen IV and glycoproteins that acts as a molecular sieve, blocking medium-sized proteins.
• Slit pores / filtration slits between the podocytes (specialised cells of Bowman's capsule whose foot processes wrap around the capillaries) — fine gaps that further restrict passage of larger plasma proteins.

Molecules up to about 69 kDa (water, glucose, urea, ions, amino acids, small proteins) pass into the Bowman's capsule; larger proteins and blood cells are retained in the blood.

Understanding the Question

This is an "outline" question worth 3 marks. The mark scheme offers any 3 of 5 credited points — so a strong answer covers 3–4 of the key points clearly. You need to give a brief, structured account of how the glomerulus and Bowman's capsule work together to perform the first stage of urine formation.

Approach

Think about four things: (1) the process (ultrafiltration), (2) the driving force (high blood/hydrostatic pressure in the glomerulus), (3) what is filtered (water and small solutes move out), and (4) the barrier it crosses (the three-layered filtration barrier — naming any two of the three layers earns a mark).

Step-by-Step Reasoning

1. The glomerulus and Bowman's capsule together form the site of ultrafiltration. Blood is filtered under pressure to produce glomerular filtrate, which then flows into the proximal convoluted tubule.
2. The driving force is the high hydrostatic pressure in the glomerular capillaries, created because the afferent arteriole is wider than the efferent arteriole. This pressure is higher than in normal tissue capillaries.
3. As a result, water and small solutes (e.g. urea, glucose, sodium ions, amino acids) are forced out of the blood in the glomerulus and into the Bowman's capsule, while blood cells and plasma proteins are retained in the blood.
4. The fluid must pass through a filtration barrier with three layers. Any two of the following earn a mark: the fenestrations in the capillary endothelium, the basement membrane, and the slit pores / filtration slits between the podocytes of the Bowman's capsule.

A complete answer would also note that the Bowman's capsule collects the filtrate and channels it into the proximal convoluted tubule for the next stages of urine formation.

Key Takeaways

• The glomerulus and Bowman's capsule together perform ultrafiltration — the first step of urine formation.
• High blood pressure in the glomerular capillaries (because the efferent arteriole is narrower than the afferent) is the driving force.
• Water and small solutes are forced through a three-layered barrier (fenestrations, basement membrane, slit pores) to form the glomerular filtrate.
• The Bowman's capsule collects the filtrate and channels it into the proximal convoluted tubule.

Common Mistakes

• Calling it "filtration" without specifying ultrafiltration (which is what produces the glomerular filtrate; the word "ultra" indicates that small molecules pass while larger ones are retained).
• Saying "blood is filtered" without explaining WHY (i.e., the high pressure).
• Confusing the glomerulus with the loop of Henle (the loop is involved in water reabsorption, not filtration).
• Saying the filtrate is forced out by "blood pressure in general" — it specifically refers to the high pressure in the glomerular capillaries.
• Omitting the structure of the barrier — the mark scheme rewards naming two of the three layers (fenestrations, basement membrane, slit pores).

Things to Be Careful About

• The mark scheme requires "hydrostatic pressure" or "blood pressure" — not just "pressure".
• The filtration barrier has THREE layers; naming any TWO earns the mark.
• The Bowman's capsule is the collecting funnel — without it, the filtrate has nowhere to go.
• Use the precise term podocytes for the cells of Bowman's capsule that form the slit pores.

Background Concept

The mammalian kidney can produce urine that is more or less concentrated than the blood, depending on the body's state of hydration. The maximum concentration it can produce depends on:

• The length of the loop of Henle (longer loops allow more water to be reabsorbed by the countercurrent multiplier, producing a steeper osmotic gradient in the medulla).
• The permeability of the collecting duct to water (controlled by ADH — when ADH is high, more water is reabsorbed via aquaporin-2 channels).

The U:P ratio compares the maximum solute concentration of the urine to the solute concentration of the blood plasma. A ratio of 1:1 would mean the urine is the same concentration as plasma (no water reabsorption from the collecting duct). A ratio of, say, 16:1 means the urine is 16 times more concentrated than the blood — the kidney has reabsorbed almost all the water from the filtrate.

Understanding the Question

This is a "suggest" question worth 3 marks. The command word "suggest" means you must use the data in Table 6.1 to make a reasoned biological interpretation, not just state a fact. The question asks what the U:P ratio shows about each species' ability to tolerate a water shortage.

Approach

Look at the data:

• Beaver: 1.7:1 (lowest, lives in rivers/lakes)
• Human: 4.5:1 (medium, variable habitat)
• Camel: 8.0:1 (high, desert)
• Rat: 9.0:1 (high, variable)
• Kangaroo rat: 16.0:1 (highest, desert)

The pattern is clear: species in dry habitats (desert) have the highest U:P ratios; species in wet habitats (rivers/lakes) have the lowest. A higher U:P means the animal can make more concentrated urine, so it loses less water in urine and can survive on less drinking water.

Step-by-Step Reasoning

1. A high U:P ratio means the species can produce more concentrated urine relative to its blood plasma. More water has been reabsorbed from the filtrate (especially in the collecting duct, under the influence of ADH), so the animal loses less water in its urine.
2. A high U:P ratio means the animal can tolerate a shortage of water. Species with a high U:P ratio (kangaroo rat 16:1, rat 9:1, camel 8:1) can survive in environments where water is scarce, because their kidneys conserve water very effectively.
3. A high U:P ratio indicates adaptation to a dry environment. The kangaroo rat (16:1) and camel (8:1) are desert animals, with long loops of Henle that allow them to produce very concentrated urine.
4. Conversely, a low U:P ratio means the species needs a plentiful supply of water. The beaver (1.7:1) lives in rivers and lakes — it has no need to conserve water because it is always surrounded by it. Its short loop of Henle and short collecting duct make it impossible to produce highly concentrated urine.

A complete answer would also note that the U:P ratio reflects the maximum concentrating ability of the kidney, which in turn depends on the length of the loop of Henle and the habitat the species evolved in.

Key Takeaways

• The U:P ratio is a measure of how much the kidney can concentrate urine compared to the blood plasma.
• High U:P → lots of water reabsorbed, small volume of concentrated urine → animal conserves water and can tolerate water shortage.
• Low U:P → little water reabsorbed, large volume of dilute urine → animal cannot tolerate water shortage and needs a watery environment.
• Desert mammals (kangaroo rat, camel) have high U:P ratios; aquatic mammals (beaver) have low U:P ratios.

Common Mistakes

• Just stating the values from the table without interpreting them.
• Saying "desert animals have higher U:P ratios because they drink more water" — the opposite is true; they conserve water.
• Failing to refer to the data in the table (the question says "with reference to Table 6.1").
• Confusing U:P with the volume of urine — a high U:P means the urine is more concentrated (less water per unit of solute), not that there is more urine.
• Saying the U:P ratio shows the amount of water in the urine — it shows the concentration of solutes relative to the plasma, which is a proxy for how much water has been retained by the body.

Things to Be Careful About

• Always state the relationship in the correct direction: "high U:P means more concentrated urine means more water reabsorbed means better at conserving water".
• The mark scheme accepts any of the credited points; the key idea is the link between a high U:P and water conservation / tolerance of water shortage / adaptation to dry environments.
• Reference the data — quoting the actual values and species from the table (kangaroo rat 16:1, beaver 1.7:1, etc.) makes the answer concrete and shows you have used the table.

Background Concept

The evolutionary relationships among species can be reconstructed by comparing their DNA (or protein) sequences. The underlying principle is:

• Species that diverged recently from a common ancestor have had less time for random mutations to accumulate, so their DNA sequences are more similar.
• Species that diverged long ago have had more time for mutations to accumulate, so their DNA sequences are more different.

By counting the nucleotide (or amino acid) differences between pairs of species and applying computer-based phylogenetic methods, scientists can construct an evolutionary tree (phylogeny) that shows the order and timing of divergences. Standard software includes sequence-alignment tools and BLAST (Basic Local Alignment Search Tool), which compares a query sequence against sequence databases.

Understanding the Question

Fig. 7.3 shows a phylogenetic tree of three Spea species:

• S. multiplicata diverged from the lineage leading to the other two species about 30 million years ago.
• S. hammondii and S. bombifrons share a more recent common ancestor (about 15 million years ago), so they are each other's closest relatives.

The question asks how analysis of DNA allowed these evolutionary relationships to be determined. The candidate must explain the method (DNA sequencing and comparison), the logic (more similarity = more recent common ancestor), and how this produces the relationships shown in the tree.

Approach

Walk through the steps: (1) sequence the DNA of each species; (2) compare the sequences using bioinformatics tools; (3) count differences/similarities; (4) interpret fewer differences as evidence of a more recent common ancestor; (5) apply this to the three Spea species to recover the tree shown.

Step-by-Step Reasoning

1. Determine the DNA sequence for each of the three species (the same gene or set of genes is sequenced across all species so that the sequences can be directly compared).
2. Use bioinformatics tools (e.g. BLAST, sequence databases, alignment software) to align the sequences and count nucleotide/base differences (or similarities) between each pair of species.
3. Interpret the data: the pair with the fewest differences has the most recent common ancestor — they are the most closely related. The pair with the most differences diverged longest ago.
4. Apply to the three species: S. hammondii and S. bombifrons have the fewest genetic differences, so they share the most recent common ancestor (about 15 million years ago). S. multiplicata differs more from both of them, so it diverged earlier (about 30 million years ago).
5. This produces the tree in Fig. 7.3: S. hammondii and S. bombifrons are sister species, with S. multiplicata as the outgroup.

Key Takeaways

• DNA sequence comparison is a powerful tool for reconstructing evolutionary relationships.
• More similar DNA sequences indicate a more recent common ancestor.
• Bioinformatics tools (BLAST, alignment software) automate the comparison and statistical analysis.

Common Mistakes

• Saying "DNA fingerprinting" or "DNA profiling" — these are forensic techniques for identifying individuals, not for comparing species.
• Stating that all three species share the same DNA — they share many genes but differ at variable sites.
• Reversing the logic: the most similar species are the most closely related, not the most different.
• Failing to mention bioinformatics/software — modern phylogenetics relies on computational analysis.

Things to Be Careful About

• "Fewer differences = more closely related" must be stated explicitly.
• The specific example (S. hammondii and S. bombifrons being the most similar) links the answer to the tree shown.
• Bioinformatics/database/software (e.g. BLAST) is a creditable point — a generic "comparison" is less specific and may not score.

Background Concept

When the gene cannot simply be cut out of donor DNA with restriction enzymes, the genetic engineer needs an alternative way to obtain a working copy. Two standard approaches are taught at A-level:

1. cDNA (complementary DNA) synthesis – Many eukaryotic genes contain introns. If a gene is cut directly from genomic DNA and inserted into a bacterial plasmid, the bacterium cannot splice out the introns, so a functional protein is not made. To avoid this, the gene is copied from the mRNA, in which introns have already been removed by splicing. The enzyme reverse transcriptase (isolated from retroviruses such as HIV) reads the mRNA template and assembles a complementary single DNA strand. A second DNA strand is then made by DNA polymerase, giving a double-stranded cDNA copy of the gene that contains only coding (exon) sequence.

2. Chemical (artificial) gene synthesis – If the protein's amino-acid sequence is known, or the gene sequence is known from a database, the gene can be assembled nucleotide-by-nucleotide in the laboratory using a DNA synthesiser. Activated nucleotides are joined in the order dictated by the target sequence, producing short single-stranded oligonucleotides that are then ligated together to form the complete gene. This approach is useful for short genes and for designing modified sequences (e.g. adding restriction sites or codons preferred by the host organism).

Understanding the Question

The question asks for an outline of two methods – a brief description of each is enough, not full experimental detail. The command word is "outline", so the candidate should provide the principle of each method, including the key enzyme (for cDNA) and the source of information (for chemical synthesis). Three marks are available; the mark scheme accepts any three from a list of four creditable points.

Approach

For each method, give a one-sentence outline plus any additional detail that matches the marking points:

• For cDNA: name the starting material (mRNA), name the key enzyme (reverse transcriptase).
• For chemical synthesis: state that nucleotides are joined, and that the sequence is known from a database (or derived from a known amino acid sequence).
  Covering all four marking points in your answer guarantees the full 3 marks, since only 3 of the 4 are required.

Step-by-Step Reasoning

Method 1 – cDNA from mRNA (marking points 1 and 2):

• The gene is not cut from DNA; instead, mature mRNA for the protein is extracted from cells of the donor organism. This mRNA already lacks introns.
• Reverse transcriptase uses this mRNA as a template to synthesise a single complementary DNA strand, producing an RNA–DNA hybrid.
• The mRNA is then degraded and DNA polymerase synthesises the second DNA strand, yielding a double-stranded cDNA copy of the gene.

Method 2 – Chemical synthesis (marking points 3 and 4):

• The required nucleotide sequence is determined from a database (or back-translated from a known amino-acid sequence).
• A machine (DNA synthesiser) chemically joins the nucleotides in the correct order, building the gene strand by strand.
• Short overlapping oligonucleotides are joined with DNA ligase to form a complete, double-stranded gene.

Key Takeaways

• cDNA synthesis is the route of choice when the gene contains introns and the host is a prokaryote, because it bypasses the splicing problem.
• Chemical synthesis is best for short, well-characterised genes and for producing designer sequences (codon-optimised, restriction-site added).
• The two methods together mean a gene is obtainable even when no DNA from the donor is available.

Common Mistakes

• Stating only "use an enzyme to make the gene" without naming reverse transcriptase – this loses the specific mark.
• Confusing reverse transcriptase (RNA → DNA) with DNA polymerase (DNA → DNA).
• Saying the gene is "made from amino acids" – amino acids are not the building blocks of DNA; the gene is built from nucleotides.
• Omitting the database/known-sequence step in the chemical-synthesis method.

Things to Be Careful About

• "Any three from" means the examiner will pick the best three points; you do not need all four, but giving all four is the safest strategy.
• Be explicit about which starting molecule is used (mRNA for cDNA; known sequence for chemical synthesis).
• Use precise terminology: reverse transcriptase, cDNA, mRNA, nucleotide – not vague terms like "the gene is made in a lab".

Background Concept

Two enzymes that often appear together in questions about gene technology but which do very different jobs:

• DNA ligase forms phosphodiester bonds between adjacent nucleotides that are already base-paired. In genetic engineering, it seals the nick between a gene inserted into a plasmid and the plasmid DNA, joining the sugar-phosphate backbones so the recombinant plasmid is covalently closed. It is also used in vivo to join Okazaki fragments on the lagging strand of replication.
• DNA polymerase adds free activated DNA nucleotides (dNTPs) one at a time to the 3′-OH end of a growing strand, using an existing strand as a template. In genetic engineering it is used in PCR (with Taq polymerase) to amplify a gene and in cDNA synthesis to make the second strand after reverse transcriptase.

The two enzymes are therefore complementary, not interchangeable: ligase joins already-existing pieces of DNA, while polymerase builds a new strand by adding nucleotides.

Understanding the Question

The candidate is asked to tick or cross each box so that every row shows whether the named enzyme performs that role. Three marks are available, one per row. No explanation is required, but understanding the wording of each role is essential to avoid a wrong tick.

Approach

Read each role in turn and ask: which enzyme actually does this?

1. "Joins two sections of sugar-phosphate backbone" – this is the textbook definition of DNA ligase. → ligase ✓, polymerase ✗.
2. "Adds a gene to a plasmid" – the gene is inserted into an open plasmid and the backbones are sealed; that is ligase's job. → ligase ✓, polymerase ✗.
3. "Adds free activated DNA nucleotides to a polynucleotide" – this is the textbook definition of DNA polymerase. → ligase ✗, polymerase ✓.

Step-by-Step Reasoning

• Row 1 (joins sugar-phosphate backbones): DNA ligase catalyses the formation of a phosphodiester bond between the 3′-OH of one nucleotide and the 5′-phosphate of the next. DNA polymerase cannot perform this sealing role; it requires a pre-existing 3′-OH to extend and does not ligate nicks between separate fragments. → ✓ ligase, ✗ polymerase.
• Row 2 (adds a gene to a plasmid): Once the gene is inserted into the cut plasmid, ligase seals the two nicks in the sugar-phosphate backbone, producing a closed recombinant plasmid. DNA polymerase is not used to insert a gene. → ✓ ligase, ✗ polymerase.
• Row 3 (adds free activated DNA nucleotides): DNA polymerase extends a primer by adding complementary activated dNTPs to the 3′ end. DNA ligase does not add nucleotides; it only joins existing ones. → ✗ ligase, ✓ polymerase.

Key Takeaways

• DNA ligase joins pre-existing DNA fragments; DNA polymerase builds a new strand from free nucleotides.
• A common exam trap is to confuse the two – they are not redundant.
• In practice, both enzymes are used during a typical cloning workflow: ligase to seal the recombinant plasmid, polymerase (Taq) during PCR to amplify the inserted gene.

Common Mistakes

• Ticking DNA polymerase for "adds a gene to a plasmid" – polymerase cannot seal the two backbones.
• Ticking DNA ligase for "adds free activated DNA nucleotides" – ligase does not catalyse phosphodiester bond formation between a free nucleotide and a strand; it joins adjacent nucleotides that are already in place.
• Ticking both boxes in a row "to be safe" – this would be wrong on every row and scores zero.

Things to Be Careful About

• Make sure the tick or cross is unambiguous; use a clear ✓ or ✗ rather than a vague mark.
• Read each role carefully – "adds a gene to a plasmid" is a process in which ligase is the final step, not a description of how the gene is inserted physically.
• DNA polymerase used here refers to the family of DNA-dependent DNA polymerases, including Taq polymerase; the same answer applies regardless of which polymerase is used.

Background Concept

The polymerase chain reaction (PCR) is a method of amplifying a specific DNA sequence in vitro. It was developed by Kary Mullis in 1983 and is now a fundamental tool in molecular biology, forensics, medical diagnostics and evolutionary studies. The reaction requires:

• the template DNA containing the target sequence,
• two short DNA primers (one for each strand) that flank the target region and define where amplification begins,
• free activated deoxyribonucleotides (dNTPs) – the building blocks,
• a heat-stable DNA polymerase, almost always Taq polymerase, isolated from the thermophilic bacterium Thermus aquaticus that lives in hot springs,
• a buffer with $\text{Mg}^{2+}$ ions as a cofactor.

The mixture is cycled through three temperatures, each held for a short time, in an automated thermocycler. The three temperatures correspond to the three chemical steps of DNA replication in vivo – strand separation, primer binding and strand extension – carried out sequentially in a single tube.

Understanding the Question

The candidate is asked to state the three temperatures (with approximate values) and outline what happens at each step. The command word "outline" means a brief description, not a full mechanism. The mark scheme rewards the temperature range plus a short description of the event at that temperature for each of the three steps, giving one mark per step.

Approach

Treat each temperature as a step in the cycle:

1. High temperature (denaturation step): the DNA double helix is melted; H-bonds between complementary bases break and the two strands separate.
2. Intermediate temperature (annealing step): cooled just enough to allow short primers to hydrogen-bond to their complementary sequences on each single-stranded template.
3. Working temperature (extension step): warm enough for the polymerase to work efficiently; the polymerase adds dNTPs to the 3′ end of each primer, extending it to make a new complementary strand.

Note the temperature ranges given in the mark scheme (and use whichever value is supplied in the syllabus). State the range, not a single value, because the optimal temperature depends on the primers and the polymerase used.

Step-by-Step Reasoning

• Step 1 – Denaturation ($90\text{–}98\,^\circ\text{C}$): heating to near boiling breaks the hydrogen bonds holding the two antiparallel strands of the template DNA together. The double helix unwinds and the two strands separate. This exposes the bases so that primers can bind in the next step. Without denaturation, the primers could not access the template.

• Step 2 – Annealing ($50\text{–}65\,^\circ\text{C}$): the mixture is cooled so that the primers can hydrogen-bond (base-pair) with their complementary sequences flanking the target region. Each primer binds to one strand: one primer to the 3′ end of the sense strand's complement, the other to the 3′ end of the antisense strand's complement. The temperature is critical: too hot and the primers will not stay bound; too cool and they will bind non-specifically.

• Step 3 – Extension ($68\text{–}75\,^\circ\text{C}$): the temperature is raised to the optimum for Taq polymerase. Working from each primer, the polymerase adds complementary dNTPs to the 3′ end of the primer, synthesising a new DNA strand in the 5′ → 3′ direction. The new strand runs from the primer towards the other primer, so each cycle doubles the number of target DNA molecules (one new strand from each template strand, and the original strands also act as templates in subsequent cycles).

A complete cycle takes 1–3 minutes; the cycle is repeated 25–35 times, giving $2^{n}$ copies of the target after $n$ cycles.

Key Takeaways

• Three temperatures, three steps: denature, anneal, extend.
• The temperatures can be remembered as ~95 °C, ~55 °C, ~72 °C, but the mark scheme accepts a small range around each.
• Taq polymerase is essential because it survives the high denaturation temperature; ordinary polymerases would be denatured themselves.
• The amplification is exponential because each new strand becomes a template in the next cycle.

Common Mistakes

• Stating the temperatures without saying what happens at each – the mark scheme requires an event, not just a number.
• Confusing the order of steps (e.g. putting extension before annealing).
• Saying "primers join the DNA" – primers do not join two strands; they bind to a single-stranded template.
• Naming "DNA polymerase" without the qualifier "Taq" or "heat-stable" – the mark scheme credits the specific polymerase.
• Giving temperatures that are too tight (e.g. exactly 72 °C only) – examiners usually allow a small range.

Things to Be Careful About

• Use the temperature ranges quoted by the syllabus; if you are unsure of an exact value, give the range to be safe.
• The mark scheme accepts slightly different ranges; what matters is the order (high → low → medium) and that each event is matched to the correct step.
• A common error is to confuse primer binding (annealing, the cool step) with strand extension (the warm step). The polymerase works at the higher of the two lower temperatures, not the lowest.

Background Concept

Under the three-domain / multi-kingdom classification used by CIE, every lichen such as Usnea is a stable, mutualistic association between two very different organisms: a photosynthetic partner from kingdom Protoctista (a green alga or a cyanobacterium in some lichens) and a fungal partner from kingdom Fungi (the mycobiont that gives the lichen its shape). Recognising the kingdom-level features of each partner explains why a lichen behaves the way it does — photosynthesis by the alga, absorption of water and minerals by the fungal hyphae.

The kingdom Protoctista is a 'catch-all' group: it includes any eukaryote that is not a fungus, plant or animal. So members may be unicellular (e.g. Amoeba, Plasmodium) or multicellular (e.g. seaweeds such as Laminaria), photosynthetic or heterotrophic, free-living or parasitic. Despite this diversity, all share eukaryotic features (true nucleus, membrane-bound organelles, 80S ribosomes, often cilia/flagella with the 9+2 microtubule arrangement).

Kingdom Fungi (e.g. Rhizopus, Saccharomyces, Penicillium) shares the eukaryote features but has its own diagnostic combination: a cell wall of chitin (not cellulose, as in plants), a body made of hyphae forming a mycelium, heterotrophic nutrition (saprotrophs secrete extracellular enzymes, parasites absorb from a living host), storage of glycogen (animal-like) rather than starch, and reproduction by spores (often produced in huge numbers in sporangia or on fruiting bodies).

Understanding the Question

Part (a) is an 'outline' question worth 4 marks. 'Outline' here means state the main defining features — you do not need to write an essay, but you must cover enough of each kingdom to show you know what makes them distinct. The mark scheme lets you give up to three points on Fungi (max 3) and asks for a single example showing the variability of Protoctista. One mark is also available for stating that both are eukaryotic.

Approach

Pick four clear, distinct, mark-scheme-friendly points. The safest set is:

1. Both are eukaryotic (general point, with a single example of a eukaryotic feature such as a true nucleus or mitochondria).
2. Protoctista – a clear statement of their variability (e.g. some are photosynthetic, others heterotrophic; some single-celled, others multicellular).
3. Fungi – chitin cell wall.
4. Fungi – one other Fungi-only feature (e.g. hyphae/mycelium, saprotrophic/heterotrophic nutrition, multinucleate, glycogen storage, spore reproduction).

Step-by-Step Reasoning

• Mark 1 (eukaryotic): Lichens are made of two eukaryotic partners, so this single point scores regardless of which kingdom you then describe.
• Mark 2 (eukaryotic example): Quote a specific eukaryotic structure (true nucleus, mitochondria, 80S ribosomes, endoplasmic reticulum) so the examiner sees you mean eukaryotic in the cell-biology sense.
• Mark 3 (Protoctista variation): The kingdom is defined almost by what it is not. The clearest single mark-winning point is to state that some are photosynthetic (algae) and some are heterotrophic (protozoa), or to mention the mix of unicellular and multicellular body plans.
• Marks 4 (Fungi): The Fungi section of the mark scheme is essentially a menu — chitin cell wall, hyphae/mycelium, multinucleate, heterotrophic/saprotrophic/parasitic, glycogen storage, spore reproduction. Any three of these would earn three marks; the question only needs one more after you have stated the shared eukaryote point and the Protoctist variation point.

Key Takeaways

• Protoctista is the most heterogeneous of the five kingdoms — it is easier to define by exclusion than by a single shared feature, but the photosynthetic/heterotrophic split and the unicellular-to-multicellular range are reliable marks.
• Fungi are never defined by photosynthesis. The five 'go-to' fungal features are: chitin wall, hyphae/mycelium, saprotrophic/heterotrophic nutrition, glycogen storage, spore reproduction.
• Saying 'eukaryotic' without giving a named eukaryotic structure only earns one of the two available 'eukaryote' marks — add a structure for the second.

Common Mistakes

• Writing 'cell wall' without specifying chitin. Plants have cellulose cell walls; the wrong polysaccharide loses the mark.
• Stating that fungi 'have chloroplasts' or 'are photosynthetic'. They are not — they are heterotrophs.
• Saying Protoctista 'reproduce sexually' without contrasting with asexual reproduction, or describing them only as 'animal-like' (protozoa) and forgetting the plant-like algae.
• Listing bacterial features (no nucleus, 70S ribosomes, peptidoglycan wall) — these are not eukaryotic and score 0.

Things to Be Careful About

• The mark scheme caps Fungi at 3 marks — you cannot pick up all four marks just from Fungi features; you need at least one Protoctista and one shared eukaryote point.
• Use the precise CIE terms: 'hyphae' (not 'fibres'), 'mycelium' (not 'mesh'), 'saprotrophic' (not 'decomposer' — that is a role, not a kingdom feature), 'glycogen' (not 'starch').
• 'Multinucleate' is preferred over 'has many cells' — fungal hyphae are typically coenocytic (one cytoplasm with many nuclei), not divided into separate cells by septa.

Background Concept

In monohybrid inheritance, a single gene controls a trait and exists as two alleles: a dominant allele (expressed in heterozygotes) and a recessive allele (only expressed in homozygotes). By convention, the dominant allele is given a capital letter and the recessive allele the corresponding lowercase letter.

When two heterozygous individuals are crossed (Aa × Aa), the offspring genotypes follow the ratio 1 AA : 2 Aa : 1 aa. The corresponding phenotypic ratio is 3 dominant : 1 recessive, because both AA and Aa express the dominant phenotype.

The peppered moth Biston betularia is the classic example of natural selection. In unpolluted environments, the typical (pale, speckled) form is camouflaged on lichen-covered tree bark. During the industrial revolution, soot pollution killed the lichens and blackened the bark, so the melanic (dark) form became better camouflaged and increased in frequency — a textbook case of directional selection.

Understanding the Question

Two melanic European moths were crossed and produced 41 melanic and 15 typical offspring — approximately a 3:1 ratio. The question asks us to construct a genetic diagram to explain these results, using the symbols A and a.

Because both parents show the melanic phenotype but some of their offspring show the typical phenotype, both parents must carry the recessive allele. The melanic phenotype is therefore dominant, and both parents are heterozygous (Aa).

Approach

Assign the dominant melanic allele the symbol A and the recessive typical allele the symbol a. Construct a Punnett square to show the four possible offspring genotype combinations from an Aa × Aa cross. Translate each genotype into its phenotype and state the resulting ratio.

Step-by-Step Reasoning

1. Identify parental phenotypes: melanic × melanic.
2. Deduce parental genotypes: Aa × Aa (both must be heterozygous because typical offspring appeared).
3. Identify the gametes each parent can produce: A and a, in equal proportions.
4. Construct the Punnett square: combining the gametes gives four offspring genotypes — AA, Aa, Aa, and aa.
5. Translate each genotype into a phenotype: AA and Aa are melanic (because A is dominant); aa is typical.
6. State the phenotypic ratio: 3 melanic : 1 typical.

This matches the observed 41:15 ratio (approximately 2.73:1) — close to the expected 3:1 with some statistical variation expected in a sample of only 56 offspring.

Key Takeaways

• A 3:1 ratio from a single cross involving two parents with the dominant phenotype indicates that both parents are heterozygous.
• Use A for the dominant allele and a for the recessive allele.
• A Punnett square visually displays all possible offspring genotype combinations from a cross.
• Always state both the offspring genotypes and phenotypes in the final ratio.

Common Mistakes

• Using AA × Aa or AA × AA as the parental cross — neither could produce typical offspring.
• Forgetting to list the gametes produced by each parent.
• Listing only one offspring genotype instead of all four.
• Confusing the genotypic ratio (1:2:1) with the phenotypic ratio (3:1).
• Stating that the melanic allele is recessive because the typical form is more common in the population.

Things to Be Careful About

• The dominant allele is assigned the capital letter regardless of which phenotype is more common.
• The Punnett square must show both parents' gametes on the appropriate edges.
• Offspring should be listed with all four genotypes visible.
• The 3:1 phenotypic ratio is a probability; small samples can deviate slightly.

Background Concept

The dominance relationship between two alleles of the same gene is determined by which allele is expressed in the phenotype of heterozygous individuals. The dominant allele masks the expression of the recessive allele. The key signature of a dominant allele is that two parents with the dominant phenotype can produce offspring with the recessive phenotype — this can only happen if both parents carry the recessive allele (i.e. both are heterozygous).

Understanding the Question

In a separate experiment, two melanic North American moths were crossed and produced 10 typical and 31 melanic offspring — also approximately a 3:1 ratio. The question asks what can be concluded about the melanic allele in moth populations on both continents, given that a similar ratio was observed in European moths in part (a).

Approach

Apply the same logic as in part (a). Two melanic parents producing some typical offspring means the melanic allele must be dominant. Because the same outcome was obtained on a different continent, the conclusion generalises to both populations.

Step-by-Step Reasoning

1. Two melanic North American moths were crossed.
2. Their offspring were 31 melanic and 10 typical (≈ 3:1).
3. The appearance of typical offspring from melanic parents confirms that melanic is the dominant phenotype.
4. The same outcome was observed in European moths in part (a), confirming that the melanic allele is dominant in both populations.

Key Takeaways

• A 3:1 ratio from a monohybrid cross is the signature of two heterozygous parents with a dominant phenotype.
• Similar genetic outcomes across populations support similar underlying genetics.

Common Mistakes

• Saying the allele is "more common" in the population — the question is about dominance, not frequency.
• Concluding that the allele is "the same" in both populations — the question only asks about dominance.
• Saying the allele is dominant in European moths but recessive in North American moths — the data shows the same outcome in both.

Things to Be Careful About

• The question asks specifically about dominance, not about whether the alleles are identical.
• The conclusion applies to both continents because the same result was observed in each population.

Background Concept

The Lincoln index (also called the Petersen method or mark-recapture method) is a standard technique in ecology for estimating the size of an animal population. The procedure involves:

1. Capture a sample of animals ($n_1$), mark them in a way that does not harm them or affect their behaviour, then release them back into the population.
2. Wait for a suitable period to allow the marked animals to mix with the unmarked population.
3. Capture a second sample ($n_2$) and count how many are marked ($m_2$).
4. Apply the formula:

The principle is that the proportion of marked individuals in the second sample should reflect the proportion of marked individuals in the whole population.

The method assumes:

• No births, deaths, immigration, or emigration between samples.
• Marks do not fall off or affect survival or catchability.
• Marked and unmarked individuals mix randomly.
• Marking does not make individuals more or less likely to be recaptured.

Understanding the Question

A light trap was used to estimate the population size of Biston betularia in a woodland. The data collected were:

• Night 1: 24 moths captured and marked with a small spot of harmless paint.
• Night 2: 29 moths captured, of which 8 showed a spot of paint.

Use the Lincoln index formula provided to calculate the estimated population size, showing working.

Approach

Identify each variable in the formula from the data:

• $n_1 = 24$ (number captured and marked on night 1)
• $n_2 = 29$ (total number captured on night 2, both marked and unmarked)
• $m_2 = 8$ (number of marked individuals recaptured on night 2)

Substitute into the formula and calculate.

Step-by-Step Reasoning

1. Identify the variables from the data: $n_1 = 24$, $n_2 = 29$, $m_2 = 8$.
2. Apply the formula:

1. Substitute the values:

1. Calculate:

   • Numerator: $24 \times 29 = 696$
   • Division: $696 \div 8 = 87$

2. State the answer: population size $\approx 87$.

Key Takeaways

• The Lincoln index estimates animal population size using capture-mark-recapture data.
• The second sample size ($n_2$) is the total captured, not just the unmarked.
• The result is an estimate, not an exact count.
• The formula relies on several assumptions that may not be perfectly met in real studies.

Common Mistakes

• Using $n_2 = 21$ (the unmarked individuals) instead of 29 (the total captured) — this gives a very wrong answer.
• Dividing in the wrong order (e.g. $m_2 \times n_2 \div n_1$).
• Not showing working.
• Stating the wrong unit (the population is just a number of individuals).

Things to Be Careful About

• $n_2$ is the total number captured in the second sample (both marked and unmarked).
• $m_2$ is specifically the number of marked individuals in the second sample.
• The result is the estimated total population.
• If $m_2 = 0$ (no marked recaptured), the formula gives an undefined result.