Biology 9700/42 — May/June 2025

Background Concept

Carrier 4 in Fig. 1.1 is the last carrier of the electron transport chain — it passes electrons to oxygen, the final electron acceptor, reducing O₂ to water. The carriers of the ETC work in series: each one is reduced by the carrier before it and oxidised by the carrier after it. If any carrier in the chain is blocked, every carrier "upstream" of it cannot be re-oxidised, and the chain grinds to a halt. The reduced coenzymes NADH and FADH₂ also cannot be re-oxidised, so they accumulate. The Krebs cycle and link reaction depend on a supply of oxidised NAD⁺ and FAD, so they too stall. The cell then has no way to re-oxidise NADH and is forced to rely on glycolysis, which must regenerate NAD⁺ anaerobically (to lactate in mammals, to ethanol + CO₂ in yeast and plant cells under hypoxia). ATP yield collapses from ~30 per glucose to 2 per glucose.

Understanding the Question

The stem tells the candidate that cyanide ions (CN⁻) bind to and inactivate carrier 4 — the terminal carrier that normally hands electrons to oxygen. The question is open-ended "suggest and explain", so the candidate must reason out the cascade of consequences, not just state one effect. There are four marks, so at least four distinct, linked ideas are required.

Approach

Work step by step through the system: what does carrier 4 do? what happens when it stops? how does that affect the proton gradient? what does that do to ATP synthesis? what does that do to the supply of oxidised NAD/FAD? what does that do to the Krebs cycle? what does the cell do instead? Each link in the chain is a potential marking point.

Step-by-Step Reasoning

• Step 1 — ETC stops. Carrier 4 normally reduces O₂ to H₂O. Cyanide blocks this, so electrons have nowhere to go. The whole chain becomes reduced (all carriers carry extra electrons).
• Step 2 — No proton pumping. With no electron flow, there is no energy to pump protons. Protons already in the intermembrane space leak back to the matrix, so the proton gradient collapses.
• Step 3 — ATP synthesis halts. Without a proton gradient, no protons pass through ATP synthase, and oxidative phosphorylation stops. ATP can no longer be made at this stage.
• Step 4 — Coenzymes accumulate. Reduced NAD and reduced FAD cannot be re-oxidised because their electrons have no path to oxygen. They build up in the matrix; the supply of NAD⁺ and FAD runs out.
• Step 5 — Krebs cycle and link reaction stop. These two stages release hydrogen atoms to NAD⁺ and FAD. With no oxidised coenzymes available, the dehydrogenations cannot proceed, so both stages halt.
• Step 6 — Glycolysis continues anaerobically. Glycolysis still produces a small amount of ATP by substrate-level phosphorylation, and regenerates NAD⁺ by reducing pyruvate to lactate (in animals) or to ethanol + CO₂ (in yeast and some plant tissues). This is anaerobic respiration.
• Step 7 — Energy crisis. ATP yield falls from ~30 to 2 per glucose. Energy-requiring processes (active transport, biosyntheses, muscle contraction) fail, and cyanide poisoning is rapidly fatal.

Key Takeaways

• The ETC is a serial chain: blocking one carrier halts the whole chain.
• Cyanide is dangerous precisely because it shuts off aerobic ATP production; cells can only fall back on the much-less-efficient anaerobic pathway.
• This is the same principle behind many other respiratory poisons (e.g. carbon monoxide, azide, antimycin A) — they all target a specific carrier in the ETC.

Common Mistakes

• Saying only that "ATP production stops" without explaining the mechanism (loss of proton gradient, etc.).
• Saying "oxygen cannot be used" without linking this to the ETC and the proton gradient.
• Saying "the cell dies because there is no oxygen" — cells die because there is no ATP, not because there is no O₂ per se.
• Failing to mention the build-up of reduced NAD/FAD and the stoppage of the Krebs cycle, which CIE marks credit for.

Things to Be Careful About

• "Suggest and explain" demands both: the candidate should propose a consequence AND link it to the cyanide binding. A bald statement of "ATP production falls" without the link back to the ETC scores nothing.
• CIE does not require the candidate to name "cytochrome oxidase" — the question uses the figure's numbering, so the answer should refer to carrier 4, not to the biochemical name.
• "No / less" wording in the mark scheme means either a complete absence or a reduction is acceptable, depending on the candidate's reasoning.

Background Concept

A selective advantage is any heritable trait that increases an organism's chances of survival or reproduction in its environment, thereby raising its fitness. In plants, common selective advantages of seed structures include:

• aiding seed dispersal (wind, animal attachment, self-burial),
• avoiding predation by herbivores or seed predators,
• improving anchorage of the seed in a suitable germination site,
• enhancing resource acquisition by the germinating seed or seedling.

Awns are long, slender, bristle-like appendages that project from the tip of grass seeds. In wild grasses they are a very common feature, and several functions have been documented.

Understanding the Question

Wild rice commonly has awns on its grains. The question asks for one selective advantage that awns might confer in the wild (compared with having no awn).

Approach

Think about what awns can physically do — they project from the seed, are rigid and bristle-like, and can interact with the environment (wind, soil, animals). Consider how each of these might help the plant survive or reproduce better than a plant lacking awns.

Step-by-Step Reasoning

• Seed dispersal: long projecting awns catch the wind, attach to passing animal fur, or move by hygroscopic twisting along the ground. This moves seeds away from the parent plant, reducing competition with the parent and with siblings and helping to colonise new habitats.
• Protection: the bristly structure physically deters small herbivores and seed predators from eating the grain.
• Anchoring: when seeds fall to the ground, awns can twist with humidity changes and drill or wedge the seed into the soil, improving contact with moist substrate and protecting the seed from being washed or blown away.
• Resource acquisition: awns have a large surface area that may condense dew or absorb water from light rain, providing moisture for germination.

Key Takeaways

• Awns are not vestigial; they have functional value in the wild.
• Selective advantage = anything that improves survival or reproduction.
• Each suggested function links a structural feature of the awn to an ecological outcome.

Common Mistakes

• Vague answers such as "helps the plant survive" — the mark scheme requires a specific mechanism.
• Confusing the wild advantage with the disadvantage in cultivation (storage and processing difficulty).
• Saying the awns "help pollination" — awns are on the seed after fertilisation, so they are not involved in pollination itself.

Things to Be Careful About

• Only one advantage is required (1 mark), but it must be specific.
• The advantage must relate to wild conditions (dispersal, predation, anchoring), not to farming.

Background Concept

In sexually reproducing organisms, body (somatic) cells are diploid ($2n$): each cell contains two sets of chromosomes, one set inherited from the mother and one from the father. The two chromosomes in a pair — one maternal, one paternal — are called homologous chromosomes (or a homologous pair). They carry the same genes at the same loci, although the alleles they carry may differ. They are also similar in size, length, centromere position and banding pattern.

Gametes, in contrast, are haploid ($n$): they contain only one set of chromosomes. Haploid gametes are produced by meiosis. At fertilisation two haploid gametes (one from each parent) fuse to form a zygote, restoring the diploid number.

Understanding the Question

Part (a) asks for an explanation (3 marks) of two related facts about the cell produced by fertilisation:

1. Why it is diploid.
2. Why it contains homologous chromosomes.

You need to give clear, mark-scheme-ready points covering both halves of the question.

Approach

Recognise that the two ideas are linked: diploidy is the number of chromosome sets, while homologous describes the relationship between the two chromosomes in each pair. The explanation therefore needs to cover:

• where the two sets come from (gamete fusion),
• what makes the two chromosomes in a pair 'homologous' (origin and shared features).

The mark scheme accepts any 3 of 4 creditable points — select the ones you can phrase most clearly.

Step-by-Step Reasoning

Point 1 — Source of the two chromosome sets. Fertilisation joins two haploid gametes, so the resulting zygote receives chromosomes from both parents. This is what restores the diploid state characteristic of the species.

Point 2 — Two sets of chromosomes. A diploid cell therefore contains two complete sets of chromosomes — one set from the male gamete and one set from the female gamete.

Point 3 — Maternal/paternal origin of each pair. Within each pair, one chromosome is maternal in origin (came from the egg) and the other is paternal (came from the sperm). This pairing of a maternal and a paternal chromosome is what defines them as homologous.

Point 4 — Shared features of homologues. Homologous chromosomes are described as having the same genes at the same loci and being similar in size/length, centromere position and banding pattern. (This point is sometimes tested separately and is a high-yield definition to learn.)

Any three of these four points is enough to earn full marks; in a strong answer you would give all four.

Key Takeaways

• Diploid = two sets of chromosomes ($2n$); haploid = one set ($n$).
• The zygote is diploid because two haploid gametes fuse at fertilisation.
• A homologous pair = one maternal + one paternal chromosome, carrying the same genes at the same loci.
• Definitions of 'homologous' that mention only 'similar shape' are not enough — the mark scheme requires shared genes/loci or stated similarity of length, centromere position and banding pattern.

Common Mistakes

• Stating only that the cell is 'formed by fertilisation' without mentioning two haploid gametes — this misses the point about chromosome number.
• Describing homologous chromosomes as merely 'similar' without giving any of the required features (same genes/loci, same length, same centromere position, same banding pattern).
• Conflating 'haploid' with 'half the size of the organism' — haploid refers to chromosome number, not cell or organism size.

Things to Be Careful About

The mark scheme lists 'any three' of four creditable points. If you can give all four cleanly, do so — it costs nothing and protects against ambiguity. Use the precise term 'homologous' rather than 'matching', 'paired' or 'similar'.

Background Concept

Meiosis consists of two divisions: meiosis I (the reduction division) and meiosis II (which resembles mitosis). The chromosome number is halved only in meiosis I, when homologous chromosomes (paired up as bivalents during prophase I) are pulled apart and dispatched to opposite poles of the cell. Meiosis II separates sister chromatids and therefore does not further reduce chromosome number — each cell entering meiosis II is already haploid (in chromosome number), although each chromosome still consists of two chromatids.

Understanding the Question

The question asks for the name of the meiotic stage at which reduction division occurs, and a reason for that choice. The answer must combine an identifiable stage with a chromosome-behaviour explanation that justifies the choice.

Approach

Identify the division in meiosis that halves chromosome number, then pinpoint the stage within that division where the actual separation is visible. Then explain why that stage is the reduction-division moment in terms of what is being separated (homologous chromosomes, not sister chromatids).

Step-by-Step Reasoning

• Meiosis I is the reduction division because the chromosome number halves (diploid → haploid) during this division.
• Within meiosis I, the actual separation of the homologues occurs at anaphase 1, when the homologous chromosomes of each bivalent are pulled to opposite poles by the spindle. By telophase 1, the homologues have reached the poles and the cell begins to divide, so telophase 1 is also credited.
• The biological reason this stage reduces chromosome number is that the two members of each homologous pair (one maternal, one paternal) are being separated into different daughter cells, so each daughter cell receives only one of the two homologues of each pair.
• Contrast with meiosis II: here, sister chromatids separate — but each cell already has one chromosome from each homologous pair, so the chromosome number is not reduced further.

Key Takeaways

• Meiosis I = reduction division (chromosome number halves).
• Anaphase 1 / telophase 1 is where the separation of homologous chromosomes is visible.
• The reason it is 'reduction' is that homologues (not sister chromatids) are being separated.
• Meiosis II is NOT a reduction division — even though it superficially looks like mitosis and does separate chromatids, the chromosome number is already $n$ at the start.

Common Mistakes

• Writing 'anaphase' without the '1' — this could refer to anaphase 1 or anaphase 2, which are very different events. Always qualify with 1 or 2.
• Stating 'because the cell divides into two' — this is true of every anaphase/telophase, so it does not earn the mark. The reason must specifically mention homologous chromosomes or bivalents.
• Confusing anaphase 1 (homologues separate) with anaphase 2 (sister chromatids separate).

Things to Be Careful About

The mark scheme accepts either anaphase 1 or telophase 1. Either is fine, but the reason must specifically refer to separation of homologous chromosomes / bivalents, not a generic statement about cell division.

Background Concept

A gene is a length of DNA that codes for a polypeptide (or functional RNA). To express a human protein in a non-human host such as Escherichia coli or yeast, a copy of the human gene must first be obtained in a form that can be inserted into the host's DNA. There are several distinct starting points, but they all converge on the same product: a length of double-stranded DNA containing the coding sequence (without introns, if the host is a prokaryote).

Understanding the Question

Part (a) is an "outline" question worth 4 marks. The mark scheme accepts any four of the listed approaches, so the candidate should aim to give a recognisable version of as many distinct methods as possible. The stem makes clear that recombinant DNA technology needs a gene coding for the human protein, so the answers must each describe a different way of obtaining that gene.

Approach

Think of the question as: "In what different ways can a molecular biologist get hold of a copy of the human gene?" There are essentially three families of method:

1. Look it up in a database of already-known sequences.
2. Build the gene chemically (synthesise it nucleotide by nucleotide).
3. Isolate the gene from biological material — either as DNA directly (using restriction enzymes) or by extracting its mRNA and converting it back to DNA.

The mark scheme rewards any four distinct points drawn from these families, so the candidate should aim to cover as many of them as time allows.

Step-by-Step Reasoning

• Method 1 — Bioinformatics / gene library: If the gene has already been sequenced, the nucleotide (base) sequence can be looked up on a database such as GenBank, or found in a cDNA or genomic gene library. This is the most common starting point in modern molecular biology because the human genome has been sequenced.
• Method 2 — Chemical synthesis: A short gene can be made entirely in vitro by chemically joining nucleotides in the desired order, using a DNA synthesiser. This is how short, defined sequences (e.g. small genes or probes) are made.
• Method 3 — mRNA → cDNA: Cells expressing the gene (e.g. pancreatic β-cells for insulin) contain large amounts of the relevant mRNA. The mRNA is extracted, and the enzyme reverse transcriptase uses it as a template to synthesise a single strand of complementary DNA (cDNA). DNA polymerase then converts this to double-stranded DNA. The advantage of this route is that the resulting cDNA contains no introns, so it can be expressed directly in a prokaryotic host.
• Method 4 — Restriction digestion of genomic DNA: A restriction enzyme is used to cut all the DNA in a sample of human cells at its specific recognition sequence, releasing a mixture of fragments. The fragment containing the gene of interest can then be identified (e.g. by hybridisation with a labelled probe) and isolated.

Key Takeaways

• A gene for genetic engineering can be obtained from a database, by chemical synthesis, by reverse transcription of mRNA, or by cutting it out of genomic DNA with a restriction enzyme.
• The cDNA route is the only one that automatically removes introns — important if the host is a prokaryote that cannot splice.
• All routes produce a double-stranded DNA fragment that can be ligated into a vector.

Common Mistakes

• Confusing reverse transcriptase with DNA polymerase: reverse transcriptase makes DNA from an mRNA template; DNA polymerase makes DNA from a DNA template (or extends a primer).
• Saying only "extract the gene" without naming a method — this is too vague to score.
• Omitting the role of bioinformatics, which is a standard feature of modern gene isolation.

Things to Be Careful About

• The mark scheme is "any four from" — there is no need to give all eight points, but covering as many as possible maximises the chance of a full mark.
• "DNA library" and "gene library" are accepted terms; the candidate should not be put off by either wording.
• "Using restriction enzymes" is the creditable detail; simply "cutting the DNA" is too vague.

Background Concept

Before recombinant DNA technology, insulin for treating diabetes mellitus was extracted from the pancreases of pigs and cattle slaughtered for food. The amino-acid sequence of pig and cow insulin is similar to, but not identical to, that of human insulin (e.g. pig insulin differs by one amino acid). This causes clinical problems: the foreign protein can provoke an immune response, supplies were limited by the slaughter industry, and many patients (e.g. vegetarians, vegans, or members of certain religions) found the source unacceptable. Recombinant human insulin is made by inserting the human insulin gene into a microorganism (originally Escherichia coli, now often yeast) and fermenting it on an industrial scale; the protein product has the exact amino-acid sequence of human insulin.

Understanding the Question

Part (b) is a 3-mark "explain" question asking for advantages of recombinant human insulin over animal insulin. The mark scheme explicitly accepts "or reverse argument" (ORA) for animal insulin, so the candidate can frame each point either as a benefit of recombinant insulin or as a drawback of animal insulin. Any three of the listed points will score full marks.

Approach

Think about every practical and clinical difference between treating a patient with a foreign animal protein and treating them with a human-identical protein produced in fermenters. Categories to consider:

• Supply — how much is available, at what cost.
• Immunology — what does the patient's immune system do to it.
• Efficacy — how well it works and how quickly.
• Safety — what other biological material is introduced.
• Ethics — which patients find the source acceptable.

Step-by-Step Reasoning

• Supply and demand: Industrial fermentation of recombinant microorganisms can be scaled up to produce essentially unlimited insulin. Animal-derived insulin depended on the slaughter industry and was vulnerable to shortages and price fluctuations.
• Immune response: Because recombinant human insulin has the identical amino-acid sequence to human insulin, the patient's immune system is much less likely to recognise it as foreign. Pig and cow insulin, with their small sequence differences, sometimes triggered allergic responses or other side-effects.
• Speed of action and dose: Human-identical insulin binds human insulin receptors more efficiently, so it can be faster-acting and may be effective at a smaller dose.
• Tolerance: Long-term exposure to a slightly foreign protein (animal insulin) can lead to the patient developing tolerance — declining responsiveness to the same dose. Recombinant human insulin avoids this.
• Infection risk: Animal tissue carries a small risk of transmitting pathogens (e.g. prions, viruses). Recombinant insulin produced in a defined fermenter is much purer and safer.
• Ethics and religion: Insulin from pigs is unacceptable to Muslims and Jews, and to vegetarians and vegans. Recombinant insulin made by microorganisms has none of these objections.

Key Takeaways

• Recombinant human insulin is identical in sequence to endogenous human insulin, so it is better recognised, less immunogenic, and acts more reliably.
• Microbial fermentation provides a virtually unlimited, consistent supply independent of the meat industry.
• It avoids the ethical and religious objections associated with pig- and cow-derived insulin.

Common Mistakes

• Saying only "it is cheaper" or "it is purer" without elaborating on the clinical or supply consequences.
• Treating "no allergic response" as a single point but then failing to give two further distinct advantages — the mark scheme is "any three from", so breadth is needed.
• Forgetting that "faster to act" and "smaller dose" are both clinical efficacy points and are credited separately.

Things to Be Careful About

• The command word is "explain", not "state". Each point should make the advantage (and ideally why it arises) explicit, not just assert it.
• ORA (or reverse argument) means that phrasing each point as a drawback of animal insulin is also acceptable.
• The candidate does not need to cover all six points — three clear, distinct points earn full marks.

Background Concept

The word recombinant literally means "combined again". In molecular biology, recombinant DNA is any DNA molecule that has been formed by joining together DNA from two different sources. The classic example is a piece of human DNA (e.g. the insulin gene) inserted into a bacterial plasmid; the resulting plasmid is a recombinant DNA molecule because it contains DNA from two different species.

Understanding the Question

Part (c) is a 1-mark "explain" question. The candidate is being asked to justify the name "recombinant" DNA. One clear sentence that names both the two sources and the joining is sufficient.

Approach

Break the word "recombinant" down: it comes from the verb "to recombine" — to combine again or in a new way. DNA from two different origins has been combined into a single molecule, hence "recombinant". In the context of producing human insulin, the two sources are: (i) the human gene (human DNA), and (ii) the plasmid / bacterial DNA used as the vector.

Step-by-Step Reasoning

• The DNA used in the production of recombinant human insulin is not solely human and not solely bacterial — it is a hybrid.
• Specifically, the human insulin gene has been joined to a bacterial plasmid (or to other bacterial DNA used as a vector).
• Because it is a combination of DNA from two different sources, it is termed "recombinant".

Key Takeaways

• "Recombinant DNA" = DNA from two different sources joined together.
• The two sources in this question are: (a) the human insulin gene and (b) the bacterial plasmid / vector.
• This definition applies to all recombinant DNA molecules, not just insulin.

Common Mistakes

• Saying only "it is a combination of DNA" — this misses the "from two different sources" qualification that the mark scheme requires.
• Naming only one source (e.g. "DNA from bacteria") and not making it clear that the other source is human.
• Confusing "recombinant DNA" with "DNA replication" — replication produces identical copies, not combinations of different sources.

Things to Be Careful About

• This is a one-mark question demanding one precise point, not an essay.
• The mark scheme accepts either of the two phrasings: "DNA from two different sources joined together" OR "DNA from the human gene joined to DNA from bacteria / plasmid".

Background Concept

An insulin analogue is a modified version of human insulin with one or a few amino acids changed. These small changes are engineered to alter the protein's properties — for example, making it faster-acting (lispro, aspart) or longer-acting (glargine, detemir) — so that patients can better control their blood glucose around meals or overnight. Because the change is precisely defined at the protein level, the corresponding change must be made precisely at the DNA level. The genetic code is read in non-overlapping triplets of nucleotides (codons), each specifying one amino acid; a single nucleotide change can therefore change one codon, and therefore one amino acid, in the final protein.

Gene editing refers to a set of techniques that allow scientists to make specific, targeted changes to a DNA sequence at a chosen location. The most widely used is CRISPR/Cas9, in which a short guide RNA directs the Cas9 nuclease to a specific DNA sequence, where it makes a double-strand break. The cell's own repair machinery then either joins the broken ends (often introducing small insertions or deletions) or, if a template DNA is supplied, uses that template to repair the break, allowing precise nucleotide replacement.

Understanding the Question

Part (d) is a 4-mark "suggest and explain" question. The stem tells the candidate that scientists have already decided what amino acid changes are needed to make the analogue; the question is how they then construct a synthetic gene encoding that analogue by making changes to a length of DNA, and how those DNA changes give rise to the correct protein. The candidate must therefore cover both the technique (gene editing) and the mechanism (codon → amino acid → protein).

Approach

Approach the question in two stages:

1. How is the gene obtained and edited? Look up the normal human insulin gene sequence, then use a gene-editing tool (e.g. CRISPR/Cas9) to insert, delete or replace nucleotides at defined points in the sequence.
2. How does the DNA change produce the analogue? Because the genetic code is read three bases at a time, changing a nucleotide alters a specific codon; the altered codon now codes for a different amino acid; the ribosome translates the modified mRNA into the desired insulin analogue.

Step-by-Step Reasoning

• Step 1 — Obtain the starting sequence: Find the nucleotide / base sequence of the human insulin gene. This information is already in a database (e.g. GenBank) or a gene library.
• Step 2 — Make the targeted change: Use gene editing (e.g. CRISPR/Cas9) to introduce an insertion, deletion or replacement of nucleotides at the specific sites in the gene that the scientists have decided to alter. The guide RNA targets the nuclease to the precise location; a supplied DNA template carries the new sequence for a replacement.
• Step 3 — From DNA to protein: The change in nucleotide sequence changes the DNA triplet (codon) at that point. Because each codon specifies one amino acid, the new codon codes for a different amino acid.
• Step 4 — The final product: When the modified gene is expressed in Escherichia coli, the ribosome translates the modified mRNA into the desired insulin analogue, with the engineered amino acid substitution(s).
• Step 5 — AVP: Credit can be earned for any further valid detail, such as a specific feature of CRISPR/Cas9 (e.g. the role of the guide RNA, the protospacer adjacent motif, or the use of a repair template for precise replacement).

Key Takeaways

• A synthetic gene is built by starting from a known sequence and using gene editing to introduce precise nucleotide changes.
• CRISPR/Cas9 is the current method of choice for targeted gene editing because of its specificity and ease of use.
• A single nucleotide change can change a single codon, and therefore a single amino acid, in the final protein — this is the molecular basis of insulin analogue design.
• The link gene → mRNA codon → amino acid → protein structure is the central explanatory chain for this part.

Common Mistakes

• Describing gene editing in vague terms without saying what is being done at the DNA level (insertion, deletion or replacement of nucleotides).
• Failing to make the codon → amino acid link, which is the central "explain" element of the question.
• Confusing gene editing (changing an existing sequence at a specific site) with restriction enzyme cutting (which is not site-specific in the same way).
• Saying that the bacteria "learn" the new sequence, or any other teleological language — bacteria do not learn; they are transformed with a plasmid carrying the modified gene.

Things to Be Careful About

• The mark scheme is "any four from", so breadth and accuracy across the chain of reasoning is what earns marks, not a single long paragraph.
• The question is about a synthetic gene for an analogue, not about a new gene from another species, so the appropriate technique is gene editing, not cDNA synthesis or database retrieval alone.
• AVP allows credit for relevant detail of the editing tool (e.g. guide RNA in CRISPR/Cas9), which is good use of the fourth mark.

Background Concept

Guard cells are highly specialised epidermal cells that flank each stoma and control its aperture. Unlike the other epidermal cells that lie flat and tightly packed around them, guard cells have a distinctive kidney (or dumb-bell in grasses) shape, with unevenly thickened walls. Their structure is the structural basis for the way they open and close the pore: when the guard cell takes in water and becomes turgid, the thinner outer wall stretches more than the thickened inner wall, pulling the two guard cells apart and bending them so that the pore opens. Because stomata control the balance between CO₂ uptake for photosynthesis and water loss by transpiration, the special features of guard cells reflect both their ion-pumping role and their photosynthetic activity.

Understanding the Question

This is a 4-mark "describe" question asking for the structural features of guard cells. The mark scheme credits any four distinct points, so the candidate should aim to give as many precise, observable features as possible from memory. "Describe" requires factual statements of structure (and any implied function is acceptable only where the mark scheme makes it a creditable point).

Approach

Read the mark scheme as a menu of acceptable points: variable wall thickness; absence of plasmodesmata; abundance of chloroplasts/mitochondria; chloroplasts with few grana; mitochondria with many cristae; folded membrane with many transport proteins; cellulose microfibrils in radial bands. Choose the ones you can phrase most precisely.

Step-by-Step Reasoning

1. Variable wall thickness – The wall on the side of the stoma (ventral wall) is much thicker than the wall on the outside (dorsal wall). This is the key feature that allows the asymmetric bending that opens the pore. Stating "thicker on the inner side" is more precise than just "thick wall".
2. No plasmodesmata – Guard cells are isolated from neighbouring epidermal cells; ions and water must cross the cell surface membrane rather than moving through plasmodesmata. This is why active transport across the plasma membrane is so central to stomatal movement.
3. Many chloroplasts – Unlike most epidermal cells, guard cells photosynthesise actively. Photosynthesis provides ATP for the H⁺/K⁺ pumps and contributes to the blue-light response that opens stomata.
4. Chloroplasts with few grana – Guard cell chloroplasts have less well-developed thylakoid stacks (few grana), reflecting that photosynthesis here is geared to producing energy locally rather than running the full Calvin cycle.
5. Mitochondria with many cristae – A high surface area of inner mitochondrial membrane supports the large ATP demand of the H⁺ pumps that drive ion accumulation.
6. Folded / transport-protein-rich membrane – The plasma membrane is thrown into folds or is densely populated with transport proteins (H⁺-ATPases, K⁺ channels, anion channels), again to support rapid ion fluxes.
7. Cellulose microfibrils in bands – The cellulose microfibrils run radially (like belts) around the cell, so when turgor increases the cell lengthens but cannot expand sideways; this forces the two guard cells to bow apart.

Key Takeaways

• Guard cells combine photosynthetic (chloroplasts), energy-generating (mitochondria) and ion-pumping (folded, transporter-rich membrane) features.
• The unique mechanical behaviour that opens the stoma comes from the unevenly thickened wall combined with radially banded cellulose microfibrils.
• Guard cells are symplastically isolated — every solute and water movement crosses the plasma membrane.

Common Mistakes

• Saying only "kidney-shaped" or "bean-shaped" without giving the reason (variable wall thickness / radial cellulose) — these are worth marks but should not be the only points.
• Confusing guard cell chloroplasts with mesophyll chloroplasts — guard cell chloroplasts have few grana, not many.
• Forgetting to mention plasmodesmata are absent; some candidates say guard cells have plasmodesmata because they "connect" with each other.
• Listing generic epidermal features (e.g. "no chloroplasts") that are true of other epidermal cells but not of guard cells.

Things to Be Careful About

• The mark scheme allows any four, so the safest strategy is to write 4–5 different points rather than spending time elaborating one.
• Avoid vague phrasing such as "have a thick wall" — the precise description is unevenly thickened (thicker on the inside).
• Mentioning function is acceptable only where the mark scheme makes the structural point sufficient; do not pad with statements about "opening the stoma" unless paired with the structural fact.

Background Concept

Stomatal closure in response to drought is driven by the plant hormone abscisic acid (ABA). When the plant senses water stress (e.g. via root-sensed drying soil), ABA is synthesised and released, particularly from the roots and the mesophyll. ABA binds to receptors on the guard cell plasma membrane, triggering a chain of ion movements: H⁺ pumping stops, Ca²⁺ enters the cytoplasm, K⁺ (and accompanying anions) leave the guard cells, the solute concentration falls, water potential rises (becomes less negative), water leaves by osmosis, the guard cells become flaccid, and the stoma closes. This sequence converts an external stress signal into a mechanical change in guard cell shape.

Understanding the Question

Table 6.1 lists ten events labelled A–J in random order. The candidate must place nine of them (positions 2, 3, 4, 6, 7, 8, 9 are blank) in the correct sequence between D (position 1) and C (position 10). The mark scheme splits the task into two halves: events between D and F (the ABA signalling arm) and events between F and C (the ion/water/mechanical arm). To score 4 marks, both halves must be in the correct internal order.

Approach

First think of the pathway as four linked phases:

1. Stress signal and hormone release (D, G, with E for ABA release).
2. Membrane and calcium signalling (B = H⁺ pump stops; F = Ca²⁺ entry — F is fixed in position 5).
3. Ion efflux and osmotic consequences (J = K⁺ leaves; I = water potential rises; A = water leaves by osmosis).
4. Mechanical outcome (H = flaccid; C = stoma closes — C is fixed in position 10).

Use the fixed anchors (D at 1, F at 5, C at 10) to anchor each half.

Step-by-Step Reasoning

First half (between D and F):

• D = stress stimulus (water deficit).
• E = the plant releases ABA in response to the stress.
• G = ABA binds to receptors on the guard cell plasma membrane.
• B = active transport of H⁺ out of guard cells stops (because ABA signalling switches off the H⁺-ATPase).
• F (position 5, given) = Ca²⁺ moves into the cytoplasm.
  This gives the order: D → E → G → B → F.

Second half (between F and C):

• F = Ca²⁺ entry triggers downstream events.
• J = K⁺ leaves the guard cells (down their electrochemical gradient through Ca²⁺-activated anion channels).
• I = water potential of the guard cells becomes less negative (increases) because the solute concentration has dropped.
• A = water leaves the guard cells by osmosis, down the water potential gradient.
• H = the guard cells become flaccid (lose turgor).
• C (position 10, given) = the stoma closes because the flaccid guard cells no longer push it open.
  This gives the order: F → J → I → A → H → C.

Mark-scheme check:

• "E G B between D and F" — correct (E, G, B are placed in that order between D and F).
• "E G B in correct order throughout table" — correct.
• "J I A H between F and C" — correct (J, I, A, H are placed in that order between F and C).
• "J I A H in correct order throughout table" — correct.
  All four marking points earned.

Key Takeaways

• ABA is the central hormonal trigger that closes stomata during water stress.
• The closure pathway has two clearly distinct phases: a signalling phase (ABA → receptors → H⁺ pump off → Ca²⁺ entry) and a mechanical phase (K⁺ efflux → water potential rises → osmotic water loss → flaccidity → closure).
• "Water potential increases" means it becomes less negative (closer to zero), which is the same as the solute concentration falling.
• The mark scheme rewards ordering in each half, so even if you swap one step within a half you lose all marks for that half.

Common Mistakes

• Putting J (K⁺ leaves) before B (H⁺ pump off). K⁺ efflux is downstream of the H⁺ pump switching off, not upstream.
• Putting I (water potential increases) before J (K⁺ leaves). The ion efflux is the cause; the water potential change is the consequence.
• Putting H (flaccid) before A (water leaves by osmosis). Flaccidity follows water loss; you cannot be flaccid before losing the water.
• Confusing "water potential increases" with "becomes more negative". In the strict physical sense, an increase in Ψ (towards 0) corresponds to a fall in solute concentration; this is the direction that drives water out of the guard cell.
• Swapping E (ABA release) and G (ABA binds to receptors): release must precede binding.

Things to Be Careful About

• The question tests order, not detailed reasoning; the safest strategy is to memorise the canonical sequence.
• "Water potential of the guard cells increases" = it becomes less negative (less solute, more dilute sap).
• "Flaccid" is the correct mechanical term; do not write "shrink" or "collapse".
• Anchor on the fixed events (D, F, C) to reduce the chance of transposition errors.

Background Concept

The resting potential is the electrical potential difference across the membrane of a neurone when it is not firing an action potential. The inside of the axon is about 60–70 mV more negative than the outside, typically quoted as −70 mV. This potential is not produced by the membrane in isolation; it is the combined result of two factors: an ion concentration gradient set up by active transport, and a differential permeability of the membrane to those ions.

Key structures involved:

• Sodium–potassium (Na⁺/K⁺) pumps — carrier proteins in the axon membrane that hydrolyse ATP to move 3 Na⁺ out of the axon and 2 K⁺ in, against their respective concentration gradients.
• Voltage-gated and leak channels — transmembrane proteins that allow ions to diffuse down their gradients. At rest the membrane is much more permeable to K⁺ than to Na⁺ because more K⁺ leak channels are open.
• Fixed anions — large negatively charged molecules (e.g. proteins, organic phosphates) trapped inside the axon that cannot cross the membrane; they contribute directly to the net negative charge of the cytoplasm.

The key idea is that the Na⁺/K⁺ pump creates the gradients, and the differential permeability (mainly K⁺ efflux through leak channels) turns those gradients into a measurable potential difference.

Understanding the Question

Part (a) is a 6-mark "describe and explain" question about the resting potential. The command word describe requires the candidate to state what is present (the structures and processes), while explain requires the candidate to say why those structures produce a negative resting potential. Six marks means six creditable points are needed.

Approach

Build the answer in two layers:

1. Active transport layer — name the Na⁺/K⁺ pump, the ions it moves, the ratio (3 Na⁺ out : 2 K⁺ in), and the fact that it uses ATP. This generates the concentration gradients.
2. Diffusion / permeability layer — note that the membrane is more permeable to K⁺ at rest, so K⁺ diffuses out, the inside becomes more negative than the outside, and the result is around −70 mV. Mention the fixed anions as a contributor to the net negative interior.

Step-by-Step Reasoning

1. The Na⁺/K⁺ pump is a carrier protein in the axon membrane. It is the only structure that can move ions against their concentration gradients, and it does so using energy from ATP hydrolysis (active transport).
2. For each cycle it exports 3 Na⁺ out of the axon and imports 2 K⁺. This unequal movement means the pump alone makes the outside slightly more positive than the inside — but the bigger effect is that it builds up very high extracellular Na⁺ concentration and very high intracellular K⁺ concentration.
3. These gradients are electrochemical gradients: there is both a concentration difference and a charge difference driving each ion across the membrane.
4. At rest the membrane is much more permeable to K⁺ than to Na⁺, because many more K⁺ leak channels are open than Na⁺ leak channels.
5. K⁺ therefore diffuses out of the axon down its concentration gradient, carrying positive charge out.
6. Because more positive charge leaves (K⁺) than enters (a small amount of Na⁺), the inside becomes negative relative to the outside.
7. Fixed anions inside the cell (negatively charged proteins and organic phosphates that cannot cross the membrane) attract K⁺ back in, but the pump keeps replacing the K⁺ that has leaked out — so the negative interior is maintained dynamically.
8. The combined result is a stable resting potential of about −60 mV to −70 mV.

Key Takeaways

• The Na⁺/K⁺ pump creates the concentration gradients; differential permeability (mostly K⁺ efflux) converts them into a potential difference.
• The 3:2 stoichiometry of the pump contributes directly to a net loss of positive charge from the axon.
• The membrane is never completely impermeable to Na⁺; a small inward Na⁺ leak continually opposes the resting potential, which is why the pump must keep working.
• Fixed anions inside the axon are essential to explaining why the inside is negative.

Common Mistakes

• Saying "sodium ions diffuse in" instead of "sodium–potassium pump moves sodium ions out". Diffusion alone cannot maintain the gradient against leakage.
• Omitting the fact that ATP is required (active transport).
• Forgetting the 3 Na⁺ : 2 K⁺ ratio; this ratio is one of the marking points and explains the net loss of positive charge from the cell.
• Stating "membrane is impermeable to sodium" — it is not; it is less permeable to Na⁺ than to K⁺.
• Confusing the direction of ion movement during the resting state with the action potential (Na⁺ in / K⁺ out happens during the action potential, not at rest).

Things to Be Careful About

• "Describe and explain" demands both the what and the why. A list of structures without an explanation of how they generate the −70 mV will lose marks.
• Use the term electrochemical gradient (not just "concentration gradient") when referring to the combined electrical and chemical driving forces.
• The phrase "more K⁺ moves out than Na⁺ moves in" is the standard way to express the differential-permeability argument; it is one of the explicit marking points.

Background Concept

The five (now often six) kingdom classification places organisms into broad groups based on fundamental features such as cell type (prokaryotic vs eukaryotic), mode of nutrition, body organisation and cell wall composition. The kingdom Fungi is one of these groups and includes moulds, yeasts, mushrooms, rusts and the fungal partner of lichens.

Key features that unite members of Fungi:

• They are eukaryotes, so their cells have a true nucleus and membrane-bound organelles.
• They lack chlorophyll and chloroplasts, so they cannot make their own food. Instead they obtain nutrients by heterotrophic means: secreting enzymes externally and absorbing soluble products (saprophytic nutrition on dead matter) or feeding on living hosts (parasitic nutrition).
• Their cell walls are made of chitin (the same polymer found in insect exoskeletons), not cellulose as in plants or peptidoglycan as in bacteria.
• The fungal body is usually a network of branching filaments called hyphae, collectively forming a mycelium. Hyphae may be divided by septa or be coenocytic, and individual cells can be multinucleate.
• Fungi reproduce by means of spores, produced sexually or asexually.
• Although most fungi are multicellular, some (such as Saccharomyces) are unicellular.

Understanding the Question

This is a "state/describe" type question worth 4 marks. The candidate must recall and list characteristic features of the kingdom Fungi. There is no single correct list — any four valid features from the accepted pool earn the four marks.

Approach

Read the question's command word ("outline") and select the four most distinctive features that, taken together, mark fungi apart from plants, animals and protoctists. A strong answer contrasts fungi with plants (no chloroplasts, chitin not cellulose, heterotrophic not autotrophic) and notes their structural organisation (hyphae/mycelium, spores, possibly multinucleate or unicellular).

Step-by-Step Reasoning

1. Eukaryotic — distinguishes fungi from bacteria and from the prokaryotic cell type. Many candidates lose this mark by saying "cell has a nucleus" without the word eukaryotic, which is the precise term the mark scheme wants.
2. No chlorophyll / no chloroplasts — places fungi firmly in the heterotrophic, not autotrophic, group. It explains why the fungal partner in a lichen must rely on the alga for photosynthate.
3. Heterotrophic / saprophytic / parasitic — describes the mode of nutrition. A specific example of how the fungus feeds (e.g. secreting extracellular enzymes) is creditworthy but the umbrella term is enough.
4. Cell wall of chitin — diagnostic chemical feature. Avoid the common error of saying "cellulose" (that is a plant feature).
5. Reproduce by spores — a reproductive feature, often paired with hyphae/mycelium.
6. Hyphae / mycelium — describes the body organisation. Many exam answers mention mycelium but forget that it is made of hyphae; the mark scheme wants at least one of the two terms.
7. Multinucleate cells — a finer structural point, often missed. Many fungal hyphae are coenocytic, with several nuclei sharing a common cytoplasm.
8. Unicellular or multicellular — covers both the yeast-like forms and the filamentous moulds/mushrooms.

Key Takeaways

• Fungi are eukaryotes with chitin cell walls, no chlorophyll, and heterotrophic (saprophytic or parasitic) nutrition.
• Their body is a mycelium of hyphae; they reproduce by spores.
• Knowing the contrasting features of Fungi, Plantae, Protoctista and Animalia is essential for taxonomic questions.

Common Mistakes

• Writing "cell wall of cellulose" — that is the plant kingdom feature, not fungi.
• Saying "they photosynthesise" — fungi do not; they lack chlorophyll.
• Calling them "prokaryotes" because they are small and microscopic — they are eukaryotes with a true nucleus.
• Confusing hyphae with rhizoids (a plant root-hair-like structure).
• Listing only three features and trying to make the fourth by repeating an idea in different words — the mark scheme requires four distinct points.

Things to Be Careful About

• Use precise terminology: eukaryotic, chitin, saprophytic, hyphae, mycelium, coenocytic / multinucleate.
• The mark scheme accepts any four features, so choose the ones you can express most accurately; do not pad a weak list with imprecise statements.

Background Concept

A mutualistic (or mutual) symbiosis is a close, long-term association between two different species in which both partners benefit. A lichen is the textbook example: a fungus (most often an ascomycete) and a photosynthetic partner — the photobiont — here the green alga Trebouxia sp.

The fungus contributes:

• A body of tightly packed hyphae that gives the lichen shape, mechanical support, and the ability to cling to substrates such as bark or rock.
• A means of absorbing and retaining water and dissolved mineral ions from rainwater, dew and the substrate.
• Protection from desiccation, intense light and grazing.
• A supply of CO₂ released by its own respiration (and trapped in the thallus), which the alga can use in photosynthesis.

The alga (photobiont) contributes:

• Photosynthate — sugars, amino acids and other organic compounds produced by photosynthesis. Because the fungus cannot make its own food, this is its main carbon (and energy) source. The alga also releases O₂, which the fungus needs for aerobic respiration.

Each partner therefore occupies a niche that the other cannot fill alone: the alga provides fixed carbon, the fungus provides the structure and the inorganic supply.

Understanding the Question

This is a "suggest" question worth 2 marks — one mark for what the fungus gains, one for what the alga gains. The verb suggest means the examiner will accept any biologically sensible reciprocal benefit drawn from the structure of a lichen.

Approach

Think of the lichen as two organisms, each with a "deficit" the other can fill:

• The fungus cannot make its own organic food ⇒ the alga supplies it.
• The alga is small and has no means of attachment or of gathering water/minerals ⇒ the fungus supplies it.
  The two marks correspond to one benefit for each partner.

Step-by-Step Reasoning

1. Fungus gains organic compounds / nutrients from the alga. This is the principal benefit, because the fungus is heterotrophic and cannot photosynthesise. The mark scheme also accepts "food" or "oxygen" (the alga releases O₂ during photosynthesis, which the fungus uses in respiration).
2. Alga gains support, water, minerals and/or CO₂ from the fungus. The fungal hyphae absorb water and dissolved ions from the environment and hold the alga against the substrate; the alga uses these for photosynthesis. CO₂ released by fungal respiration is recycled to the alga. "Protection" from desiccation and grazing is also creditworthy.

Key Takeaways

• Lichens are a mutualism: the fungus provides structure, water and minerals; the alga provides organic carbon (and O₂).
• "Suggest" means the marking is generous — any reasonable reciprocal benefit is accepted.
• The two partners occupy complementary niches and could not survive in the same habitat on their own as readily.

Common Mistakes

• Giving only one benefit, e.g. just what the fungus gains — that scores one mark, not two.
• Writing the benefits in the wrong direction (e.g. saying the alga gets food from the fungus) — the mark scheme is direction-specific.
• Using vague terms like "the fungus helps the alga" without naming what is exchanged. Always be specific: organic compounds, water, CO₂.
• Saying the alga "gets protection from the fungus" alone — the alga also needs inorganic resources, not just safety.

Things to Be Careful About

• Each mark is tied to a specific direction of benefit. Be sure the answer clearly attributes one benefit to the fungus and one to the alga.
• The mark scheme is generous (any one of several options for each side scores the mark), so pick the precise terms you know are correct: organic compounds / food for the fungus, water / minerals / CO₂ / support for the alga.

Background Concept

Biodiversity describes the variety of living organisms in an area. It has three components: genetic diversity, species diversity and ecosystem diversity. For a single habitat (a field), the relevant measure is species diversity — a combination of the number of different species (species richness) and how evenly individuals are spread across those species (evenness). Sampling must therefore record both how many species are present and roughly how abundant each is, and the sample must be representative of the whole area, not biased by where the investigator chooses to look.

A frame quadrat is a square frame, usually $0.25\ \text{m}^2$ or $1\ \text{m}^2$, that delimits a known area of ground. By counting/recording everything inside the frame, the investigator gets a quantitative value (abundance, percentage cover, ACFOR scale, or frequency) that can be compared between places or times. Combining many random quadrats across the field gives an unbiased estimate of biodiversity, which can be expressed as species density, species frequency, or as a diversity index such as Simpson's Index of Diversity (D):

where $n$ is the number of individuals of each species and $N$ is the total number of individuals of all species.

Understanding the Question

The command word outline requires a brief, structured account — not a full lab protocol. The question is testing whether the candidate knows the practical steps that make a quadrat survey valid: standardising the area, placing the frame randomly, identifying what's inside it with a key, recording abundance/cover, repeating to get a large sample, and processing the data into a diversity measure. Any four from the mark scheme's list of seven points will earn full marks.

Approach

Select the most important features of a valid quadrat survey that are NOT obvious from the term "quadrat" alone. Mentioning the standard area, random placement, use of a key, recording species counts/cover, and repeats/large sample size captures all of the marks-scheme's first six points; the seventh (Simpson's) can be added as the analysis step. The order should reflect a logical sequence: choose a frame → place it randomly → identify and count what's inside → repeat → analyse.

Step-by-Step Reasoning

1. Standardised area of the frame — All quadrats placed must be the same size so that density/abundance values are comparable. The mark scheme specifically wants the idea of a "known/same/standard area/size".
2. Random placement — Lay out a grid over the field, generate pairs of random coordinates (e.g. with a random number generator or random-number tables) and place the frame at those coordinates. This removes observer bias (the tendency to place the quadrat where plants look interesting). Setting out a grid is the mark-scheme-accepted alternative when coordinate pairs are generated.
3. Identify with a key — A dichotomous or taxonomic key is used to identify each plant species found inside the quadrat; without it, the candidate cannot convert observations into species data.
4. Record the response variable — Either the number of individuals of each species (abundance) or the percentage cover of each species. ACFOR (Abundant, Common, Frequent, Occasional, Rare) and the Braun-Blanquet scale are accepted alternatives.
5. Repeat / large sample size — A single quadrat is not enough; the frame is placed many times in randomly chosen positions so the mean values are reliable. The mark scheme's "large sample size / repeats" point is satisfied here.
6. Calculate a biodiversity measure — The combined results are used to calculate a meaningful number: species density (number of species per unit area) or Simpson's Index of Biodiversity (D) (which combines richness and evenness).

Key Takeaways

• A frame quadrat gives a quantitative measure of biodiversity only when it is of standard area, placed randomly, and used many times.
• Recording should capture BOTH which species and how many of each (richness + evenness).
• A key is essential for species identification; without it, the data have no biological meaning.
• The raw data are turned into a meaningful value (Simpson's D, species density) only after sufficient repeats.

Common Mistakes

• Forgetting the random placement — stating simply "place the quadrat on the ground" without explaining where and how chosen fails to remove observer bias and loses a mark.
• Recording presence/absence only — noting "which species are present" without a quantitative measure (count, cover, ACFOR) means evenness is lost and Simpson's index cannot be calculated.
• No mention of repeats — a single quadrat is a sample of $0.25\ \text{m}^2$, not of "the field"; without repeats the survey is not representative.
• Forgetting the key — naming species from memory is not acceptable practice and the mark scheme explicitly rewards the use of a key.
• Confusing Simpson's Index with Simpson's Index of Diversity — candidates often quote the raw $\sum (n/N)^2$ form rather than the $1 - \sum (n/N)^2$ diversity form.

Things to Be Careful About

• "Outline" demands concise, structured points — not a long prose paragraph.
• The mark scheme uses ";" between alternatives within a single point (e.g. "abundance / percentage cover / ACFOR / Braun-Blanquet"), so any one of these is acceptable.
• "Large sample size" and "repeats" are the same mark-scheme point — do not double-count them as two separate marks.
• When stating the area, "known" or "standard" or "same" is acceptable; do not write "big" or "appropriate".

Background Concept

Aerobic respiration in mammals — glycolysis, the link reaction, the Krebs cycle and oxidative phosphorylation — uses oxygen as the final electron acceptor in the electron transport chain and produces carbon dioxide and water. When oxygen is in short supply, cells can still run glycolysis (which does not need oxygen) but the link reaction, Krebs cycle and electron transport chain cannot operate. To keep glycolysis running, the pyruvate produced must be dealt with: it accepts hydrogen from reduced NAD, regenerating NAD so that glycolysis can continue, and is itself converted to lactate. This is anaerobic respiration in mammals:

Aerobic respiration releases far more ATP per glucose than glycolysis alone, so anaerobic respiration is a short-term emergency system. The build-up of lactate in active muscles is associated with the oxygen debt (also called excess post-exercise oxygen consumption, EPOC): after strenuous activity, the animal continues to breathe heavily to oxidise the lactate (mostly by the liver, via the Cori cycle — lactate → pyruvate → glucose) and to restore ATP, phosphocreatine and oxygen stores (haemoglobin and myoglobin).

Marine mammals such as grey seals Halichoerus grypus are exceptional divers. While underwater, they cannot breathe, and circulation is partly restricted to essential organs (the dive response / bradycardia), so working muscles become hypoxic and rely on anaerobic respiration, generating lactate. Crucially, this lactate is largely trapped in the muscles and is not flushed into the general circulation until the seal surfaces and normal circulation is restored — which is why the blood lactate peak on a graph usually appears after the dive, not during it.

Understanding the Question

Fig. 10.2 plots blood lactate concentration against time. Before the dive, the concentration is low and stable at about $2\ \text{mmol dm}^{-3}$. During the labelled dive (35–45 min) the blood lactate is still low (the seal is using anaerobic respiration inside the muscles, but circulation to muscles is restricted). When the dive ends at 45 min, blood lactate spikes sharply to about $25\ \text{mmol dm}^{-3}$ by ~48 min — this is the lactate being flushed out of the previously oxygen-starved muscles into the general circulation. After the peak, the concentration falls steadily back towards the resting value by 80 min.

The question asks for reasons for these changes during and after the dive, in terms of the biochemistry of lactate production and removal. It is a 4-mark "suggest" question, so the answer should pick out the four most important creditable points.

Approach

• Recognise the two phases the mark scheme is testing: lactate production (during / as a result of the dive) and lactate removal (after the dive).
• For production, link "no oxygen" to "anaerobic respiration" and to "pyruvate reduced to lactate by reduced NAD".
• For removal, link "oxygen available again" to "aerobic respiration", to "lactate oxidised to pyruvate", and to processing by the liver (with optional credit for oxygen debt / EPOC).
• Make sure each point is a complete marking point — state the cause and the biochemical consequence.

Step-by-Step Reasoning

1. During the dive — why blood lactate rises (after the dive). Early in the dive the seal is still using oxygen bound to myoglobin in its muscles, so aerobic respiration continues and the blood lactate stays low. As the dive continues the muscle cells run out of oxygen, so the rate of aerobic respiration falls. The cells switch to anaerobic respiration to keep producing some ATP. In anaerobic respiration, the pyruvate produced by glycolysis is reduced by reduced NAD (the cofactor must be regenerated so that glycolysis can keep going), and this reduction converts pyruvate into lactate. The lactate accumulates inside the muscle cells and is released into the blood once circulation is restored — hence the rise seen in the graph at the end of the dive.

2. After the dive — why blood lactate falls. The seal surfaces and breathes, so oxygen is again available to its tissues. The liver (and to some extent the muscles) now respire aerobically. Lactate is oxidised back to pyruvate, which can then enter the link reaction and the Krebs cycle, with the reducing equivalents feeding the electron transport chain. In addition, much of the lactate is transported in the blood to the liver, where it is converted first to pyruvate and then to glucose (gluconeogenesis) — the Cori cycle — and the glucose can be returned to the muscles.

3. Why the curve still decreases even after the obvious peak. The seal continues to breathe heavily for some time after the dive, taking in extra oxygen. This extra oxygen is needed to oxidise all the lactate that built up and to replenish the ATP, phosphocreatine and oxygen stores that were used up during the dive. This is the oxygen debt or excess post-exercise oxygen consumption (EPOC), and the mark scheme credits it as an additional valid point (AVP).

4. Interpreting the graph shape. The fact that the blood lactate peak occurs after the dive (and not during it) is itself evidence that the seal was holding lactate inside poorly-perfused muscles while underwater. The rapid rise at 45 min is the moment circulation to the muscles is restored and the trapped lactate floods into the general circulation. The slow decline afterwards reflects the time taken for the liver and aerobic respiration to clear it.

Key Takeaways

• Anaerobic respiration in mammals = pyruvate is reduced to lactate by reduced NAD, regenerating NAD so that glycolysis can continue.
• Lactate builds up when oxygen supply to a tissue is inadequate for aerobic respiration.
• Once oxygen is available again, lactate is oxidised back to pyruvate and respired aerobically, or converted to glucose in the liver (Cori cycle).
• The "oxygen debt" / EPOC describes the continued elevated oxygen consumption after exercise/diving to clear lactate and restore energy stores.
• In diving mammals, the blood lactate peak is delayed until the dive ends because lactate is trapped in muscles whose blood supply is restricted during the dive.

Common Mistakes

• Stating only "the seal is anaerobic during the dive" without naming the pathway: it is the conversion of pyruvate to lactate by reduced NAD that is the marking point.
• Saying "the seal cannot breathe underwater, so it suffocates" — the seal has adaptations (myoglobin, bradycardia, peripheral vasoconstriction) and does not suffocate; the relevant point is that the muscles become anaerobic, not the whole animal.
• Confusing lactate with carbon dioxide or with ethanol. Only lactate is made in mammalian anaerobic respiration; ethanol is the yeast / plant anaerobic product.
• Saying "lactate is broken down" without specifying the oxidation to pyruvate or the role of the liver — both are required to earn the available marks.
• Saying "lactate is removed by the kidneys" — it is processed mainly by the liver, not excreted by the kidneys (although a small amount can be excreted in urine).
• Forgetting to mention that the rise in blood lactate is delayed until after the dive (because the muscles are poorly perfused during the dive itself).

Things to Be Careful About

• Each marking point needs the reason as well as the observation — "oxygen is not available, so anaerobic respiration occurs" not just "anaerobic respiration occurs".
• Use the precise term lactate (or lactic acid), not "acid" or "pyruvate".
• For "after the dive", the key concept is the return to aerobic respiration — make the link to oxygen explicitly.
• The graph clearly shows that the peak is after the dive (≈48 min, well after the 45 min dive end), so any answer that claims the peak is "during" the dive is misreading the figure.
• AVP is reserved for an additional valid point, e.g. oxygen debt / EPOC; do not use it as a substitute for the standard marking points.

Background Concept

Conservation is the active management of a species or habitat to reduce the risk of extinction. It is driven by the recognition that biodiversity has intrinsic value and also provides ecosystem services (food, tourism, nutrient cycling, etc.). For an endangered species, conservation biologists identify the threats acting on it and design interventions that either remove the threat, increase the population, or both.

For marine mammals such as seals, the main threats are:

• Direct exploitation — historical and ongoing hunting for fur, blubber and meat; by-catch in fishing gear; disturbance by boats and tourists.
• Habitat loss / degradation — coastal development, pollution (plastics, persistent organic pollutants, oil spills, noise), and the loss of haul-out sites and breeding beaches.
• Prey availability — overfishing removing the fish that seals depend on.
• Climate change — warming seas, sea-ice loss, ocean acidification, and shifts in the distribution of prey species.

Effective conservation of an endangered species usually combines in-situ measures (protecting the species where it lives — reserves, legal protection, threat reduction) with ex-situ measures (breeding and maintaining animals outside their natural habitat — zoos, aquaria, gene banks / "frozen zoos"), and is supported by legislation, research, monitoring, education, and international cooperation (e.g. CITES, IUCN, the Marine Mammal Protection Act).

Understanding the Question

Part (b) is a 3-mark "suggest" question. The question stem says "some seal species are classified as endangered on the IUCN Red List of Threatened Species™" and asks for ways in which seal species may be conserved. The mark scheme accepts any three valid conservation measures from a wide list. The candidate is not asked to prioritise or to discuss any one measure in depth — the answer should simply name three distinct, credible conservation actions.

Approach

• Think of conservation in two broad categories: actions that directly protect / increase the seal population (e.g. breeding programmes, reserves, bans on hunting) and actions that remove the underlying threat (e.g. reduce fishing, reduce pollution, reduce climate change).
• Pick three distinct points rather than three rewordings of the same idea — each must be a creditable, separate marking point.
• Match your wording to the mark scheme's expected terms (e.g. "captive breeding programme", "marine reserve", "ban on hunting") to be safe.

Step-by-Step Reasoning

1. Captive breeding programmes / frozen zoos. Small, vulnerable populations can be augmented by breeding seals in zoos or aquaria and releasing the offspring into the wild. Gametes and embryos can also be stored in "frozen zoos" for future use. This addresses the problem of small, declining populations.

2. Marine reserves / protected areas. Designating stretches of coastline and sea as reserves gives seals protected haul-out sites and breeding beaches, free from disturbance, hunting, fishing and certain forms of pollution. Reserves are an in-situ form of conservation.

3. Bans on hunting and trade. Seals have been hunted for their fur and blubber for centuries. A legal ban (often backed by international agreements such as CITES, which controls trade in endangered species) reduces direct mortality.

4. Education and awareness. Public awareness campaigns and stakeholder engagement (e.g. involving local fishing communities) help build support for conservation measures and reduce demand for seal products.

5. Research. Studying seal populations — their size, distribution, behaviour, genetics, disease status, and the threats they face — provides the evidence base for management decisions (e.g. where to place a reserve, whether a population is recovering).

6. Reduce ocean / sea pollution. Plastic waste, chemical pollution, oil spills, and noise pollution from shipping all harm seals and their prey. Pollution-control measures directly benefit the marine ecosystem.

7. Reduce fishing / by-catch. Seals are often killed accidentally in fishing nets. Technical fixes (e.g. acoustic deterrents, modified nets) and reductions in overfishing protect both the seals' food supply and the seals themselves.

8. Reduce climate change. Sea-ice loss, sea-temperature rise and ocean acidification all affect seal habitat and prey distribution. Cutting greenhouse-gas emissions is a long-term conservation measure.

Any three of these are valid and would earn full marks.

Key Takeaways

• Conservation combines in-situ measures (reserves, legal protection, threat reduction) with ex-situ measures (captive breeding, gene banks).
• Threats to marine mammals include hunting, by-catch, habitat loss, pollution and climate change — each can be tackled with a specific conservation response.
• Conservation is supported by research, monitoring, education, legislation, and international cooperation (CITES, IUCN).
• A good "suggest ways in which…" answer gives distinct points rather than reworded versions of the same idea.

Common Mistakes

• Giving vague answers such as "look after them" or "protect them" — these do not describe a concrete conservation measure and do not score.
• Repeating the same point in different words, e.g. "ban hunting" and "stop people killing seals" — the mark scheme treats these as one point.
• Confusing ex-situ measures (zoos, captive breeding) with in-situ measures (reserves) — both are valid but they are different marking points.
• Stating a problem rather than a solution, e.g. "pollution harms seals" without saying what is done about it.
• Confusing seal conservation with general "save the planet" statements that are not specific to seals.

Things to Be Careful About

• "Suggest" is an open command word — the candidate is free to choose any reasonable answer, but the chosen answer must be a concrete conservation action.
• The question is worth 3 marks, so three distinct points are needed.
• The mark scheme wording gives specific accepted terms (e.g. "marine reserves", "captive breeding programmes", "ban on hunting / trade"); using the same or similar language is the safest route to full credit.
• "Education" and "research" are both credited, but they are different marking points — make sure they are clearly distinct if both are offered.