Biology 9700/42 — February/March 2025

Background Concept

A respirometer is a piece of laboratory apparatus used to measure the rate of oxygen uptake (and, indirectly, carbon dioxide release) of a small organism such as a blowfly larva, germinating seed, or woodlouse. It typically consists of a sealed chamber containing the organism, connected via a capillary tube to a coloured fluid. As the organism respires, the change in gas volume inside the chamber moves the fluid along the capillary, and the displacement per unit time gives the rate of oxygen uptake.

Understanding the Question

Part (b) introduces the idea of using RQ to identify the substrate being respired and asks you to name the apparatus used to determine the RQ of an organism such as a blowfly larva.

Approach

Recall the standard piece of apparatus for measuring gas exchange in small organisms in the A Level Biology syllabus.

Step-by-Step Reasoning

The RQ is calculated from the ratio of CO₂ produced to O₂ consumed. To obtain both quantities, the apparatus must measure oxygen uptake while also accounting for the carbon dioxide released. The standard item of equipment for this is a respirometer.

Key Takeaways

• A respirometer measures the volume change caused by respiration in a closed system.
• It is the standard apparatus for RQ determination of small organisms in the A Level syllabus.

Common Mistakes

• Writing 'spirometer' (an instrument used in human physiology to measure lung volumes, not gas exchange of small organisms).
• Writing 'gas analyser' or 'oxygen meter' — these are not the apparatus credited on the mark scheme.

Things to Be Careful About

Make sure you spell the word correctly — a misspelling such as 'respirrometer' is risky on a written paper.

Background Concept

In a respirometer, the organism takes in O₂ and releases CO₂. If both gases were free in the chamber, the two volume changes would partially cancel out and the net fluid movement would not equal the oxygen uptake. The standard solution is to include a chemical that absorbs only the CO₂ (commonly soda lime, or potassium hydroxide solution on filter paper), so that the only gas changing in volume is oxygen.

Understanding the Question

The question asks why chemicals such as soda lime or KOH are placed inside the respirometer during an RQ measurement.

Approach

Think about which gas is responsible for each component of the RQ measurement and which one must be eliminated to obtain a clean reading of oxygen uptake.

Step-by-Step Reasoning

1. The organism consumes O₂ (volume decreases) and produces CO₂ (volume would increase).
2. To measure the volume of O₂ taken in, the CO₂ must be removed from the air in the chamber.
3. Soda lime and KOH react with (absorb) CO₂ but not with O₂.
4. Therefore the only net change in gas volume is the decrease due to O₂ uptake, which moves the manometer/capillary fluid in proportion to the oxygen consumption.

Key Takeaways

• Carbon dioxide must be absorbed in a respirometer so the volume change observed is due solely to oxygen uptake.
• Soda lime or KOH are the standard absorbents used.

Common Mistakes

• Saying only 'to absorb gases' — you must specify CO₂, not O₂.
• Saying 'to provide oxygen' — these chemicals do not release oxygen, they absorb CO₂.

Things to Be Careful About

It is the carbon dioxide that is absorbed, not the oxygen. The mark scheme specifically credits 'to absorb / remove carbon dioxide'.

Background Concept

When a respiratory substrate is oxidised in aerobic respiration, oxygen is consumed and carbon dioxide and water are released. The full balanced equation must conserve every element (C, H and O) on both sides. For an organic acid such as malic acid (C₄H₆O₅) the equation takes the same form as for a carbohydrate or lipid, but the stoichiometry of oxygen is different because the substrate already contains some oxygen atoms within its structure.

Understanding the Question

You are given a partly written equation for the aerobic respiration of one molecule of malic acid (C₄H₆O₅) and asked to fill in the number of molecules of O₂ required.

Approach

Balance the equation element by element. Start with carbon (easiest), then hydrogen, then oxygen (the trickiest because oxygen appears in the substrate as well as in O₂ and in the products).

Step-by-Step Reasoning

1. Carbon: 4 carbons on the left must produce 4 CO₂ — already given, so carbon is balanced.
2. Hydrogen: 6 H on the left, 3 H₂O on the right gives 6 H, so hydrogen is balanced.
3. Oxygen: 5 oxygens in malic acid + 2x oxygens in xO₂ on the left; 8 from 4 CO₂ + 3 from 3 H₂O = 11 on the right.
   So 5 + 2x = 11 → 2x = 6 → x = 3.
4. Therefore 3 molecules of O₂ are required.

Key Takeaways

• The oxygen in the substrate itself contributes to the products, so fewer O₂ molecules are required than the total oxygen in the products.
• A useful shortcut: total O atoms in products = 11; substrate already supplies 5; the remaining 6 must come from O₂, i.e. 3 molecules.

Common Mistakes

• Forgetting that the substrate contains its own oxygen atoms and so treating the equation as if 11 O₂ were required.
• A common incorrect answer is 4 or 5.

Things to Be Careful About

Double-check the count: 4 × 2 + 3 × 1 = 11 O atoms in the products; the substrate contributes 5, leaving 6 → 3 O₂.

Background Concept

The respiratory quotient (RQ) is defined as the volume (or number of molecules) of CO₂ released divided by the volume (or number of molecules) of O₂ consumed during aerobic respiration. It is a useful diagnostic because different respiratory substrates have characteristic RQ values (carbohydrate ≈ 1.0, lipid ≈ 0.7, protein ≈ 0.8, organic acids can be higher or lower than 1).

Understanding the Question

You are asked to use the RQ formula shown in Fig. 1.1 together with the balanced aerobic respiration equation for malic acid from (c)(i) to find the RQ of malic acid, to two decimal places.

Approach

Read off the number of CO₂ molecules produced and the number of O₂ molecules consumed from the balanced equation, then divide.

Step-by-Step Reasoning

1. From (c)(i), the equation is C₄H₆O₅ + 3O₂ → 4CO₂ + 3H₂O.
2. CO₂ produced per malic acid = 4 molecules.
3. O₂ consumed per malic acid = 3 molecules.
4. RQ = 4 / 3 = 1.333…
5. Rounded to two decimal places: 1.33.

Key Takeaways

• RQ for an organic acid can exceed 1.0 because the substrate is already partially oxidised — it contains C–O bonds but still releases a relatively large amount of CO₂ per O₂ taken in.
• A value of 1.33 indicates that, per molecule of O₂, more CO₂ than usual is being released, consistent with a substrate that is more oxidised than a carbohydrate.

Common Mistakes

• Inverting the ratio (writing 3/4 = 0.75). The RQ is CO₂ produced ÷ O₂ taken in, in that order.
• Quoting 1.3 instead of 1.33 — the question specifically asks for two decimal places.

Things to Be Careful About

Use the value of O₂ from (c)(i). If you made an error there, you may still obtain a mark here via error carried forward (ecf) provided your final value follows correctly from your (c)(i) answer.

Background Concept

The respiratory quotient (RQ) reveals which substrate is being oxidised, because different substrates give different CO₂:O₂ ratios. Carbohydrate (RQ ≈ 1.0) yields one CO₂ per O₂; lipid (RQ ≈ 0.7) yields less CO₂ per O₂ because its hydrocarbon chains already contain a high H:C ratio; protein (≈ 0.8) lies in between. A shift in the RQ of an animal therefore indicates a shift in the substrate mix being respired.

Torpor is a short-term, controlled reduction in metabolic rate and body temperature. Energy demand is greatly reduced, but the animal must still keep respiring to stay alive, so it draws on its stored energy reserves.

Understanding the Question

Fig. 1.3 plots the RQ of a deer mouse between 06:00 and 22:00, with a period of torpor marked between roughly 08:00 and 17:00. You are asked to describe the trend during torpor and to suggest an explanation. The marking scheme awards three marks for: a description of the trend, a data quote, and an explanation in terms of substrate.

Approach

Read the start and end values of RQ during the torpor window from the graph, describe the overall change, and link that change to the substrate the mouse is now using (using Table 1.2 in the question as a reference).

Step-by-Step Reasoning

1. At the start of torpor (about 08:00) the RQ is approximately 0.9 (close to 0.95 in places, but ~0.9 is a fair read).
2. By the end of torpor (about 17:00) the RQ has fallen to approximately 0.75.
3. The overall trend is therefore a decrease in RQ during torpor (with small fluctuations, but the general direction is downwards).
4. The end value (~0.75) is close to the RQ of lipid (0.7) and protein (0.8), and well below the RQ of carbohydrate (1.0). This implies the mouse has switched from using mainly carbohydrate to using mainly lipid (and some protein) during torpor.
5. A plausible biological explanation: easily mobilised carbohydrate / glycogen stores become depleted over the hours of torpor, so the mouse progressively oxidises stored lipid (and possibly some protein) to meet its reduced energy needs.

Key Takeaways

• A drop in RQ below 1.0 indicates that lipid (or protein) is contributing more to respiration than carbohydrate.
• Stored carbohydrate is the most readily mobilised fuel but is limited in quantity; lipid stores are vast but require more time to mobilise and oxidise.
• Torpor, by reducing energy demand, prolongs the use of these reserves.

Common Mistakes

• Saying only that the RQ 'decreases' without quoting values from the graph — the mark scheme explicitly requires a data quote.
• Saying the mouse 'is respiring fat because it is hungry' — you must link the RQ value (0.75) to the substrate table (lipid 0.7, protein 0.8) explicitly.
• Confusing torpor with hibernation (torpor is a daily/short-term state; hibernation is seasonal).

Things to Be Careful About

The trend is described using two figures from the graph (start and end of torpor). Choose values you can actually see on the curve. Small fluctuations around the trend are normal — describe the overall direction, not every wiggle.

Background Concept

Torpor is energetically expensive to enter and leave (it requires reheating the body, which costs ATP and therefore food reserves), so an animal should only enter torpor when the benefits clearly outweigh the costs. For a small endotherm such as the deer mouse, the costs of staying active and the benefits of becoming inactive both change with the season.

Key seasonal pressures on a small nocturnal mammal:

• Winter: low ambient temperature → high heat loss; food (insects, seeds) is scarcer; short day length → less time to forage. Conserving energy by being inactive during the day is very advantageous.
• Summer: high daytime temperatures → risk of overheating and dehydration; food is more abundant so a long rest is less costly; remaining hidden in a burrow during the day conserves water and reduces exposure.
• Year-round: small mammals suffer high predation; being hidden and still during the day reduces the chance of being seen by diurnal predators.
• Breeding season: females caring for young cannot afford to be inactive for long; torpor is generally suppressed.

Understanding the Question

The question states that deer mice enter torpor only at certain times of the year, and asks you to suggest reasons why. The mark scheme accepts any two reasonable suggestions, and the strongest answers link a specific seasonal factor (cold, food, water, predators) to a clear advantage of being in torpor.

Approach

For each potential advantage, name the season or environmental condition, state the pressure the mouse faces, and explain how torpor helps.

Step-by-Step Reasoning

1. Winter / cold weather: small mammals lose heat rapidly because of their high surface area to volume ratio. When food is also scarce, reducing metabolic rate (torpor) cuts both the energy needed to keep warm and the food needed to fuel that energy. So torpor is favoured in winter.
2. Hot, dry summer days: being active in a burrow during the day avoids overheating and reduces evaporative water loss; the mouse is anyway more active at night.
3. Predation: being still and concealed during daylight, when many predators hunt, reduces the risk of being detected.
4. AVP: e.g. during the breeding/lactating period, females must forage often to feed their young and so do not enter torpor.

Key Takeaways

• Torpor is a seasonally timed adaptation to environmental pressures on energy, water, temperature and predation.
• The most commonly credited reasons are: cold + food shortage (winter) and water/heat conservation (summer); predator avoidance and breeding constraints are also acceptable.

Common Mistakes

• Saying only 'to save energy' without specifying the season or the environmental pressure.
• Saying torpor helps the mouse 'find food' — torpor is a state of inactivity; the benefit is not feeding but reducing the need to feed.

Things to Be Careful About

The question asks for reasons why torpor occurs only at certain times of the year, so link your answer to a specific time of year or seasonal condition. Generic 'energy saving' statements without a seasonal anchor will not earn the mark.

Background Concept

Speciation is the formation of new species from an existing one. There are two main modes:

• Allopatric speciation: a population becomes geographically isolated (e.g. by a mountain range, ocean or river) and the separated groups evolve independently until they can no longer interbreed.
• Sympatric speciation: a new species arises within the same geographical area as the parent population, without any physical barrier. Separation is driven instead by behavioural, ecological or genetic isolation (e.g. different mating calls, different food sources, polyploidy in plants).

The defining feature of sympatric speciation, therefore, is that the diverging populations continue to share the same geographical range while reproductive isolation develops.

Understanding the Question

You are asked to use Fig. 2.2 to explain why the speciation occurring in Northeast Atlantic orca is described as sympatric. Fig. 2.2 shows circles (Type 1) and squares (Type 2) plotted in the Northeast Atlantic. The command word explain here means you should use the figure to justify the term 'sympatric'.

Approach

Look at the symbols on the map and ask: are Type 1 and Type 2 found in the same places, or in completely different places? If they overlap, there is no geographical separation, and this is the hallmark of sympatric (as opposed to allopatric) speciation.

Step-by-Step Reasoning

1. Inspect Fig. 2.2 — both Type 1 (circles) and Type 2 (squares) are plotted in broadly the same waters around Iceland, the Faroe Islands, the northern UK, and the coast of Norway.
2. There is no physical barrier (land, ice, current system shown) dividing the ranges of the two types.
3. Because the populations co-occur geographically, any reproductive isolation between them cannot be due to distance or physical separation — it must instead be due to behavioural or ecological differences.
4. This is exactly the criterion for sympatric speciation, hence the term applies.

Key Takeaways

• Sympatric = 'same homeland' (Greek syn = together, patria = homeland).
• The geographical overlap between Type 1 and Type 2 in Fig. 2.2 is the visual evidence for sympatric speciation.
• Sympatric speciation still requires a mechanism of reproductive isolation — that mechanism is explored in (a)(ii).

Common Mistakes

• Saying 'because they live in the same ocean' — too vague. Cite specific overlapping regions shown on the map (Iceland / Faroes / UK / Norway).
• Confusing sympatric with allopatric: remember allopatric requires geographical isolation, sympatric does not.
• Writing 'because they don't interbreed' — that is the consequence of speciation, not the reason the term 'sympatric' is used.

Things to Be Careful About

• Use the figure as evidence — do not give a textbook definition of sympatric speciation in place of a figure-based explanation.
• The mark is awarded for stating that there is no geographical separation between the two types.

Background Concept

For sympatric speciation to occur, the two diverging populations must become reproductively isolated even though they share the same range. Reproductive isolation can occur before mating (prezygotic) or after mating (postzygotic). In animals, the most common route in sympatric speciation is behavioural isolation — differences in signals or habits that prevent mating even when individuals meet.

Behavioural isolation can involve:

• Differences in courtship displays or mating calls so individuals do not recognise each other as potential mates.
• Differences in breeding season / time of day of mating.
• Differences in habitat use within the same area (e.g. feeding in shallow vs deep water).
• Differences in diet or foraging behaviour that reduce encounters between the groups.

Understanding the Question

You are told that Type 1 orca feed mainly on fish and Type 2 feed mainly on seals / marine mammals, and that the two types share the same Northeast Atlantic range. You are asked to suggest behavioural separations (i.e. plausible behaviours that would keep the two types from interbreeding) that contribute to sympatric speciation. Suggest means propose reasonable possibilities — you do not need to know them as established fact.

Approach

Think about the kinds of behaviours that, if they differed between two populations in the same area, would prevent them from meeting at mating time. Then ask which of these could plausibly vary between orca types that already differ in diet.

Step-by-Step Reasoning

1. Mating calls / courtship: if the two types produce different acoustic signals or perform different courtship displays, individuals may not recognise each other as mates. Orca are known to have distinct dialects within pods, so it is plausible that Type 1 and Type 2 calls differ.
2. Breeding timing: if one type mates at a different time of year, the chances of cross-mating are reduced. Even slight seasonal differences in oestrus would suffice.
3. Group size / social structure: orca populations vary in pod size. Type 1 might travel in larger family groups, Type 2 in smaller pods or as solitary individuals. Different social structures reduce interactions between the types.
4. Diet / hunting behaviour: because Type 1 and Type 2 hunt different prey in different ways (fish vs seals), they spend time in different micro-habitats and use different hunting techniques. This further reduces the chance of mixed groups meeting, reinforcing isolation.

Any three of these four marking points (or other plausible behaviours) score full marks.

Key Takeaways

• Sympatric speciation requires behavioural or ecological isolation, not physical isolation.
• In social mammals such as orca, differences in group size, diet, and mating calls are realistic isolating mechanisms.
• Prezygotic isolation (especially behavioural) is the most common route to sympatric speciation in animals.

Common Mistakes

• Giving only one or two points when three are required.
• Writing vague statements such as 'they behave differently' — be specific (which behaviour?).
• Confusing behavioural with geographical or mechanical isolation.
• Omitting the link back to mating / not interbreeding.

Things to Be Careful About

• The mark scheme accepts either the general behaviour (e.g. 'different group sizes') or a specific example (e.g. 'groups vs solitary'). Either scores.
• Three independent points are needed; restating the same idea twice does not earn extra marks.

Background Concept

Phenotypic differences between closely related populations (or subspecies) are the raw material on which natural selection, genetic drift, or sexual selection act. Identifying consistent morphological differences is the classical first step in taxonomy and in detecting incipient speciation.

Understanding the Question

Fig. 2.4 shows side-view diagrams of Type B, Type C and Type D orca on the same scale (0–7 m scale bar). You are asked to state one way in which Type D differs from both Type B and Type C — a single, specific, observable difference from the figure.

Approach

Compare the Type D diagram with the Type B and Type C diagrams. Look at:

• Eye patch (size and shape)
• Dorsal fin (size and shape)
• Body / head shape
• Colour patches on the back, sides and underside

Pick one difference that is clearly shown in the diagram and unique to Type D.

Step-by-Step Reasoning

1. In Fig. 2.4, the eye patch of Type B is large and roughly horizontal.
2. The eye patch of Type C is smaller and slanted.
3. The eye patch of Type D is very small and narrow — clearly smaller than in either B or C.
4. Other differences visible: Type D has a bulbous / rounded head, no black 'cape' on the back, a small grey patch just behind the dorsal fin, and a taller, more pointed dorsal fin. Any one of these is acceptable.

Key Takeaways

• Phenotype = observable characteristics (morphology, colour, behaviour).
• Distinct, consistent morphological differences between populations are evidence that divergence is occurring.

Common Mistakes

• Giving a difference that is not shown in the figure (e.g. internal anatomy).
• Stating a size difference without giving a feature (size of what?).
• Listing more than one difference when only one is asked for.

Things to Be Careful About

• State = a single, direct answer — no explanation needed.
• The difference must distinguish Type D from both B and C, not just one of them.

Background Concept

Genetic drift is the random (chance) change in allele frequency from one generation to the next, with no involvement of natural selection. Its effects are strongest in small populations, where chance events have a proportionally large effect on the gene pool.

The founder effect is a special case of genetic drift: when a new population is founded by a small number of individuals separated from the parent population, the founders carry only a sample of the original alleles — by chance, some alleles may be over-represented and others absent entirely. Subsequent generations inherit this biased gene pool, so the new population diverges genetically (and eventually phenotypically) from the original.

Understanding the Question

You are told Type D orca occur offshore in the Southern Ocean, geographically separate from the inshore Type B and Type C populations. You are asked to suggest how genetic drift (including the founder effect) could lead to phenotypic differences between Type D and the other types. Suggest means you need to construct a plausible chain of reasoning, not recall a single fact.

Approach

Identify the sequence of events the mark scheme rewards:

1. Isolation of a small group from the parent population.
2. Small starting population / small gene pool.
3. Chance / founder effect acting on this small gene pool.
4. Change in allele frequency / loss of alleles.
5. Phenotypic consequences.

Step-by-Step Reasoning

1. Isolation: a small group of orca became separated from the larger population and migrated to the offshore Southern Ocean region (Fig. 2.3 shows Type D is geographically separate from B and C).
2. Small starting population: the founding group was small, so only a sample of the alleles present in the original population was carried into the new region.
3. Small gene pool / low genetic diversity: with fewer individuals, the new population started with less genetic variation than the parent population.
4. Founder effect / chance: by chance, the founders did not carry all the alleles from the parent population — some were absent from the start. Subsequent generations are built from this biased sample.
5. Change in allele frequency: because the population remains small, genetic drift continues to change allele frequencies at random each generation; some alleles are lost, others become fixed.
6. Phenotypic differences: the resulting change in allele frequencies produces the visible phenotypic differences seen in Fig. 2.4 (eye-patch size, head shape, dorsal fin, etc.).

Key Takeaways

• Genetic drift is most powerful in small populations and is driven by chance, not selection.
• The founder effect is genetic drift initiated by a small colonising group — it is the most likely mechanism behind the divergence of an offshore orca population.
• Drift causes changes in allele frequency, which in turn produce phenotypic change.

Common Mistakes

• Confusing genetic drift with natural selection — drift is random; selection is non-random (favours beneficial alleles).
• Saying 'inbreeding caused the differences' — inbreeding may occur in a small population, but the mark scheme's mechanism is founder effect / chance, not inbreeding per se.
• Forgetting to mention the small initial population size — without this, drift has little effect.
• Giving only one or two points when three are required.

Things to Be Careful About

• The question says 'including the founder effect', so founder effect specifically must be mentioned (or implied via 'chance' / small founding population).
• Drift affects allele frequencies, not individuals — be precise about the level at which the change occurs.

Background Concept

The polymerase chain reaction (PCR) is an in-vitro method for amplifying a specific region of DNA. Each cycle has three temperature-controlled steps:

1. Denaturation (≈$95\ \text{°C}$): the double-stranded DNA is heated so the two strands separate by breaking the hydrogen bonds between complementary bases.
2. Annealing (≈$55\ \text{°C}$): short single-stranded DNA primers bind (anneal) to the flanking regions of the target sequence on each single strand.
3. Extension (≈$72\ \text{°C}$): a DNA polymerase adds free DNA nucleotides to the 3' end of each primer, synthesising a new complementary strand.

Because each new strand acts as a template in the next cycle, the number of target DNA molecules doubles per cycle, giving exponential amplification.

The DNA polymerase used must remain active after being heated to ≈$95\ \text{°C}$. Ordinary human DNA polymerases would denature. Taq polymerase, isolated from the thermophilic bacterium Thermus aquaticus, is naturally heat-stable.

Understanding the Question

The question asks you to describe AND explain the role of Taq polymerase — that is, what it does and why it is suitable for PCR. Four marks are available, so four clear, distinct points are needed (one mark each). Pick points from the mark scheme's list of eight; any four that are biologically correct will earn the marks.

Approach

Start with the basic identification of what Taq polymerase IS (an enzyme, a DNA polymerase), then describe what it DOES in the cycle (synthesise new strands, bind primers, add nucleotides, act at the extension step), and finally explain WHY it is suitable (thermostable at $95\ \text{°C}$, optimum at $72\ \text{°C}$, can be reused). Linking the property of thermostability to the practical advantage — no need to add fresh enzyme after every denaturation step — is the key explanatory link that examiners look for.

Step-by-Step Reasoning

• MP1 — Type of enzyme. Taq polymerase is a DNA polymerase. This is the foundational identification; everything else flows from it.
• MP2 — What it does. It synthesises a new (complementary) DNA strand, using the single-stranded template exposed after denaturation.
• MP3 — Where on the DNA it acts. It binds to the primers that have already annealed to the target sequence, and adds nucleotides to the 3' end of each primer during the extension step. It cannot start a strand from scratch — it needs a primer with a free 3'-OH.
• MP4 — Temperature optimum. It works optimally at about $72\ \text{°C}$, matching the extension temperature of a standard PCR programme.
• MP5 — Thermostability. It is not denatured by the ≈$95\ \text{°C}$ denaturation step. This is the property that makes it the polymerase of choice.
• MP6 — Why thermostability matters. Because it survives the denaturation step, a single addition of enzyme at the start of the reaction can be reused for all 25–35 cycles, which would otherwise be impossibly laborious and expensive.

Any four of these earn the four marks. Stronger answers combine several, e.g. identifying the enzyme, naming what it does, stating the optimum temperature, and linking thermostability to repeated reuse.

Key Takeaways

• Taq polymerase is the heat-stable DNA polymerase used in PCR.
• It extends primers (cannot initiate synthesis) at the extension step of each cycle.
• Its thermostability is what makes automated thermal cycling practical.

Common Mistakes

• Calling it "a polymerase" without specifying DNA polymerase.
• Saying it "makes copies of DNA" without specifying that it synthesises a complementary strand using a template.
• Saying the enzyme "unwinds the DNA" or "separates the strands" — this is done by the high temperature, not the enzyme.
• Saying it adds nucleotides to either end of the primer (it must be the 3' end).
• Forgetting to link thermostability to re-use across many cycles.

Things to Be Careful About

• The mark scheme accepts $70\ \text{–}75\ \text{°C}$ for the optimum temperature.
• "Thermostable" is the required word; "heat-resistant" or "works at high temperatures" are less precise but usually still credited if the meaning is clear.
• A common trap is to describe PCR generally instead of focusing on Taq polymerase specifically — keep the answer centred on the enzyme.

Background Concept

The MT-RNR1 gene encodes a mitochondrial ribosomal RNA. A specific mutation in this gene predisposes carriers to aminoglycoside-induced ototoxicity — gentamicin (an aminoglycoside) damages the hair cells of the cochlea and causes irreversible deafness in anyone carrying even a single copy of the mutant allele. Screening before treatment is therefore essential, and the screening is done with PCR followed by gel electrophoresis.

In this test, primers are designed to amplify a region of MT-RNR1 that changes length when the mutation is present (the mutation creates or removes a sequence, so the amplified fragment is shorter in mutant DNA). On a gel, longer DNA fragments stay near the wells, while shorter DNA fragments travel further towards the anode.

Understanding the Question

The question gives you a completed gel (Fig. 3.1) with three children's DNA samples. You have to decide which child(ren) carry the mutation and therefore must NOT be given gentamicin. Note that this is a one-mark question, but the answer has two children in it, so reading the pattern correctly is essential.

Approach

Translate each child's band pattern into a genotype:

• one band at the "top" position = homozygous normal
• one band at the "bottom" position = homozygous mutant
• two bands (one at each position) = heterozygous

Then apply the rule: anyone with at least one mutant allele must not receive gentamicin. Only the homozygous normal child (Child 1) is safe.

Step-by-Step Reasoning

1. Identify the gel setup. DNA migrates from top to bottom, so the band nearer the top has travelled less far, and the band further down has travelled further. Smaller/shorter DNA fragments travel further.
2. Translate band position to allele type.
   • The higher band (less far migrated) = the longer PCR product = the normal allele.
   • The lower band (further migrated) = the shorter PCR product = the mutant allele.
3. Read each lane.
   • Child 1: one band, at the higher (normal) position → homozygous normal → safe to give gentamicin.
   • Child 2: one band, at the lower (mutant) position → homozygous mutant → must NOT be given gentamicin.
   • Child 3: two bands, one at each position → heterozygous → carries the mutant allele → must NOT be given gentamicin.
4. Apply the rule. Children carrying even a single mutant allele cannot receive gentamicin, so the answer is Children 2 and 3.

Key Takeaways

• A single band on a gel can mean homozygous for whichever allele that band represents.
• Two bands at the two diagnostic positions means heterozygous.
• For a dominant adverse drug reaction, heterozygotes must also be excluded from the drug — carriers are not safe.

Common Mistakes

• Saying only Child 2 (forgetting the heterozygote Child 3).
• Saying only Child 3 (forgetting that a single mutant band also indicates the mutation).
• Misreading the gel direction — assuming the lower band represents the larger fragment.

Things to Be Careful About

• This is autosomal (mitochondrial) inheritance: the mutation is dominant in its effect on drug response, so a heterozygote is just as much at risk as a homozygote.
• A one-mark question with two names in the answer is a common Cambridge trap: the mark is only awarded if BOTH names are correct, so don't list them separately across multiple sentences.

Background Concept

Gel electrophoresis separates DNA fragments (or proteins) by applying an electric field across a gel. Two principles govern DNA migration:

1. Charge. Every nucleotide in DNA carries a phosphate group in the sugar–phosphate backbone, giving the whole molecule a uniform negative charge per unit length. So all DNA fragments — whatever their size — are pulled in the same direction by the electric field, towards the positive electrode (anode). The wells are loaded at the cathode (negative) end.
2. Size. The agarose (or polyacrylamide) gel is a mesh of pores. Long DNA fragments are slowed by the mesh and so migrate more slowly; short DNA fragments slip through more easily and migrate faster. The result is that distance travelled is inversely related to fragment length.

A molecular-weight marker (DNA ladder) is run alongside the samples so that band positions can be converted to approximate sizes in base pairs (bp).

Understanding the Question

The question asks you to explain the mechanism of gel electrophoresis as it applies to the specific pattern in Fig. 3.1. That means you must state the underlying principles AND link them to what is actually shown: a single upper band in lane 1, a single lower band in lane 2, and both bands in lane 3. Four marks are available, so aim for three to four developed points.

Approach

Lay out the logic in this order:

1. State the principle of separation (size-based).
2. State WHY DNA migrates at all (its charge and the electric field).
3. State the direction of movement (cathode → anode) and the consequence of size (shorter = further).
4. Apply it to this specific experiment: the normal allele's PCR product is longer → less far; the mutant allele's PCR product is shorter → further; therefore Child 1 is homozygous normal, Child 2 homozygous mutant, Child 3 heterozygous.

Step-by-Step Reasoning

• MP1 — Separation by size. Gel electrophoresis separates DNA fragments according to their length (or mass). Fragments of different sizes are resolved into distinct bands.
• MP2 — DNA is negatively charged. The phosphate groups in the sugar–phosphate backbone give DNA an overall negative charge.
• MP3 — Direction of movement. The negatively charged DNA is attracted to the positive electrode (anode) and therefore migrates from the wells (cathode end) towards the anode.
• MP4 — Size determines distance. Smaller/shorter DNA fragments experience less resistance from the gel matrix and so migrate faster and further. Conversely, longer fragments travel less far in the same time.
• MP5 — Application to Fig. 3.1.
  • The PCR product from the normal allele is longer → travels a shorter distance → upper band.
  • The PCR product from the mutant allele is shorter → travels a greater distance → lower band.
  • Child 1 (one upper band only) is homozygous normal; Child 2 (one lower band only) is homozygous mutant; Child 3 (both bands) is heterozygous.

Key Takeaways

• Gel electrophoresis separates DNA by size because of the sieving effect of the gel.
• All DNA moves the same direction (to the anode) because the charge-to-mass ratio is uniform.
• Shorter fragments travel further in a given time — this is the principle that allows alleles of different lengths to be distinguished.

Common Mistakes

• Saying DNA moves towards the negative electrode — DNA is negative and so is repelled by the negative electrode, attracted to the positive one.
• Saying larger fragments move faster because they "have more charge" — all DNA has the same charge per unit length, so charge does not differ between fragments.
• Failing to link the principle to the actual bands in Fig. 3.1 (so the answer reads like a generic description of electrophoresis with no link to MT-RNR1).
• Confusing "size" with "mass" — both are accepted, but do not say "shape" or "density".

Things to Be Careful About

• The mark scheme phrases the size point as "smaller DNA fragments migrate faster/further" (or, in reverse argument, "longer fragments migrate slower/less far"). Either wording earns the mark, but be consistent.
• The explanation must include BOTH the direction of movement (charge-based) AND the size-based separation; missing one of these forfeits a mark.
• For part (iii) you do NOT need to re-state which child is which genotype — that was part (ii). Focus on the mechanism.

Background Concept

Natural selection acts on heritable variation in a population. If individuals carrying a particular allele have greater survival or reproductive success in a given environment, that allele's frequency in the population increases over generations. A situation in which heterozygotes have higher fitness than either homozygote is called heterozygote advantage; the classic textbook example is the sickle-cell allele (HBB^S) in malarial regions, and the C282Y allele of HFE is sometimes presented as a parallel example in relation to iron deficiency.

A recessive disease allele can be maintained at relatively high frequency if, in heterozygous form, it confers a survival benefit that outweighs the disadvantage of the rare homozygous-recessive individuals who do become ill.

Understanding the Question

The stem says the C282Y allele has been maintained in European populations even though homozygotes develop a serious disease. The question asks the candidate to suggest how the allele has been maintained. The expected answer is heterozygote advantage driven by a low-iron diet during the early farming era, with a secondary point that in modern, iron-rich diets the allele is effectively neutral in heterozygotes.

Approach

1. Identify the change in diet (low iron) as a new selection pressure.
2. State the advantage conferred on heterozygotes.
3. State the consequence in terms of survival and reproduction.
4. State the consequence in terms of allele frequency, and name the process as natural selection.
5. (Optional) Note the modern situation in which the allele is effectively neutral in heterozygotes.

Step-by-Step Reasoning

• Low dietary iron acts as a selection pressure: people with insufficient iron become ill and may be less able to survive and reproduce.
• Heterozygotes (Hh) absorb more iron from their food than HH individuals (the stem says this explicitly), so they are less likely to become iron deficient.
• Therefore heterozygotes have a greater chance of surviving and reproducing in a low-iron environment — this is heterozygote advantage.
• Because more heterozygotes passed on the h allele, the frequency of h in the population increased over generations. This is natural selection.
• In modern, iron-rich diets the C282Y allele is essentially neutral in heterozygotes (they get enough iron but rarely suffer the disease, because disease occurs only in homozygotes). An allele that is not strongly selected against can persist in a population for many generations.

The candidate should also be aware that the question allows additional credit (AVP) for valid points such as: the allele may also confer protection against other diseases, or that the allele frequency in present-day populations reflects past selection rather than current selection.

Key Takeaways

• Recessive disease alleles can be maintained by heterozygote advantage if the heterozygote phenotype is beneficial in some environmental condition.
• A change in environment (here, the shift to farming) can create a new selection pressure that favours an allele previously rare or neutral.
• Allele frequencies in a modern population reflect the cumulative effect of selection over many generations; an allele currently neutral may be common because of past advantage.

Common Mistakes

• Stating only that heterozygotes 'have an advantage' without specifying the advantage (more iron absorbed, less likely to be iron deficient).
• Failing to name natural selection or to link survival/reproduction to allele frequency.
• Claiming the allele is maintained by mutation alone — mutation introduces new alleles but cannot by itself maintain a high frequency.
• Confusing heterozygote advantage with the disease being harmless in heterozygotes (it is not 'harmless' — heterozygotes have a measurable phenotype, increased iron absorption).

Things to Be Careful About

• The mark scheme credits four points, and the answer should make the chain from selection pressure → heterozygote advantage → survival/reproduction → allele frequency explicit.
• A named example of the heterozygote advantage (e.g. 'less likely to be iron deficient') is required for full credit.

Background Concept

Allele frequencies in modern populations reflect their evolutionary history. A new mutation occurs in one individual in one place; over time it can spread through migration (gene flow) into neighbouring populations, becoming diluted as it does so. The further a population is from the origin of the mutation, the lower its frequency tends to be — unless selection or further migration events have altered the pattern. Geographical barriers (mountains, seas) and political borders restrict gene flow, while differences in environment and in medical care alter the selection pressure acting on the allele.

Understanding the Question

Fig. 4.1 shows the percentage of people with one C282Y allele in different European countries. The map shows a clear gradient, with the highest percentage in Ireland (10.1%) and generally lower percentages in countries further away. The candidate must suggest explanations for these differences.

Approach

Read the map for the spatial pattern and then propose a small number of plausible explanations consistent with that pattern: origin, migration, barriers, diet and medical care.

Step-by-Step Reasoning

1. The C282Y mutation is thought to have first occurred in Ireland. The highest percentage in Ireland is therefore consistent with the mutation having arisen there and been present in the local gene pool for longest.
2. The percentage generally falls with distance from Ireland (e.g. UK 8.1%, Norway 7.5%, France 6.1%, Spain 2.5%, Italy 2.0%). This pattern is consistent with migration of carriers out of Ireland into neighbouring countries, with allele frequency being diluted by mixing with non-carrier populations the further the gene flow spreads.
3. Migration is not uniform: geographical barriers (e.g. the Alps, the Mediterranean) and political borders have slowed gene flow into some regions, producing local differences in allele frequency. For example, southern European countries (Italy, Greece) have very low frequencies.
4. Diet and medical care vary between countries. In countries with a low-iron diet, the heterozygote advantage described in part (c)(i) maintains the allele at higher frequency; in countries where haemochromatosis is well-managed (e.g. routine blood-letting), the homozygous-recessive individuals survive and reproduce, so the allele is not strongly selected against and can persist.

Key Takeaways

• Patterns of allele frequency across a continent often reflect a combination of origin, migration, gene flow, geographical isolation, and selection.
• Reading a distribution map correctly involves identifying the spatial pattern (here, a fall with distance from Ireland) and offering several plausible mechanisms rather than a single explanation.
• Migration dilutes allele frequencies as it spreads them; barriers preserve local differences.

Common Mistakes

• Offering only one explanation when the question asks for suggestions (the mark scheme expects up to three independent points).
• Failing to link the highest frequency in Ireland to the proposed origin of the mutation.
• Ignoring the role of barriers and borders in slowing gene flow.
• Suggesting without justification that differences are due to 'different races' — this is not a biological mechanism and would not be credited.

Things to Be Careful About

• The mark scheme credits three independent points; aim to make all three clearly distinct.
• 'Migration' alone is too vague — it is the direction of migration (out of Ireland into the rest of Europe) and its limitation by barriers that explain the observed pattern.

Background Concept

Blood glucose concentration is kept within narrow limits (≈ 4–6 mmol dm⁻³) by the antagonistic hormones insulin and glucagon, both secreted by the islets of Langerhans in the pancreas. Insulin is released by β-cells when blood glucose rises above the set point; it is the only blood-glucose-lowering hormone. It acts on body cells to increase glucose uptake and on the liver to store the excess as glycogen (glycogenesis).

Understanding the Question

Fig. 6.1 traces what happens after a carbohydrate-rich meal: blood glucose rises above the set point, the pancreas releases a hormone P, P has two effects — increasing glucose uptake by certain cells and promoting conversion of glucose to a storage product in liver cells — and blood glucose falls back. You must name the hormone P, the type of cell Q, and the storage product R.

Approach

Read each box carefully and match it to the role of insulin:

• Released by the pancreas in response to high blood glucose → insulin.
• Stimulates glucose uptake in fat cells AND another cell type → insulin increases glucose uptake in muscle cells (and, indirectly, adipocytes); in the liver it mainly promotes glycogen synthesis, so the missing Q cell type is the muscle cell (liver is mentioned separately in the next branch).
• Liver cells convert glucose to a storage polysaccharide → glycogen.

Step-by-Step Reasoning

1. The pancreas releases P when blood glucose rises. Of the two pancreatic hormones, insulin is the one released in response to high blood glucose, so P = insulin.
2. The diagram states that P stimulates glucose uptake in fat cells and Q cells. Insulin increases glucose uptake from the blood into muscle cells (and adipose tissue); the partner of fat cells given here is muscle cells (mark scheme also accepts liver cells here).
3. The diagram states that P stimulates liver cells to convert glucose to R. Insulin promotes glycogenesis in hepatocytes, so R = glycogen.

Key Takeaways

• Insulin = the only blood-glucose-lowering hormone; secreted by β-cells.
• Insulin's main targets: muscle, adipose tissue (↑ glucose uptake) and liver (↑ glycogenesis, ↓ gluconeogenesis).
• Glycogen is the body's short-term glucose store, predominantly in liver and skeletal muscle.

Common Mistakes

• Writing "P = glucagon" — glucagon is the raising hormone; the diagram shows the decrease of blood glucose.
• Naming R as glucose, fat or starch — only glycogen is the human storage polysaccharide made from glucose in the liver.
• Putting "liver" for Q — the mark scheme accepts muscle or liver, but the next branch already names the liver, so muscle is the cleaner answer.

Things to Be Careful About

Q must be a CELL TYPE (muscle cells / liver cells), not an organ or a tissue. Spelling "glycogen" correctly is required — "glucagon" is the antagonist and will not earn the mark.

Background Concept

Homeostasis maintains the internal environment around a set point. The two broad types of control are:

• Negative feedback — the response reverses the initial change, restoring the variable towards its set point (e.g. blood glucose, temperature, CO₂).
• Positive feedback — the response amplifies the initial change (e.g. oxytocin in childbirth, action-potential depolarisation).

Understanding the Question

In Fig. 6.1, a rise in blood glucose triggers a response (insulin release) that lowers blood glucose — i.e. the response opposes the stimulus. This is the textbook definition of one specific kind of control loop, and you must name it.

Approach

Identify the direction of the response relative to the stimulus: stimulus = ↑ glucose; response = ↓ glucose. Because the response opposes (negates) the stimulus, the answer is negative feedback.

Step-by-Step Reasoning

1. The set point is the target value (≈ 5 mmol dm⁻³ glucose).
2. The stimulus (above set point) is detected by pancreatic β-cells.
3. The effector response (insulin) reduces blood glucose back towards the set point.
4. The correction is in the opposite direction to the deviation → negative feedback.

Key Takeaways

• Nearly all homeostatic control in mammals is by negative feedback.
• The hallmark feature is that the effector response opposes the original stimulus.

Common Mistakes

• Writing "positive feedback" because the system is "active" — the question asks about the type of control, and a stimulus–response that reverses itself is negative.
• Vague answers such as "homeostasis" or "self-regulation" — the mark requires the exact term negative feedback.

Things to Be Careful About

Don't write "negative feedback loop" unless the mark scheme credits it; "negative feedback" alone is sufficient.

Background Concept

The Sumatran tiger (Panthera tigris sumatrae) is a subspecies of tiger native to the Indonesian island of Sumatra and is listed as Critically Endangered on the IUCN Red List. Population sizes of endangered species are routinely monitored, and changes in size over time are summarised as a mean rate of change:

A negative result indicates a decrease. The mean rate does not reveal whether the decrease has been steady or has accelerated/decelerated — that requires looking at the shape of the graph — but it gives a useful single number to summarise a long-term trend.

Understanding the Question

Fig. 7.2 plots the estimated number of wild Sumatran tigers at 10-year intervals from 1970 to 2020. We are told to find the mean rate of decrease between 1970 and 2020, in units of $\text{year}^{-1}$ (tigers per year, ignoring the sign).

The command word is calculate, so we must show working and a numerical answer with units.

Approach

1. Read the values at the two endpoints (1970 and 2020) as carefully as the grid allows.
2. Take the difference (later minus earlier; the question asks for decrease, so subtract the smaller from the larger).
3. Divide by the time interval (50 years).
4. Quote the answer with the correct unit, $\text{year}^{-1}$.

Step-by-Step Reasoning

Step 1 — Read Fig. 7.2.
The plotted point in 1970 sits very close to the 1000 gridline, slightly above it. The mark scheme accepts a reading of 1010. The plotted point in 2020 lies just below the 500 line; the mark scheme accepts 470.

Step 2 — Compute the decrease.

Step 3 — Divide by the time interval.

Step 4 — Attach the unit.
The question specifies $\text{year}^{-1}$, giving a mean loss of 10.8 tigers per year over the 50-year window.

Key Takeaways

• A mean rate of change is always (change in y) ÷ (change in x).
• Always quote the unit the question specifies.
• For "decrease" questions the sign convention is up to you — CIE accepts the magnitude with $\text{year}^{-1}$, or a negative value; just be consistent.

Common Mistakes

• Dividing 50 ÷ 540 instead of 540 ÷ 50, giving 0.0926 year⁻¹.
• Using only one of the endpoint values (e.g. just reading off 2020 alone).
• Forgetting to include the unit $\text{year}^{-1}$ (the unit is part of the mark).
• Using the wrong interval, e.g. 40 years between 1980 and 2020.

Things to Be Careful About

• Read both endpoints carefully — small reading errors here are amplified by the division.
• The mark scheme awards 1 mark for correct working even if the final value is wrong, so always show the substitution.
• Note that the answer is given to 3 significant figures; do not round prematurely.

Background Concept

Conservation of an endangered species is rarely achieved by a single measure; biologists combine in-situ actions (protecting animals and habitat where they naturally live) with ex-situ actions (managing populations outside the wild, such as in zoos). The question excludes captive breeding, so the marks must come from other measures — chiefly in-situ, but also legal, political and educational interventions.

Key categories of intervention for large, wide-ranging mammals like the Sumatran tiger are:

1. Legal — bans on hunting, trade and poaching, with enforcement.
2. Habitat-based — protected areas, national parks, corridors connecting fragmented forests.
3. Monitoring — radio collars, camera traps, population surveys.
4. Social — education and awareness to reduce demand for tiger parts and to gain local support for conservation.
5. Other — veterinary intervention, CITES listings, anti-poaching patrols, eco-tourism revenue.

Understanding the Question

We are asked to outline (give a brief explanation of) three conservation measures for Sumatran tigers other than captive breeding. Each mark is for a distinct, sensible measure; vague statements such as "help the tigers" score nothing.

Approach

Work through the categories above and pick three measures, each phrased in a way that is concrete enough to earn a mark. Aim for variety — three habitat-based points, for example, would be weak.

Step-by-Step Reasoning

• Measure 1 — Ban hunting/poaching/trade. Tigers are killed for skins, bones and traditional medicines. A legal ban removes the supply side, and the mark is strengthened by mentioning enforcement / strict penalties, which actually makes the ban work.
• Measure 2 — National parks/protected reserves/habitat protection. Sumatran tigers live in lowland and montane forest. Without protecting the forest, the prey base (wild pig, deer) collapses and tigers are killed when they venture near villages. Protected areas plus habitat corridors are the main in-situ tool.
• Measure 3 — Tracking devices/monitoring. GPS collars or camera traps allow rangers to locate tigers, identify poaching hotspots, and estimate population sizes — essential for assessing whether the conservation measures are working.
• (Optional additional) — Education/awareness campaigns target the local and international demand for tiger products; veterinary care treats injured or snared wild tigers; CITES listings restrict international trade.

Key Takeaways

• Conservation is multi-pronged: legal, habitat, monitoring, social and veterinary.
• Ex-situ work (zoos, captive breeding, IVF) is only one part; in-situ measures are at least as important.
• An "outline" requires a brief description, not just a buzzword — "ban hunting" alone is enough but adding why it works improves the answer.

Common Mistakes

• Mentioning captive breeding or zoos — explicitly excluded.
• Writing vague statements such as "look after them" or "stop killing them".
• Repeating the same idea in different words (e.g. "ban hunting" then "stop poaching") — only one mark is awarded.

Things to Be Careful About

• Stay within the scope: Sumatran tigers, not generic wildlife conservation.
• Do not invent impractical ideas (e.g. "move all tigers to Antarctica").
• "Awareness" alone is weak; pair it with what the audience should do, or accept it as the mark scheme does.

Background Concept

The rate of the light-dependent stage of photosynthesis depends on light intensity until some other factor (usually CO₂ concentration or temperature) becomes limiting. On a graph of photosynthetic rate against light intensity the curve therefore rises (light is limiting) and then flattens (light is saturating and another factor is limiting). The height of the plateau reflects the activity of the dark stage (Calvin cycle) and the capacity of the electron transport chain.

The rate of oxygen production is a direct measure of the rate of the light-dependent stage, because the O₂ released by photolysis of water at photosystem II is stoichiometric with the electrons entering the electron transport chain.

Understanding the Question

The stem describes an investigation in which normal and mutant tobacco plants (mutants have more chlorophyll b) had their rate of O₂ production measured across a range of light intensities, with other variables held constant. Part (b)(i) asks the candidate to describe what the graph shows — meaning the trend, the comparison between the two curves, and at least one numerical comparison taken from the figure. The mark scheme awards 1 mark for the shared shape (rise then plateau), 1 mark for the comparison (mutant > normal), and 1 mark for a paired data quote.

Approach

Read the axes first, then describe the shape common to both curves, then state the difference, then quote a paired value to support the difference. The "describe" command word requires trends, not explanations — explanations of why belong in (b)(ii).

Step-by-Step Reasoning

• Shared shape (1 mark): Both curves start at the origin (no light → no O₂), rise steeply at low light intensities, and then level off at higher light intensities. This is the classic light-saturation curve: at low light, light is the limiting factor, so the rate increases linearly; at higher intensities some other variable (CO₂, temperature) becomes limiting, and the rate plateaus.
• Comparison (1 mark): Throughout the range where light is non-zero, the mutant curve lies above the normal curve. This means the mutant has a higher rate of photosynthesis at any given light intensity. The mutant also plateaus at a higher absolute rate, so the difference is preserved (and slightly magnified) at saturating light.
• Comparative data quote (1 mark): The mark scheme requires a paired value at the same light intensity. Reading the graph at 1000 lux (the right-hand edge), the mutant curve plateaus at approximately 280 mmol mg⁻¹ hour⁻¹ and the normal curve at approximately 165 mmol mg⁻¹ hour⁻¹. Either the absolute difference (~115 mmol mg⁻¹ hour⁻¹) or a ratio (mutant ≈ 1.7× normal) is acceptable. A second example: at 400 lux, the mutant is approximately 200 mmol mg⁻¹ hour⁻¹ and the normal is approximately 110 mmol mg⁻¹ hour⁻¹.

The three points together — shape, comparison, data quote — earn all 3 marks.

Key Takeaways

• A "describe the graph" question always needs three ingredients: trend in each line, comparison between the lines, and a numerical quote.
• "Increases then levels off" identifies a light-saturation response — light is the limiting factor on the rising part, another variable on the plateau.
• The plateau height depends on the dark (light-independent) stage and the chloroplast's capacity, not directly on the amount of pigment — although more pigment can raise the curve on the rising part too.

Common Mistakes

• Describing only one of the two curves and ignoring the other.
• Saying "the rate increases" without mentioning the plateau — this loses the most easily earned mark.
• Offering an explanation (e.g. "because there is more chlorophyll b") in a "describe" question. The mark scheme wants observation, not causation here.
• Quoting a value without units, or quoting values that are not at the same light intensity (the comparison must be paired).

Things to Be Careful About

• Read both axes carefully: x is light intensity in lux; y is rate in mmol mg⁻¹ hour⁻¹. Units must appear in the data quote.
• The two curves are very close at the lowest light intensities, so the comparison "the mutant is always higher" is more accurate for light intensities above about 50 lux.
• Significant figures: two or three significant figures are appropriate when reading off a printed graph.

Background Concept

The two stages of photosynthesis are linked by the products of the light-dependent stage:

• The light-dependent stage (on the thylakoid membranes) uses light energy to photolyse water (releasing O₂) and to generate ATP and reduced NADP.
• The light-independent stage / Calvin cycle (in the stroma) uses ATP and reduced NADP to reduce CO₂, first to triose phosphate (TP), then to hexose sugars and to RuBP (which is regenerated).
• The hexose produced is the raw material for almost every other organic molecule in the plant: through respiration it provides ATP; through biosynthetic pathways it is converted into amino acids, lipids, cellulose, starch and many other compounds.

Chlorophyll b, as an accessory pigment, broadens the wavelengths of light that can be captured. More chlorophyll b means more total light energy absorbed per unit time, which means a faster rate of the light-dependent stage, which means more ATP and reduced NADP are produced per unit time.

Understanding the Question

The stem has established that mutant tobacco plants have more chlorophyll b and grow faster than normal plants. Part (b)(ii) asks the candidate to suggest explanations for the faster growth. The mark scheme allows any four of eight creditable points, which together form a single logical chain from the extra pigment to the observed phenotype.

Approach

Build the answer as a stepwise causal chain. Each step is one mark, and a well-constructed chain will hit four or five marks almost automatically. Start with the pigment → light-absorption link (the question's premise), then move into the light-dependent stage, then into the Calvin cycle, then into the fates of the hexose produced.

Step-by-Step Reasoning

• Step 1 — "absorbs more light": The mutant has more chlorophyll b, an accessory pigment. More pigment molecules → more total light energy captured per unit time. This is the starting point of the chain.
• Step 2 — "faster rate of photosynthesis / light-dependent stage": The captured energy drives the light-dependent stage, so a higher rate of light capture gives a faster light-dependent stage. This is the visible result on Fig. 9.1 (mutant curve higher than normal curve).
• Step 3 — "more ATP and reduced NADP produced": Photophosphorylation and the reduction of NADP⁺ at photosystem I both depend on the light energy delivered to the reaction centres; more captured energy means more ATP and more reduced NADP per unit time. These two products are the bridge between the two stages.
• Step 4 — "more turns of the Calvin cycle": The Calvin cycle requires ATP (for the reduction of GP to TP) and reduced NADP (as the reducing agent); the rate of the cycle is therefore limited by the supply of these two molecules. More ATP and reduced NADP allow more CO₂ to be fixed per unit time.
• Step 5 — "more triose phosphate (TP) produced": TP is the immediate three-carbon product of the Calvin cycle. More CO₂ fixed per unit time means more TP produced per unit time.
• Step 6 — "more hexose for respiration": TP is the substrate for hexose synthesis (two TP → one hexose, e.g. glucose or fructose). More TP → more hexose. Some of this hexose enters glycolysis, the link reaction, the Krebs cycle and oxidative phosphorylation, releasing ATP that powers cell division and growth.
• Step 7 — "more energy for growth": ATP from respiration of the extra hexose fuels the active processes of growth — cell division, active loading of the phloem, synthesis of new macromolecules.
• Step 8 — "more synthesis of amino acids / proteins / lipids / cellulose / starch": Hexose is also a precursor for biosynthesis. Amino acids are made from intermediates of glycolysis and the Krebs cycle combined with nitrogen from the soil; cellulose and starch are polymers of glucose; lipids are built from acetyl-CoA. A larger hexose pool therefore supports faster synthesis of the structural and functional components of new cells and tissues.

Any four of these eight steps will earn full marks, but a complete chain (steps 1 → 3 → 4 → 5 → 7, for example) is the strongest answer and easiest to write under exam pressure.

Key Takeaways

• A "suggest" question with a multi-part mark scheme is usually best answered as a chain — every link is a mark.
• The two stages of photosynthesis are linked specifically by ATP and reduced NADP; these are the two molecules whose increased supply drives a faster Calvin cycle.
• Hexose produced in photosynthesis is not just an energy store; it is the carbon skeleton for almost every other class of biological molecule a plant makes.
• Linking a phenotype (faster growth) back to a pigment (more chlorophyll b) is a typical CIE question structure: the chain is photosynthesis → biosynthesis → growth.

Common Mistakes

• Stopping at "more photosynthesis happens, so the plant grows faster" — this loses most of the marks because it skips the intermediate biochemistry.
• Saying "more glucose is made so the plant has more energy" without distinguishing between glucose as a respiratory substrate (which yields ATP) and glucose as a carbon skeleton (which yields biomass). CIE marks are awarded for both ideas, separately.
• Forgetting to mention ATP and reduced NADP by name — these are the precise terms the mark scheme credits.
• Confusing triose phosphate (TP, a 3-carbon sugar) with hexose (a 6-carbon sugar) or with RuBP. The chain should be GP → TP → hexose → other molecules.
• Saying "the mutant grows faster because it has more chlorophyll" without explaining the role of the extra chlorophyll b in extending the wavelength range or capturing more light.

Things to Be Careful About

• "Suggest" is a softer command word than "explain", but the mark scheme here still requires precise biological terms (ATP, reduced NADP, TP, hexose). Vague answers such as "more energy is produced" do not earn the same marks.
• The maximum mark is 4 even though 8 points are listed — the mark scheme says "any four from". A candidate who lists all 8 should still cap at 4, and an examiner will simply pick the strongest 4.
• Do not claim chlorophyll b is itself used in biosynthesis; it is a pigment, not a substrate.
• "Growth" requires both energy (ATP) and materials (biosynthesis); an answer that mentions only one of the two is incomplete.