Biology 9700/41 — May/June 2025

Background Concept

Respiration is the controlled release of energy from energy-rich organic molecules. Many different substrates can be respired, but the most common entry points into the pathway are the carbohydrates glucose and the three-carbon sugar phosphates (triose phosphates and fructose phosphates) produced during glycolysis. Lipid breakdown releases glycerol and fatty acids; fatty acids are broken down by β-oxidation into acetyl (2-carbon) units that enter as acetyl coenzyme A. Amino acids, after deamination, also feed carbon skeletons into the pathway.

Understanding the Question

The command word is name, so a short, accurate response is all that is required — no description is needed. Two molecules must be supplied, each worth one mark, and any two from the mark scheme's accepted list are acceptable. The parent stem makes clear we are looking for molecules that enter the respiration pathway (i.e. substrates that feed in), not for every molecule produced along the way.

Approach

Recall which substrates connect directly to glycolysis or the link reaction. Glucose and its phosphorylated derivatives (glucose phosphate, fructose bisphosphate, triose phosphate) are the most fundamental and form the basis of glycolysis. For two marks, the simplest, most secure answer is the substrate (glucose) and the immediate glycolytic intermediate (triose phosphate).

Step-by-Step Reasoning

• Glucose (phosphate) is the starting substrate for glycolysis; in some textbooks it is shown entering the cell as glucose and being immediately phosphorylated to glucose 6-phosphate.
• Triose phosphate is the three-carbon sugar phosphate produced when fructose 1,6-bisphosphate is split, and it is the substrate that donates a high-energy phosphate group directly to ADP (substrate-level phosphorylation) to form ATP.
• Other acceptable answers include fructose bisphosphate, glycerol (from triglycerides), fatty acids, and amino acids, all of which can be channelled into the pathway at various points.

Key Takeaways

• Carbohydrate substrates enter glycolysis; triose phosphate is the immediate ATP-yielding intermediate.
• Lipid and protein breakdown products (fatty acids, glycerol, amino acids) can also be respired once converted into appropriate entry-point molecules (e.g. acetyl CoA, pyruvate).

Common Mistakes

• Listing products of respiration (e.g. pyruvate, CO₂) rather than substrates that enter the pathway. Pyruvate is produced by glycolysis; it is not a substrate that "enters" the pathway in the sense of the question.
• Writing only "glucose" without the phosphate option — fine in itself, but two distinct entries are required for two marks.
• Naming ATP, ADP or Pi as energy-rich substrates. They are intermediates/energy carriers, not the energy-rich molecules the question is about.

Things to Be Careful About

• The mark scheme accepts glucose, fructose, triose phosphate, glycerol, fatty acids and amino acids. Anything outside this list (e.g. sucrose, starch, triglycerides) is too far upstream and would not be credited as a molecule that directly enters the pathway.

Background Concept

ATP (adenosine triphosphate) consists of three parts visible in Fig. 1.1: an adenine base (a double-ring nitrogenous base with an –NH₂ group), a ribose sugar (a five-carbon sugar with two –OH groups on the ring), and a chain of three phosphate groups attached to carbon 5 of the ribose. Note the negative charges shown on the oxygens of each phosphate group — these like charges repel one another, making the bonds between adjacent phosphates (the phosphoanhydride bonds) unstable and therefore high in energy.

ATP is called the energy currency of the cell because it stores energy in a form that is portable, transferable, and immediately usable. When the terminal (γ) phosphate bond is hydrolysed, energy is released that can be coupled to energy-requiring processes such as active transport, biosynthesis, and muscle contraction.

Understanding the Question

The command word is explain, which means each marking point must be a feature plus the reason it matters. Four independent points are required. The mark scheme lists five possibilities (plus AVP), so any four earn full marks. The parent stem reminds us that ATP is the energy currency of the cell, so the answer must explain why ATP is suitable to play this role universally.

Approach

Think about what any energy currency needs: a way of releasing energy on demand, a way of being recharged, portability, and the ability to interact with the reactions that spend energy. Map each of these ideas to a specific property of ATP shown in the diagram.

Step-by-Step Reasoning

1. Hydrolysis releases a usable amount of energy. Removing the terminal phosphate to form ADP + Pi releases about $30.5\ \text{kJ mol}^{-1}$ — small enough to avoid damaging the cell but large enough to drive many reactions.
2. The reaction is reversible. Because the energy released on hydrolysis is modest, the reaction has a small energy of activation in both directions, and ATP can be resynthesised from ADP and Pi during respiration. This makes it a currency rather than a fuel — it circulates and is reused.
3. Small and soluble. The molecule is small enough to diffuse freely through the cytosol to wherever it is needed, so it can be supplied rapidly to energy-consuming processes in any part of the cell.
4. AVP — coupling to other reactions. The phosphate group can be transferred to another molecule (phosphorylation), raising the energy of that molecule and allowing it to take part in a reaction that would otherwise be energetically unfavourable. The repulsion between the negatively charged phosphates (visible on Fig. 1.1) is what makes the terminal bond so reactive.

Key Takeaways

• ATP is the universal energy currency because of four key properties: modest energy release, reversibility, solubility, and ability to phosphorylate other molecules.
• The negative charges on the phosphates are crucial — they create the unstable, high-energy bonds that make ATP reactive.
• Being a currency (recyclable) is just as important as being an energy source (fuels are consumed, currency is reused).

Common Mistakes

• Stating only that "ATP releases energy" without saying how much, or that it is "high energy" without describing what this means for the cell.
• Confusing ATP (an energy currency) with glucose (an energy store or fuel). Glucose must be broken down over many steps; ATP releases energy in a single, coupled step.
• Saying ATP "contains energy" without explaining how it releases it (through hydrolysis of a phosphate bond).
• Writing about the structure of ATP without linking each feature to its functional consequence — "it has three phosphates" alone is not an explanation.

Things to Be Careful About

• The mark scheme uses both "$\text{kJ mol}^{-1}$" and the value 30.5. Quoting the energy value is the cleanest way to score that marking point.
• "Reversible" must be qualified: the mark scheme accepts either "reversible reaction", "high turnover" or "can be regenerated" — any of these is fine.
• The question awards 4 marks for any four from five options, so there is no need to include all five.

Background Concept

There are two ways in which ATP is made during respiration:

• Substrate-level phosphorylation: a phosphate group is transferred directly from a phosphorylated substrate (an intermediate in the pathway) onto ADP, forming ATP. The energy has already been "stored" in that intermediate during an earlier step.
• Oxidative phosphorylation (chemiosmosis): ATP is made by ATP synthase, driven by a proton gradient across the inner mitochondrial membrane that is generated by the electron transport chain.

In a single turn of glycolysis, ATP is made twice by substrate-level phosphorylation (at two different steps). In each turn of the Krebs cycle, one ATP (or GTP, which is then converted to ATP) is also made by substrate-level phosphorylation.

Understanding the Question

The command word is identify; the question asks for the type of phosphorylation that occurs in both glycolysis and the Krebs cycle. Since both stages make ATP directly from a phosphorylated intermediate, the answer is the same for both. The contrast the question is hinting at is with oxidative phosphorylation, which happens on the inner mitochondrial membrane (not in the cytosol or matrix at the named steps).

Approach

Recall that in glycolysis the steps that produce ATP involve 1,3-bisphosphoglycerate and phosphoenolpyruvate; in the Krebs cycle succinyl CoA → succinate is the equivalent step. In every case, a phosphate is transferred from a substrate to ADP. That is, by definition, substrate-level phosphorylation.

Step-by-Step Reasoning

• The term to remember is substrate-level phosphorylation (also written "substrate-linked phosphorylation" by the mark scheme).
• It is specifically not oxidative phosphorylation, which occurs at the inner mitochondrial membrane and depends on oxygen.
• One mark for the correct term; do not add extra, uncreditable material.

Key Takeaways

• Substrate-level phosphorylation produces a small number of ATP directly from pathway intermediates; oxidative phosphorylation produces the much larger bulk of ATP and depends on the electron transport chain and a proton gradient.
• Glycolysis and the Krebs cycle both yield ATP by substrate-level phosphorylation; only the electron transport chain uses oxidative phosphorylation.

Common Mistakes

• Writing "oxidative phosphorylation" — this is wrong for the stages named in the question and loses the mark.
• Writing "photophosphorylation" — this is the photosynthesis equivalent, not the respiration one.
• Writing "substrate phosphorylation" without the "-level" or "-linked" qualifier — accepted in conversation, but the mark scheme requires the full term.

Things to Be Careful About

• The mark scheme accepts "substrate-level" or "substrate-linked". Either is fine; "substrate phosphorylation" on its own is risky.

Background Concept

A respirometer is a simple piece of apparatus used to measure the rate at which a small organism takes up oxygen during aerobic respiration. It relies on the fact that in aerobic respiration, an organism consumes $\text{O}_2$ and releases an equal volume of $\text{CO}_2$.

If both gases were released and absorbed equally, there would be no net change in the volume of gas inside the sealed tube, and so no way to detect respiration. Sodium hydroxide (or another $\text{CO}_2$ absorbent such as soda lime) removes the $\text{CO}_2$ as soon as it is produced, leaving only the decrease in volume due to $\text{O}_2$ uptake. This change in gas volume produces a tiny decrease in pressure inside the tube, which is amplified and visualised by a coloured drop of liquid in a narrow horizontal capillary tube.

Understanding the Question

Part (a) asks the candidate to explain — point by point — how the apparatus shown in Fig. 2.2 actually produces a measurement of the rate of respiration. The apparatus consists of a horizontal boiling tube containing woodlice, separated by wire mesh from sodium hydroxide pellets. The right-hand end of the boiling tube is closed by a rubber bung through which a graduated capillary tube passes; a coloured drop of liquid sits in the capillary tube. The question wants the chain of reasoning that links "respiring woodlice" → "drop moving" → "rate of respiration".

Approach

Work through the chain in the order the events actually happen:

1. Respiration: $\text{O}_2$ in, $\text{CO}_2$ out.
2. The $\text{CO}_2$ is absorbed by the NaOH, so the only net change is loss of $\text{O}_2$.
3. The gas (and therefore pressure) inside the sealed tube falls.
4. Atmospheric pressure pushes the drop of liquid inwards (towards the woodlice).
5. The distance moved in a known time gives a rate; the volume of $\text{O}_2$ used = cross-sectional area of capillary × distance moved.

Step-by-Step Reasoning

• NaOH absorbs the $\text{CO}_2$ produced by respiration. The wire mesh keeps the woodlice away from the pellets, preventing them from touching the corrosive NaOH while still allowing the gas in the tube to circulate freely.
• $\text{O}_2$ consumed by aerobic respiration is no longer replaced, so the number of gas molecules inside the sealed boiling tube decreases.
• This produces a small reduction in pressure inside the tube relative to atmospheric pressure outside.
• Atmospheric pressure pushes the drop of liquid in the capillary tube towards the boiling tube (i.e. to the left in the figure). The narrower the capillary, the greater the distance moved for a given volume change, so the apparatus is sensitive.
• The distance the drop moves in a set time is recorded, allowing calculation of a rate:

• To convert this to a true volume of $\text{O}_2$ used per unit time, multiply the distance moved by the cross-sectional area of the capillary tube ($V = \pi r^2 \times d$, where $d$ is the distance moved). This volume is the $\text{O}_2$ consumed in that time interval, i.e. the rate of aerobic respiration.

Key Takeaways

• A respirometer measures $\text{O}_2$ uptake, not $\text{CO}_2$ release, because the $\text{CO}_2$ is removed by an alkali.
• The liquid drop in the capillary tube is a sensitive manometer: it amplifies the small pressure/volume change into a measurable distance.
• A valid rate must always be expressed as volume of $\text{O}_2$ per unit time (often per unit mass of organism, e.g. $\text{mm}^3\,\text{min}^{-1}\,\text{g}^{-1}$).
• Standard controls: a parallel tube set up identically but with dead woodlice (or no woodlice) — to compensate for any physical effects of temperature or pressure change that are unrelated to respiration.

Common Mistakes

• Stating that the drop moves because the woodlice "give off heat" or "use energy" — this is wrong; it is the fall in gas pressure, not temperature change, that moves the drop.
• Saying the drop moves "outwards" or "to the right" — because gas is being consumed (not produced), the drop must move towards the woodlice.
• Failing to state the role of NaOH; without it, $\text{O}_2$ uptake and $\text{CO}_2$ release would approximately cancel and the drop would not move.
• Writing "rate = distance" without dividing by time.
• Not converting distance to a volume — to compare experiments using different capillary tubes, a true volume is required.

Things to Be Careful About

• In a Cambridge practical question you may be expected to give a worked equation for the volume of gas used.
• Woodlice must be given an equilibration period after being placed in the tube so that handling (and any change in temperature/activity) does not affect the first reading.
• The tube must be airtight (well-greased bung, no gaps around the capillary) for the pressure change to drive the drop predictably.
• Always run a control tube without woodlice to allow for any change in atmospheric pressure or temperature during the experiment.

Background Concept

Aerobic respiration is catalysed by a long sequence of enzymes — those of glycolysis in the cytoplasm and those of the link reaction, Krebs cycle and oxidative phosphorylation in the mitochondria. Like all enzymes, these catalysts have an optimum temperature (around 35–40 °C for many mammalian enzymes) and their activity depends on the kinetic energy of the molecules involved. Below the optimum, the rate of an enzyme-catalysed reaction falls because the molecules have less kinetic energy and collide less frequently, and because a smaller fraction of collisions have enough energy to exceed the activation energy of the reaction.

Understanding the Question

Part (b) asks the candidate to compare the rate of respiration of woodlice at $5\,^{\circ}\text{C}$ with that at $20\,^{\circ}\text{C}$ and to explain the difference. The mark scheme demands both a prediction (which temperature gives the higher rate) and a reason linked to enzymes or kinetic energy.

Approach

Identify that woodlice are poikilothermic (their body temperature follows the surrounding temperature), so a change in environmental temperature directly alters the kinetic energy of the enzymes and substrates inside their cells. Apply the general rule for the effect of temperature on enzyme activity below the optimum.

Step-by-Step Reasoning

• At the lower temperature ($5\,^{\circ}\text{C}$), the kinetic energy of the enzyme and substrate molecules is reduced.
• Molecules therefore move more slowly and collide less often.
• A smaller proportion of those collisions have sufficient energy to overcome the activation energy of the enzyme-catalysed reaction.
• The rate of each step of respiration (glycolysis, link reaction, Krebs cycle) — and therefore the overall rate of oxygen uptake — is lower.
• The drop of liquid in the respirometer would therefore move a shorter distance in a given time at $5\,^{\circ}\text{C}$ than at $20\,^{\circ}\text{C}$.

Key Takeaways

• For any enzyme-controlled process, a temperature below the optimum gives a slower rate because of reduced kinetic energy and fewer effective collisions.
• A "predict and explain" question requires both parts: a clear comparative statement and a biological reason.
• Woodlice are ectotherms, so the temperature of their environment directly determines the temperature inside their cells.

Common Mistakes

• Saying the rate is higher at $5\,^{\circ}\text{C}$ (incorrect — colder temperatures slow enzyme activity below the optimum).
• Giving only the prediction without an explanation: the mark scheme requires a linked cause (kinetic energy, collisions, or named respiration steps).
• Stating that "the enzymes are denatured" at $5\,^{\circ}\text{C}$ — denaturation requires much higher temperatures; cold simply reduces activity, it does not destroy the enzyme.
• Vague answers such as "the woodlice are cold and so are less active" — this is not biological reasoning; the question demands an enzyme-based explanation.

Things to Be Careful About

• The mark scheme is strict: it wants the keyword "kinetic energy" or "enzyme–substrate collisions" (or an explicit reference to a named respiration step). A general phrase such as "the enzymes don't work as well" will not score.
• Note the temperature range used ($5\,^{\circ}\text{C}$ versus $20\,^{\circ}\text{C}$): both are well below the typical denaturation temperature, so the comparison is about kinetic energy only, not about denaturation.

Background Concept

Sex-linked (X-linked) inheritance describes genes carried on the X chromosome. A recessive X-linked allele, such as the one for haemophilia, is normally only expressed in males (who have just one X chromosome) or in homozygous females. For a female to express a recessive X-linked trait, she must inherit the recessive allele on both of her X chromosomes — one from her mother and one from her father. The father must therefore also show the trait (or be hemizygous for the recessive allele, in which case he will express it because he has no second X chromosome to carry a dominant normal allele).

A dominant X-linked allele is different: only one copy is needed for the trait to be expressed. An affected heterozygous female (X^D X^d) will pass the dominant allele to roughly half her offspring regardless of the father's genotype, and the daughter can be affected even though her father is unaffected.

Understanding the Question

Fig. 3.1 shows a two-generation pedigree. In generation 1 the mother (1) is affected and the father (2) is unaffected. In generation 2 they have four children: an affected daughter (3), an unaffected daughter (4), an affected son (5), and an unaffected daughter (6). The question is already given the information that the allele is X-linked, and is asking for one other conclusion about the inheritance pattern, supported by evidence from the pedigree.

The command words are "conclusion" and "evidence" — the mark scheme rewards a single clear statement of the conclusion (the word "dominant") plus the supporting logic drawn from the pedigree.

Approach

Test the two possibilities (dominant vs. recessive) against the pedigree:

• If the X-linked allele were recessive, the affected daughter (3) would need X^p X^p. She must have received one X^p from each parent. From the father (2), who is unaffected, the only X chromosome he can pass on carries the normal allele (X^P). So he could not pass on X^p. This contradicts the pedigree — therefore the allele cannot be recessive.
• If the X-linked allele were dominant, the affected daughter only needs one copy of the mutant allele. She can have received the mutant X from her affected mother and a normal X from her unaffected father. This is consistent with the pedigree.

Step-by-Step Reasoning

1. The mother (1) is affected, so her genotype is X^A X^? (at least one mutant allele, written as A assuming dominant).
2. The father (2) is unaffected, so his genotype is X^? Y (one normal X chromosome and a Y).
3. Daughter 3 is affected. She has genotype X^A X^? — she received one X from each parent.
4. From the father, she must have received the X chromosome, which carries the normal allele (since the father has no disease).
5. The mutant allele she shows must therefore have come from the mother.
6. With one mutant allele from the mother and one normal allele from the father, she is heterozygous, yet she expresses the disease. This is only possible if the mutant allele is dominant to the normal allele.
7. If the allele were recessive, daughter 3 would need two copies of the mutant allele (one from each parent), and so the father would have to be affected — which he is not.

Key Takeaways

• A dominant X-linked trait can pass from an affected mother to a daughter whose father is unaffected, because the daughter needs only one copy of the mutant allele.
• For a recessive X-linked trait, an affected daughter must have an affected father, because she must have received the recessive allele on his X chromosome.
• A single pedigree can therefore often rule out one of the two possibilities (dominant vs. recessive) just by looking at father–daughter combinations.

Common Mistakes

• Saying the allele is recessive because the disease is rare — incorrect; dominance and rarity are independent.
• Stating the conclusion but not providing the supporting evidence drawn from the pedigree (e.g. simply saying "the allele is dominant because the disease is X-linked").
• Failing to explain why the recessive possibility is ruled out (i.e. the father would have to be affected).

Things to Be Careful About

• The conclusion must be a single, clear genetic statement ("dominant"). Avoid vague answers like "the disease is inherited".
• The evidence must refer explicitly to specific individuals in the pedigree (individual 2 and individual 3).
• The mark scheme accepts either the affirmative argument (daughter affected, father not) or the negative argument (if recessive, daughter could not be affected).

Background Concept

The founder effect is a special case of genetic drift. When a new population is established by a very small number of individuals drawn from a larger parent population, the allele frequencies in the new population are determined mainly by the genotypes of those few founders — not by the frequencies in the parent population as a whole. By chance, some alleles that were rare in the parent population may be present in the founders at much higher frequency (or vice versa).

Once established, a small population is also strongly affected by genetic drift — random fluctuations in allele frequency from one generation to the next — and by inbreeding, which does not change allele frequency by itself but keeps rare alleles (in homozygous form) circulating in the population.

Understanding the Question

The question gives two pieces of data:

• In Europe as a whole, variegate porphyria occurs in 1 in 100 000 people.
• In South Africans of Dutch descent, it occurs in 1 in 1000 people — about 100 times more often.

The historical context is that this South African Dutch population was founded in the 1600s by approximately 100 migrants. The question asks you to suggest and explain why the disease is more common in this descendant population.

Approach

The key concept to reach for is the founder effect. Then layer in additional reasoning:

• How did the higher frequency arise in the founders? (The allele must have been carried by some of the 100 founders at a higher-than-European frequency.)
• Why has the high frequency been maintained over ~400 years? (Small population → genetic drift; inbreeding within the group.)

Step-by-Step Reasoning

1. The original European population is large, and the disease allele is rare (1/100 000).
2. A small group of ~100 Dutch people migrated to South Africa; the allele frequencies in this group are determined by chance (which alleles the founders happened to carry).
3. By chance, the founders included at least one (probably more) carrier(s) of the variegate porphyria allele, so the allele frequency in the founding population was higher than in Europe overall.
4. This is the founder effect: a new population's allele frequencies are not representative of the source population.
5. After establishment, the new South African Dutch population remained relatively small and isolated. In a small gene pool, genetic drift has a disproportionately large effect on allele frequencies, and random fluctuations are amplified.
6. The group was also highly inbred (people tended to marry within their own community), so the mutant allele was maintained in the gene pool over many generations rather than being diluted out by mixing with populations carrying the normal allele.
7. The result is that today, the descendant population of 2.5 million still shows a much higher frequency of the allele (1/1000) than the European source population (1/100 000).

Key Takeaways

• The founder effect is a genetic-drift phenomenon that operates on a small founding population; the allele frequencies in the descendants reflect those of the founders, not the source population.
• Founder effect plus a long period of relative isolation (with inbreeding) is the classic explanation for high frequencies of otherwise rare alleles in particular human populations (e.g. the high frequency of certain alleles in the Afrikaner population, the Old Order Amish, or isolated island populations).
• "Suggest and explain" questions on evolution want the mechanism named and the specific link to the data given.

Common Mistakes

• Saying only "founder effect" without explaining how the higher frequency arose or how it was maintained.
• Confusing founder effect with bottleneck effect: a bottleneck reduces population size suddenly and randomly; a founder effect establishes a new population from a small number of individuals. Both are forms of genetic drift, but the timing and mechanism differ.
• Suggesting natural selection favoured the allele — there is no evidence for any selective advantage of variegate porphyria, and the mark scheme is asking for a neutral, drift-based explanation.
• Suggesting a higher mutation rate in the South African Dutch population — not supported by the data.

Things to Be Careful About

• The mark scheme wants specific points tied to the data: small founding population, allele present in the founders, frequency higher than in the source population, genetic drift in a small gene pool, and inbreeding.
• Any 3 of these (out of the 7 listed in the mark scheme) will earn full marks, so give the strongest, most specific points.

Background Concept

A DNA microarray is a small chip carrying thousands of different single-stranded DNA probes, each representing a known gene or allele, fixed at a specific position on the chip. By exposing the chip to a sample of DNA (or cDNA), complementary sequences hybridise (form base pairs) with their matching probe. Any hybridised sample can be detected because it has been labelled, usually with a fluorescent dye.

A microarray can be used in two distinct ways:

• Genotyping / allele detection — to test whether a person carries a particular mutant allele. The probes on the chip are designed to match either the normal or the mutant sequence, and the patient's DNA (or cDNA) is tested against both.
• Gene expression profiling — to measure how actively a gene is being transcribed. mRNA is extracted from the patient's cells; the more mRNA a cell makes, the more actively that gene is being expressed.

This part of the question is specifically about gene expression (the wording is "detecting disease by analysing gene expression"), so the procedure must include extracting mRNA and converting it to cDNA.

Understanding the Question

The question is worth 4 marks and asks for a description of how a microarray detects a disease by analysing gene expression. The mark scheme rewards four distinct steps: making cDNA from mRNA, labelling the DNA, hybridising to probes, and reading fluorescence as expression.

Approach

The key is to follow the biological chain in order:
mRNA (from the cell) → cDNA (by reverse transcription) → labelled cDNA → hybridisation to probe → fluorescence = expression of the gene.

Step-by-Step Reasoning

1. mRNA extraction. Cells from the patient are broken open and the mRNA is isolated. mRNA is chosen because it is a direct read-out of which genes are being transcribed. Genes that are not expressed will produce little or no mRNA.
2. Reverse transcription. The enzyme reverse transcriptase uses the mRNA as a template to make a complementary strand of DNA, called cDNA (complementary DNA), which is single-stranded.
3. Fluorescent labelling. A fluorescent dye (often Cy3, which fluoresces green, or Cy5, which fluoresces red) is incorporated into the cDNA, either during the reverse-transcription step or afterwards.
4. Hybridisation. The fluorescently labelled cDNA is washed over the microarray. The cDNA molecules base-pair (hybridise) only with probes whose sequence is complementary. If the patient is expressing the mutant allele, cDNA corresponding to that allele will bind to its matching probe.
5. Detection. A laser scans the array; the spot(s) where cDNA has hybridised fluoresce, and the position on the array identifies which gene/allele is being expressed, while the intensity of the fluorescence indicates how strongly it is being expressed. A fluorescence pattern different from that of a healthy control indicates disease-associated expression.

Key Takeaways

• Microarrays for gene expression start with mRNA, not genomic DNA. The conversion of mRNA to cDNA via reverse transcriptase is the key step.
• The fluorescent label does not "find" the gene on its own — hybridisation by complementary base pairing is the matching step, and the label is what makes that match visible.
• A spot's position on the array identifies the gene; the intensity of fluorescence indicates how strongly the gene is being expressed.
• Microarrays for allele detection (genotyping) work slightly differently — they use short allele-specific oligonucleotide probes and test genomic DNA, not mRNA. The question specifies expression, so the mRNA → cDNA route is required.

Common Mistakes

• Skipping the reverse-transcription step and saying "DNA is added directly to the microarray". Without the mRNA → cDNA step, you are not measuring expression.
• Saying the microarray "reads the DNA sequence" or "sequences the gene". That is sequencing, not microarray analysis.
• Saying the fluorescence "detects the disease" without specifying that it is detecting expression of the disease allele. Microarrays do not detect the presence of an allele directly unless used in genotyping mode.
• Forgetting the role of complementary base pairing: the cDNA does not "stick" non-specifically to the chip; it hybridises to its complementary probe.

Things to Be Careful About

• Mark scheme explicitly credits the use of mRNA → cDNA (or ssDNA), the fluorescent label, the hybridisation step, and the interpretation of fluorescence as expression.
• A fifth "AVP" mark is available, e.g. for mentioning that the position on the chip identifies which gene is being expressed, or for noting that comparing the patient's fluorescence pattern with a healthy control reveals the disease.

Background Concept

Many rare genetic diseases are caused by different mutations in the same gene, or by mutations in different genes that produce similar symptoms. Identifying the exact mutation in a patient — for example by sequencing the relevant gene and comparing the sequence against known pathogenic variants in a database — is often the first step in making a definitive diagnosis and in selecting the most appropriate treatment.

Sequence databases such as ClinVar, OMIM, the Human Gene Mutation Database (HGMD) and locus-specific databases (LSDBs) store the exact nucleotide changes recorded in patients with particular diseases. They are used by:

• Diagnostic laboratories, to interpret a patient's sequence (is the variant seen known to cause disease?).
• Clinicians, to choose a treatment that targets the specific mutation.
• Researchers, to find out what mutations are known, identify gaps, and plan new therapies.

Understanding the Question

The question is worth 1 mark and asks for one benefit of having a database of nucleotide sequences for alleles that cause rare diseases. The mark scheme accepts any of three benefits (diagnosis, treatment selection, or global research resource). The student should give one clear, specific benefit.

Approach

Pick the most direct benefit (usually diagnosis) and state it in one or two sentences. Avoid vague answers like "it is useful for science".

Step-by-Step Reasoning

1. A patient with a suspected rare porphyria has their gene sequenced.
2. The sequence is compared with the entries in the database.
3. If a match is found, the disease is confirmed and its specific form identified — the patient can be diagnosed.
4. Knowing the exact mutation allows the correct treatment to be chosen (or developed), and the data is available to researchers worldwide as a resource for further study.

Key Takeaways

• Sequence databases turn a patient's raw sequence into a clinically meaningful result by enabling comparison with known pathogenic variants.
• The benefit is not just "having the information" but the diagnostic, treatment and research uses to which it is put.
• For rare diseases, where a single clinician may never see a case, a global database is especially valuable.

Common Mistakes

• Vague answers: "it helps doctors" or "it is useful for research" without specifying the benefit.
• Saying the database "treats" the disease — it does not; it enables the right treatment to be chosen.
• Saying the database "prevents" the disease — sequence data does not, by itself, prevent disease.

Things to Be Careful About

• One mark is available for a single clear, specific benefit. Choose one and state it precisely.

Background Concept

Congenital erythropoietic porphyria (CEP) is an autosomal recessive condition in which a substitution mutation in the gene encoding the enzyme uroporphyrinogen III cosynthase (UROS) produces an enzyme with little or no activity. Without this enzyme, the haem / porphyrin biosynthesis pathway cannot proceed normally, and porphyrin intermediates accumulate. Porphyrins are light-sensitive, so their accumulation in the skin causes the characteristic photosensitivity and skin damage of CEP.

CRISPR–Cas9 gene editing uses two key components:

• Cas9, an endonuclease that cuts both strands of DNA at a precise location.
• A guide RNA (gRNA) that base-pairs with the target DNA sequence and directs Cas9 to that exact location. The gRNA is typically about 20 nucleotides long and must be complementary to the target site; this is what gives the system its specificity.

Once Cas9 has cut the DNA, the cell's own repair machinery can be exploited: if a template DNA with the correct sequence is provided, the cell's homology-directed repair (HDR) pathway will use it to repair the cut, replacing the faulty sequence with the correct one. The result is a precise, targeted change to a single gene.

Understanding the Question

Fig. 3.2 shows part of the CEP gene before and after editing. Before editing the sequence reads TATGGCTCG; after editing it reads TACGGCTCG. The change is a single nucleotide: the third base, a T, has been replaced by a C. This is the substitution that the question tells you produces a non-functional enzyme.

The question asks you to suggest and explain how the gene-editing procedure will:

• prevent the accumulation of porphyrins (the biological outcome), and
• do this without damaging other genes (the safety/specificity issue).

Three marks are available: one for the nucleotide change, one for the functional protein outcome, and one for the gRNA specificity (with an AVP mark available for a further valid point).

Approach

Read the question in two halves:

1. How does the editing prevent porphyrin accumulation? The mutation is corrected → normal mRNA is transcribed → functional enzyme is translated → the porphyrin pathway works → no porphyrin accumulation.
2. Why doesn't it damage other genes? The gRNA is complementary to the mutant CEP allele only, so Cas9 is targeted to that one site and does not cut elsewhere in the genome.

Step-by-Step Reasoning

1. Identify the change. In Fig. 3.2, the mutant sequence contains T (third base) where the corrected sequence contains C. This single nucleotide change (T → C) restores the correct codon and hence the correct amino acid in the enzyme.
2. Restored function. With the corrected sequence, the cell's transcription and translation machinery produce a normal, functional uroporphyrinogen III cosynthase enzyme.
3. Metabolic outcome. The functional enzyme catalyses its step in the porphyrin / haem biosynthesis pathway, so porphyrin intermediates are no longer left unprocessed and no longer accumulate. The disease phenotype is therefore prevented.
4. Specificity of gRNA. The guide RNA is a short single-stranded RNA designed to be complementary in base sequence to the mutant CEP allele. Because base pairing is sequence-specific, the gRNA will only bind to that exact target sequence and not to other regions of the genome. Cas9, which is carried to the DNA by the gRNA, will therefore only cut the CEP gene.
5. Repair uses the template. Once Cas9 has cut the DNA at the CEP site, the cell's homology-directed repair mechanism uses the supplied template DNA to fill in the correct sequence, completing the correction.
6. Off-target effects avoided. Because the gRNA does not bind to other genes, Cas9 does not cut them, so other genes are not damaged by the procedure.

Key Takeaways

• The biological principle of gene therapy for a loss-of-function disease is: correct the mutation → restore the normal protein → restore the biochemical pathway → prevent the build-up of the harmful intermediate.
• The specificity of CRISPR–Cas9 comes from the guide RNA's base sequence. A 20-nucleotide gRNA is long enough to be essentially unique in the human genome, so it binds only at its intended target.
• The cell's own repair machinery (homology-directed repair, HDR) is what actually inserts the new DNA; Cas9 only makes the cut.

Common Mistakes

• Saying only that "the gene is corrected" without explaining the downstream consequence (functional enzyme → working pathway → no porphyrin accumulation).
• Saying the gRNA "recognises the disease" or "finds the faulty gene" without explaining the molecular basis of this recognition (complementary base pairing to the mutant allele).
• Confusing Cas9 (the cutting enzyme) with the gRNA (the targeting molecule) and saying Cas9 itself "recognises" the target.
• Saying the procedure "removes the mutant gene" rather than correcting it — gene editing in this protocol alters the sequence in place, it does not delete the gene.

Things to Be Careful About

• Three distinct ideas are needed: the T → C substitution, the functional enzyme outcome, and the gRNA specificity. Watch for marks that are easy to leave on the table by being too vague.
• The AVP mark is for any further valid point — e.g. that HDR uses the supplied template DNA, or that the change is at a single specific site.

Background Concept

Natural photosynthesis, summarised by the overall equation

takes carbon dioxide and water and, using light energy captured by chlorophyll in chloroplasts, builds an organic molecule (a carbohydrate) and releases oxygen. The light-dependent stage, on the thylakoid membranes, uses light to split water, generate ATP and reduce NADP, with $\text{O}_2$ released as a by-product. The light-independent stage (Calvin cycle) in the stroma uses these to fix $\text{CO}_2$ into triose phosphate. Throughout, energy is transduced from light into chemical form, and electrons flow from $\text{H}_2\text{O}$ to NADP.

The artificial system in Fig. 4.1 uses a solar panel to convert sunlight into electricity, which then drives an electrolyser reacting $\text{CO}_2$ with $\text{H}_2\text{O}$ to give $\text{O}_2$ and acetate (an organic, two-carbon molecule). The acetate feeds Chlamydomonas, which is then processed into food.

Understanding the Question

The command word is "identify" — the candidate simply needs to name shared features, with three marks available, so three points are required. The question asks for similarities, i.e. features that BOTH processes have in common. It is easy to drift into differences, which score nothing here.

Approach

Compare the inputs and outputs of the two systems and look for overlap:

• Inputs to the artificial system: sunlight, $\text{H}_2\text{O}$, $\text{CO}_2$.
• Inputs to natural photosynthesis: sunlight, $\text{H}_2\text{O}$, $\text{CO}_2$.
• Outputs of the artificial system: $\text{O}_2$, acetate (organic).
• Outputs of natural photosynthesis: $\text{O}_2$, carbohydrate (organic).
• Energy pathway: light → electrical → chemical in the artificial system; light → chemical in natural photosynthesis.
• Electron pathway: electrons flow from the solar panel through the electrolyser; electrons flow from $\text{H}_2\text{O}$ through the thylakoid electron transport chain to NADP in the natural system.

Any three matches earn the three marks.

Step-by-Step Reasoning

The mark scheme lists seven possible creditable points; any three are accepted:

1. Both use $\text{CO}_2$.
2. Both use $\text{H}_2\text{O}$.
3. Both use sunlight energy.
4. Both make an organic molecule/product.
5. Both release $\text{O}_2$.
6. Both involve energy transduction (one form to another).
7. Both involve electron flow.

A strong answer picks the three most concrete, most easily-verifiable points — usually $\text{CO}_2$, $\text{H}_2\text{O}$ and either light or $\text{O}_2$ — to make sure all three marks are earned unambiguously.

Key Takeaways

• The overall equation of photosynthesis is the easiest reference point for any "compare photosynthesis with X" question.
• "Organic molecule/product" is broader than "carbohydrate" — the artificial system makes acetate, which is organic, so this wording covers both.
• Energy transduction and electron flow are deeper similarities that show the chemistries are fundamentally linked, even though the two systems use very different machinery.

Common Mistakes

• Confusing the algae's metabolism with photosynthesis. The algae in the dark are using acetate directly (heterotrophically), not photosynthesising. They contribute to the use of the organic product, not to making it.
• Listing features only one process has, e.g. "the artificial system uses electricity" — that is a difference, not a similarity.
• Vague answers such as "both involve chemical reactions" — the mark scheme rewards the named substance ($\text{CO}_2$, $\text{H}_2\text{O}$, $\text{O}_2$) or the named concept (light energy, electron flow), not generic statements.

Things to be Careful About

• Use the specific chemical species ($\text{CO}_2$, $\text{H}_2\text{O}$, $\text{O}_2$) — not "a gas" or "water".
• "Energy transduction" is the technical term; "energy change" is a weaker but still credit-worthy paraphrase.
• Do not list more than three unless time allows; the question only requires three, so extra points are wasted effort.

Background Concept

A chloroplast is a double-membraned organelle specialised for photosynthesis. Inside, the thylakoid membrane system is folded into stacks called grana (singular: granum), connected by stromal lamellae. The thylakoid membrane carries chlorophyll and the protein complexes of the light-dependent reactions. The fluid stroma surrounding the thylakoids contains the enzymes of the Calvin cycle, starch grains (storage of carbohydrate product) and, in many algae, a pyrenoid — a proteinaceous body where the enzyme Rubisco is concentrated and around which starch is deposited.

In Chlamydomonas the single chloroplast is large and cup-shaped, occupying much of the cell. The pyrenoid is a prominent feature, embedded within the chloroplast and surrounded by starch plates/grains.

Understanding the Question

The question states explicitly that both A and B are found within the chloroplast, so the candidate is identifying sub-chloroplast structures on a transmission electron micrograph of Chlamydomonas. A points to a dense, dark, roughly circular deposit near the base of the cup-shaped chloroplast — the classic appearance of a starch granule in (or attached to) the pyrenoid. B points to the dark, layered, ribbon-like stacks running along the wall of the chloroplast — characteristic of thylakoid membranes (often arranged as grana in Chlamydomonas).

Approach

Match each label to the classic appearance of a chloroplast sub-structure and give the textbook name.

• A: dense, dark, rounded → starch grain (granule).
• B: layered, parallel, dark stripes → thylakoid membrane / lamella / granum.

Step-by-Step Reasoning

The mark scheme accepts:

• A: starch grain or starch granule (1 mark).
• B: thylakoid (membrane) / lamella / granum — any of these is acceptable (1 mark).

Both terms are simply named. The candidate does not need to add descriptions of function, but knowing the structure helps locate it on the unfamiliar micrograph.

Key Takeaways

• In a chloroplast TEM, thylakoids look like dark, stacked, ribbon-like layers; starch grains look like dense, dark, rounded deposits — often lying next to or within a pyrenoid in algae.
• "Thylakoid", "lamella" and "granum" are interchangeable in this context: a granum is a stack of thylakoid discs; a lamella is a single thylakoid disc; stromal lamellae connect grana.
• Starch grains in the chloroplast are temporary storage of the carbohydrate product of photosynthesis — they are not the same as the membrane stacks of thylakoids.

Common Mistakes

• Confusing A with the pyrenoid itself. The pyrenoid is the protein body; the dark deposit is usually a starch grain around the pyrenoid. "Starch grain" is the credit-worthy answer.
• Calling B "chloroplast" or "stroma" — too vague. The thylakoid membrane is a sub-chloroplast structure, and the question insists both A and B are within the chloroplast.
• Calling B "grana" (plural) when the diagram clearly shows a single stack — though "granum" or "grana" is credit-worthy, "grana" alone at this single spot is borderline.

Things to be Careful About

• Use the singular/plural form that matches the structure shown. "Granum" for a single stack; "thylakoid" for a single disc; "starch grain" for a single deposit.
• "Starch grain" and "starch granule" are interchangeable; either scores the mark.
• The question's image description notes the cell is a Chlamydomonas with a "cup-shaped chloroplast" — this is consistent with the structures named.

Background Concept

The Calvin cycle fixes $\text{CO}_2$ onto the 5-carbon acceptor ribulose bisphosphate (RuBP) in a reaction catalysed by the enzyme Rubisco. The unstable 6-carbon intermediate splits immediately into two molecules of glycerate 3-phosphate (GP). GP is then reduced (using ATP and reduced NADP from the light-dependent stage) to glyceraldehyde 3-phosphate, more commonly called triose phosphate (TP). TP is the first carbohydrate produced by the Calvin cycle and is the branch point for almost all other photosynthetic products:

• Some TP leaves the cycle to form hexose phosphates (e.g. glucose, fructose), which can be made into sucrose for transport or starch for storage.
• Some TP is used to regenerate the $\text{CO}_2$ acceptor RuBP (using ATP).
• Some TP is diverted into lipid synthesis (by forming acetyl-CoA) and into amino acid synthesis (by transamination of keto acids derived from TP).

Understanding the Question

The question asks for the Calvin cycle intermediate that is used to produce starch. The answer must be the 3-carbon sugar that is the direct precursor of all carbohydrates in the cycle, including starch. Only 1 mark.

Approach

Identify the immediate product of carbon fixation that is the precursor of all carbohydrates. The chain is: $\text{CO}_2$ → GP → TP → hexose phosphate → starch. The intermediate that links the Calvin cycle to starch is therefore TP.

Step-by-Step Reasoning

• Starch is a polymer of glucose.
• Glucose is made in the Calvin cycle from two TP molecules combining to form a hexose phosphate.
• The Calvin cycle intermediate directly feeding into hexose synthesis is triose phosphate.
• Hence, triose phosphate is the answer.

Key Takeaways

• Triose phosphate (TP) is the first carbohydrate made by photosynthesis and the "gateway" to all the other major classes of biological molecule.
• Starch is built from TP, but only indirectly — TP must first be combined with another TP to make a hexose, and several further reactions are needed before starch.
• The other common Calvin cycle intermediates (GP, RuBP) are not direct starch precursors.

Common Mistakes

• Saying "glucose" or "glycerate 3-phosphate (GP)" — these are wrong. Glucose is the product of further reactions on TP, and GP is the immediate product of carbon fixation, not the starch precursor.
• Saying "RuBP" — RuBP is the $\text{CO}_2$ acceptor that is regenerated, not the carbohydrate output.
• Spelling "triose phosphate" incorrectly (e.g. "triose-P", "TrioP") — at A-level, write the full name or the standard abbreviation TP.

Things to be Careful About

• The full name "triose phosphate" or the abbreviation "TP" are both credit-worthy. "GALP" (glyceraldehyde 3-phosphate) is also an accepted CIE synonym.
• Note that TP exists as a phosphate ester — "triose phosphate", not just "triose".

Background Concept

Natural photosynthesis by crop plants is famously inefficient at converting incident sunlight into chemical energy stored in food — typical values are around 1–2% for the whole plant over a growing season, because:

• Plants can only use light of wavelengths absorbed by their pigments (mainly chlorophyll a and b — blue and red light).
• Photosynthesis is limited by the slowest of light intensity, $\text{CO}_2$ concentration and temperature (Blackman's law of limiting factors).
• Plants lose fixed carbon by respiration, both day and night.
• Only a small fraction of the plant is edible (leaves, fruits, seeds, tubers — not roots, stems, woody tissue).
• Plants have a long generation time and are subject to pests, disease and weather.

The artificial system in Fig. 4.1 uses a solar panel (which is engineered to absorb a wide spectrum, although with low conversion efficiency) and an electrolyser (which can be optimised to operate continuously at a chosen $\text{CO}_2$ partial pressure, temperature and pressure) to produce acetate, which is then fed to Chlamydomonas. The algae are grown heterotrophically in the dark.

Understanding the Question

The question asks the candidate to evaluate the claim — i.e. give reasons both for and against it. The mark scheme is phrased as "any two plus one opposing arguments", meaning the candidate must offer two reasons supporting the claim and at least one counter-argument (or vice versa) for a total of three marks. The candidate should structure the answer under "true" (reasons the claim may be correct) and "not true" (reasons the claim may be incorrect).

Approach

Think systematically about where each system gains or loses energy:

• Light capture efficiency
• Energy losses at each step of the conversion chain
• Losses to respiration
• Limiting factors that constrain the rate
• Quality of the food product
• Resources and space needed

Pick the strongest two points for one side, and one strong point for the other.

Step-by-Step Reasoning

The mark scheme lists 18 possible marking points, but only three are needed. The strongest biological arguments are:

In favour of the claim (true):

1. The solar panel absorbs a wider range of wavelengths than chlorophyll, so a higher fraction of the light energy is captured (mark scheme point 2 — absorption spectrum / wavelengths used).
2. The algae grow in the dark on acetate, so the system is not subject to the limiting factors of light intensity, day length or weather (mark scheme point 8 — night / dark process; also point 6/7 — limiting factors).
3. (Optionally) Plants lose fixed carbon by respiration; algae grown heterotrophically in the dark and harvested at the end of a rapid growth cycle lose less to respiration (mark scheme point 3).

Against the claim (not true):

1. Solar panels themselves have a low efficiency (typically 15–22% for converting sunlight to electricity) — much of the light energy is lost at the very first step (mark scheme point 4/5 — stage where energy is lost; also point 1 — percentage/amount of light absorbed).
2. Additional energy inputs are required — e.g. the electrolysis reaction, mixing, pumping and maintaining the algae culture (mark scheme point 11/12).
3. The food product is algae, which is less palatable and less familiar to consumers than conventional crop food; it may also be nutritionally less complete (mark scheme point 9 — quality/nutrition/palatability of food product).
4. (Optionally) A large area of solar panels is required to capture enough light (point 16).

A strong answer gives two clear reasons for the claim and one clear reason against it, with the biological reasoning attached.

Key Takeaways

• A meaningful efficiency comparison must look at every step of the energy-conversion chain, not just one step in isolation.
• Natural photosynthesis by crop plants is limited by Blackman's law — the slowest of light, $\text{CO}_2$ and temperature caps the rate.
• "Efficiency" depends on what is being measured — light-to-food energy conversion, edible biomass per hectare, or simply grams of food per unit area.

Common Mistakes

• Only giving arguments for one side. The question asks for reasons the claim may or may not be true — both sides are required.
• Vague answers like "it's better because it's more modern" or "it's not better because it's expensive". The mark scheme rewards specific biological reasons — wavelengths absorbed, named limiting factors, energy losses at each stage.
• Forgetting to mention the biological reasons. The candidate should anchor each argument in photosynthesis, respiration or ecology, not in pure engineering.
• Saying "plants use 100% of the light they absorb, but solar panels only 20%" — plants use a much smaller fraction of incident sunlight than this, because most wavelengths are not absorbed.

Things to be Careful About

• Quote a named biological concept (limiting factor, respiration, action spectrum) for each reason.
• Where possible, give a number or a comparison (e.g. "solar panel efficiency ~20% vs crop photosynthesis ~1–2%").
• Structure the answer clearly under "true" and "not true" so the examiner can find each marking point.

Background Concept

Disruptive (diversifying) selection occurs when individuals at both extremes of a phenotypic range have higher fitness than those near the mean. The intermediate phenotype is selected against. The classic graphical signature of disruptive selection is a bimodal distribution — two peaks with a trough at the mean — because the intermediate individuals survive and reproduce less successfully than the extremes.

This contrasts with:

• stabilising selection (mean unchanged, variance reduced) and
• directional selection (mean shifts to one side).

For disruptive selection to be sustained, the environment must offer two distinct resources that favour two different phenotypes — exactly the situation in this question, where seed sizes on Santa Cruz are bimodal.

Understanding the Question

G. fortis on Santa Cruz face a food supply that is bimodal: most available seeds are either small or large, with very few of intermediate size. The current population of G. fortis on Santa Cruz also shows a bimodal bill-size distribution (Fig. 5.3). The question asks you to explain how natural selection has produced this bimodal distribution from an originally unimodal one.

Approach

• Identify the type of selection: with two favoured extremes, this is disruptive selection.
• Identify the selection pressure: the bimodal seed-size distribution on Santa Cruz.
• Explain why the intermediate phenotype is at a disadvantage: intermediate-billed birds cannot crack large seeds (bill too small) and cannot pick up small seeds efficiently (or are out-competed by the better-adapted extremes).
• State which phenotypes are favoured: both extremes (small-billed and large-billed birds).
• Describe the genetic consequence: change in allele frequencies — alleles for small and large bills increase; alleles for intermediate bills decrease.

Step-by-Step Reasoning

1. Selection pressure: food availability is bimodal (small and large seeds, few intermediates). So bill size has to match one of two resource sizes to be advantageous.
2. Intermediate disadvantage: a bird with a mid-sized bill is poorly matched to both seed types — it cannot efficiently crack large seeds and it loses the small seeds to the more efficient small-billed birds. Competition for both seed types therefore reduces the fitness of intermediate-billed birds.
3. Extreme advantage: small-billed birds efficiently eat small seeds; large-billed birds efficiently eat large seeds. Both extremes therefore have higher survival and reproductive success than the intermediate phenotype.
4. Differential reproduction: extreme-billed birds pass more alleles to the next generation; intermediate-billed birds pass fewer. Over generations, the frequency of alleles producing small bills and the frequency of alleles producing large bills both rise, while the frequency of alleles producing intermediate bills falls.
5. Outcome: the population's bill-size distribution becomes bimodal, exactly as shown in Fig. 5.3.

Key Takeaways

• A bimodal resource or habitat tends to drive disruptive selection.
• Intermediate phenotypes are at the greatest disadvantage when the resources themselves are bimodal — they are 'jacks of both trades and masters of none'.
• Disruptive selection can be a precursor to sympatric speciation if the two sub-populations become reproductively isolated.

Common Mistakes

• Naming the type of selection as 'directional' — that is a shift in one direction, not a split into two peaks.
• Naming it as 'stabilising' — the opposite; stabilising reduces variance.
• Saying only that 'birds with bigger bills survive' without also addressing the small-billed birds — disruptive selection favours both extremes.
• Failing to mention the intermediate disadvantage; this is the key feature that distinguishes disruptive from directional selection.
• Using vague terms like 'adapt' or 'evolve' without explaining the mechanism of differential survival and reproduction.

Things to Be Careful About

• Use the precise term 'disruptive (selection)' or 'diversifying (selection)'.
• The mark scheme credits any four of six points, but a complete answer hits all of them: type, selection pressure, competition, intermediate disadvantage, extreme advantage, change in allele frequency.

Background Concept

The endocrine system is one of the two major communication systems of the body (the other being the nervous system). Endocrine glands are ductless glands that secrete hormones — chemical messengers — directly into the blood. A hormone is defined as a chemical produced by an endocrine gland, transported in the blood, which acts on target cells carrying specific receptors, to produce a specific response. For a molecule to qualify as a cell-signalling molecule of the endocrine system it must therefore: (1) be made by an endocrine gland, (2) be released into the bloodstream, (3) be carried in the blood to distant target cells, (4) bind to specific receptors on those target cells, and (5) trigger a biological response.

Insulin is a peptide hormone produced by the β cells of the islets of Langerhans in the pancreas. The stimulus for its release is a rise in blood glucose concentration (e.g. after a carbohydrate-rich meal). Once released it travels in the blood and binds to insulin receptors on the plasma membrane of target cells — principally hepatocytes (liver), myocytes (muscle) and adipocytes (fat). Binding activates intracellular second-messenger cascades that increase glucose uptake (by inserting GLUT4 transporters into the membrane) and promote glycogenesis and lipogenesis, lowering blood glucose.

Understanding the Question

This is an "outline" question — the candidate must hit the key defining points of an endocrine cell-signalling molecule and apply them to insulin. The mark scheme (4 marks) allows any four from a list of creditable points. There is no graph to interpret; the answer comes from biological knowledge of how insulin works.

Approach

The most efficient way to score the four marks is to state, in order, the five classic defining features of a hormone and apply each to insulin. The examiner will accept any four, so the candidate should aim to cover as many as possible, naming the gland, the transport route, the target cells, the receptor interaction and the response.

Step-by-Step Reasoning

1. State that insulin is a hormone — this is the umbrella term that the rest of the points flesh out.
2. Name the source as an endocrine gland — the β cells of the islets of Langerhans in the pancreas. Naming the tissue shows understanding that endocrine glands are ductless.
3. Note the stimulus for release (raised blood glucose), which demonstrates that secretion is regulated rather than continuous.
4. State that insulin travels in the blood — this is what makes it endocrine rather than paracrine.
5. State that it binds to specific receptors on target cells (liver, muscle, adipose) — the receptor concept is essential because it explains why only certain cells respond.
6. State that binding triggers a response (e.g. increased glucose uptake) — closing the loop from signal to effect.

Key Takeaways

• A hormone is a chemical messenger secreted by an endocrine gland into the blood that acts on specific target cells with the appropriate receptor.
• Insulin fits every one of those criteria: β cells of pancreas → blood → insulin receptors on liver, muscle and adipose cells → increased glucose uptake.
• "Outline" means a brief statement of the main points, not a detailed essay.

Common Mistakes

• Writing "insulin is secreted by the pancreas" without specifying β cells or the islets of Langerhans — the mark scheme specifically credits the named tissue.
• Saying insulin "tells cells what to do" without mentioning a receptor — the receptor is the mark-bearing point.
• Confusing endocrine (blood-borne, distant target) with paracrine (local diffusion) signalling.
• Describing nervous-system features (e.g. "carried along neurones") — insulin is not a neurotransmitter.

Things to Be Careful About

• Use the precise term "hormone" rather than "chemical" or "messenger".
• Specify that transport is in the blood (plasma), not just "in the body".
• Receptors are on the target cell — not "in" or "near" — to make the mechanism clear.

Background Concept

All living organisms are placed in one of three domains: Archaea, Bacteria and Eukarya. The first two domains contain only prokaryotic organisms — single-celled organisms whose DNA is not enclosed within a membrane-bound nucleus. Although Archaea and Bacteria look superficially similar under the microscope (both are small, simple prokaryotes), molecular and biochemical studies have shown that they are profoundly different. Carl Woese proposed the three-domain classification in 1977, primarily on the basis of differences in ribosomal RNA (rRNA) sequences. The molecular distinctions between the two prokaryotic domains are now known to extend to their cell walls, their membrane lipids and many aspects of their metabolism.

Key distinguishing features:

• Membrane lipids — Bacterial phospholipids have fatty acids attached to glycerol via ester bonds (the same linkage as in eukaryotes). Archaeal membrane lipids are attached via ether bonds, and their isoprenoid chains can form a monolayer rather than a bilayer, giving Archaea greater stability in extreme environments.
• Ribosomal RNA — Bacteria have a single rRNA operon giving one type of ribosome. Archaea have three rRNA types and their ribosomes/proteins are more similar to those of eukaryotes (which is one reason Woese argued the three domains should be separate).
• Cell wall — Almost all Bacteria have a cell wall containing peptidoglycan (a polymer of sugars and amino acids). Archaea lack peptidoglycan; their walls are made of other polymers such as pseudopeptidoglycan, polysaccharides or proteins (S-layers).

Understanding the Question

The command word here is "describe" (worth 3 marks). The question asks for features that are used to classify prokaryotes into the two domains. The mark scheme explicitly says "with reference to the two prokaryotic domains", so candidates must name the two domains (Archaea and Bacteria) and then state the contrasting features. A common error is to give only one side of the comparison (e.g. "Archaea have ether-linked lipids") without giving the corresponding bacterial feature.

Approach

To earn all three marks, list the two domains first (this is the foundation of the comparison) and then pick the strongest three contrasting features from those on the mark scheme:

1. Membrane lipid linkage (ester vs ether)
2. rRNA type (one vs three) — or the alternative phrased as similarity to eukaryotes
3. Cell wall composition (peptidoglycan vs not peptidoglycan)

State each as a paired contrast (Bacteria: …, Archaea: …) so both sides of the comparison are credited.

Step-by-Step Reasoning

• Mark point 1 (name the domains): The two prokaryotic domains are Archaea and Bacteria. This is worth one mark and must come first to anchor the comparison.
• Mark point 2 (membrane lipids): Bacteria have ester-linked lipids, while Archaea have ether-linked (or non-ester) lipids. The mark scheme accepts "not ester-linked" as an alternative for the Archaea side.
• Mark point 3 (rRNA): Bacteria have a single type of rRNA, while Archaea have three types of rRNA. The mark scheme also accepts the alternative formulation that archaeal rRNA/ribosomes are more similar to eukaryotic ones.
• Mark point 4 (cell wall): Bacterial cell walls contain peptidoglycan; archaeal cell walls do not. "Different" or "absent" is accepted for Archaea.

Because the question is out of 3 marks, only three of the above are required. A complete answer might give all four; the strongest three to learn are membrane lipids, rRNA and peptidoglycan.

Key Takeaways

• The three domains are Archaea, Bacteria and Eukarya; only the first two are prokaryotic.
• Archaea and Bacteria are biochemically distinct: ester vs ether membrane lipids, 1 vs 3 rRNA types, presence vs absence of peptidoglycan in the cell wall.
• Always give both sides of a comparison when describing a difference between two named groups.

Common Mistakes

• Stating only the Archaea feature (e.g. "Archaea have ether-linked lipids") without the corresponding bacterial side. The mark scheme requires the contrast.
• Confusing peptidoglycan with cellulose or chitin. Cellulose is a plant cell-wall polysaccharide, chitin is in fungal cell walls — only peptidoglycan is the bacterial marker.
• Saying Archaea have no cell wall — they do, but it lacks peptidoglycan.
• Including eukaryotic features (nucleus, membrane-bound organelles) — the question explicitly restricts the answer to prokaryotic differences.

Things to Be Careful About

• Use the precise terms: "peptidoglycan", "ester-linked", "ether-linked", "rRNA".
• The mark scheme allows error-carried-forward style alternatives, so "Archaea rRNA similar to eukaryotic" also scores on the rRNA point.
• An AVP (additional valid point) is credited — e.g. differences in RNA polymerase, in flagellin structure, or in the initiator tRNA. Only mention these if you are sure of the detail.
• "Describe" does not require reasons; it requires a clear, structured statement of the features.

Background Concept

The taxonomic hierarchy places every named organism into a series of nested ranks. For CIE 9700, the principal ranks to know are:

$\text{domain} > \text{kingdom} > \text{phylum} > \text{class} > \text{order} > \text{family} > \text{genus} > \text{species}$

The binomial scientific name of a species has two parts written in italics: the genus (capital letter) followed by the specific epithet (lower-case). For example, Panthera tigris — the genus is Panthera and the specific epithet is tigris. Both parts are needed to identify the species uniquely, but only the first part of the binomial gives the genus.

The three domains are Archaea, Bacteria and Eukarya. All animals — including the tiger — are eukaryotes, so the domain is Eukarya. Within Eukarya, the five kingdoms are commonly listed as Protoctista, Fungi, Plantae, Animalia and (sometimes) Monera, with multicellular heterotrophs placed in Animalia.

Understanding the Question

The candidate is shown a photograph of a tiger and given its scientific name Panthera tigris. Table 9.1 is partially completed: phylum, class, order and family are filled in. Three entries are blank: domain, kingdom and genus. The mark scheme awards 1 mark for each correct entry, giving 3 marks total.

The genus is read directly from the binomial (Panthera). The domain (Eukarya) and kingdom (Animalia) must be recalled from the classification of an animal.

Approach

1. Genus — read the first word of the binomial Panthera tigris: it is Panthera. Note that the species epithet tigris is not the genus; only the first word is.
2. Kingdom — the tiger is a multicellular, heterotrophic, motile organism with no cell wall, no chlorophyll and no true tissues other than the standard animal tissues. This places it in Animalia.
3. Domain — any organism with membrane-bound organelles and a true nucleus (i.e. any animal, plant, fungus or protoctistan) belongs to Eukarya.

Step-by-Step Reasoning

• Panthera tigris: a tiger is a large, carnivorous mammal native to Asia. The image confirms it is a mammal (fur, four limbs, tail).
• The genus is the first word of the binomial, Panthera.
• The kingdom Animalia contains all multicellular heterotrophs that develop from a blastula and lack cell walls.
• The domain Eukarya contains all organisms whose cells have a true membrane-bound nucleus, distinguishing them from the prokaryotic Archaea and Bacteria.

Key Takeaways

• The genus is always the first word of a binomial; the second word is the specific epithet, which by itself does not give the genus.
• All animals belong to the kingdom Animalia and the domain Eukarya.
• The full hierarchy for Panthera tigris is: Eukarya > Animalia > Chordata > Mammalia > Carnivora > Felidae > Panthera > tigris.

Common Mistakes

• Writing the species epithet (e.g. tigris) in the genus box — a very common error that loses the mark. The genus is the first word of the binomial only.
• Omitting italics in the genus — the binomial must always be italicised, with a capital on the genus and a lower-case specific epithet.
• Putting "Animal" or "Mammalia" as the kingdom — the kingdom rank for a tiger is Animala (the taxonomic rank), not the common description.
• Writing "Eukaryota" or "Eukarya" — both forms are accepted, but consistency with the syllabus (Eukarya) avoids ambiguity.

Things to Be Careful About

• The genus name must be italicised and begin with a capital letter.
• The species epithet alone does not identify the genus; several species share the epithet across different genera (e.g. Felis tigris is a different, outdated classification of the tiger).
• When the question gives a binomial, read the first word for the genus; do not be distracted by the rest.
• The phylum Chordata is already supplied — do not change it. Only the blank cells should be filled in.

Background Concept

A neurone's membrane potential is the potential difference (pd) across its plasma membrane, measured in millivolts (mV), with the inside taken as the reference. Three named states appear in this question:

• Resting potential (~ –70 mV): the membrane is polarised — outside positive, inside negative. It is set up and continually maintained by the sodium–potassium pump (an active-transport carrier that uses ATP to move 3 Na⁺ out and 2 K⁺ in per cycle) and by the membrane's greater permeability to K⁺ than to Na⁺ at rest, so K⁺ leaks out faster than Na⁺ leaks in.
• Depolarisation (rising to +40 mV): a stimulus above threshold opens voltage-gated Na⁺ channels; Na⁺ rushes in down its electrochemical gradient, reversing the polarity so the inside becomes positive relative to the outside.
• Hyperpolarisation (overshoot to –90 mV): voltage-gated Na⁺ channels inactivate and voltage-gated K⁺ channels open more slowly; K⁺ continues to leave, driving the pd even more negative than the resting value before the K⁺ channels close and the Na⁺/K⁺ pump restores –70 mV.

The terminology in the mark scheme is precise: candidates must distinguish diffusion (passive, down a gradient) from active transport / pump (against a gradient, ATP-using), and must say which ion moves and in which direction.

Understanding the Question

Table 10.1 lists three pd values, and part (a) asks you to describe the events that cause each potential difference. The command word "describe" here is being used in an extended sense — examiners want you to identify the named state (resting, depolarisation, hyperpolarisation) and then explain the underlying ion movements, channels and pump activity that produce it. The 7 marks are split 3 (X) + 2 (Y) + 2 (Z) — you do not need to give an equal amount for each.

Approach

For each point, work in the same order:

1. Name the state (resting, depolarised, hyperpolarised).
2. Identify which ion channels/pumps are active.
3. State the direction of ion movement.
4. Explain the consequence for the membrane potential.

Use precise ion symbols (Na⁺, K⁺) and the words "diffuses" (for passive movement) and "active transport / pump" (for ATP-driven movement). Avoid vague phrases like "ions move".

Step-by-Step Reasoning

Point X (–70 mV):

• The Na⁺/K⁺ pump uses ATP to move 3 Na⁺ out per 2 K⁺ in. Because 3 positive charges leave for every 2 that enter, the outside accumulates positive charge relative to the inside — this is the basis of the resting potential.
• In addition, the resting membrane is more permeable to K⁺ than to Na⁺, so K⁺ diffuses out faster than Na⁺ diffuses in, reinforcing the inside-negative state.
• Net result: outside positive, inside negative, pd ≈ –70 mV. The pump does not create –70 mV alone — it maintains a steep Na⁺/K⁺ gradient that, combined with selective permeability, produces the resting pd.

Point Y (+40 mV):

• A stimulus that exceeds the threshold opens voltage-gated Na⁺ channels in the membrane.
• Na⁺ ions diffuse into the axon down both their concentration gradient (high outside) and electrical gradient (attracted to the negative inside).
• This inward positive charge reverses the membrane polarity, producing a pd of about +40 mV — this rising phase is depolarisation.

Point Z (–90 mV):

• Voltage-gated Na⁺ channels inactivate (close and become refractory), while voltage-gated K⁺ channels open (they are slower to open, so they peak as Na⁺ channels are closing).
• K⁺ ions diffuse out of the axon, carrying positive charge out faster than Na⁺ is moving in.
• The inside of the membrane becomes more negative than the resting value (–90 mV rather than –70 mV) — this is hyperpolarisation (the "undershoot").
• The K⁺ channels then close, the Na⁺/K⁺ pump restores the ion gradients, and the membrane returns to –70 mV.

Key Takeaways

• Resting potential = Na⁺/K⁺ pump + greater K⁺ permeability → inside negative (~ –70 mV).
• Depolarisation = voltage-gated Na⁺ channels open, Na⁺ diffuses in → inside positive (~ +40 mV).
• Hyperpolarisation = voltage-gated K⁺ channels open, K⁺ diffuses out → inside more negative than rest (~ –90 mV).
• The mark scheme rewards specific ion names, direction of movement and the distinction between diffusion and active transport.

Common Mistakes

• Writing "ions move" without naming the ion (Na⁺ or K⁺) or the direction — both are required for the mark.
• Saying the pump "creates" the resting potential on its own — it sets up the gradients, but selective permeability (K⁺ leak > Na⁺ leak) is what actually produces –70 mV.
• Confusing depolarisation with hyperpolarisation, or with the falling phase (repolarisation back to –70 mV).
• Writing "K⁺ enters" or "Na⁺ leaves" at any of the three points — these are wrong directions and score 0 for that point.
• Calling the –90 mV point "repolarisation" — repolarisation is the fall from +40 mV back towards –70 mV; the –90 mV value is past the resting line and is hyperpolarisation.

Things to Be Careful About

• Use the correct symbols: Na⁺, K⁺, with the plus sign as a superscript.
• State both the channel and the ion direction for full credit at Y and Z.
• The pump ratio (3 Na⁺ out : 2 K⁺ in) is a specific marking point — quoting it earns the second point at X, and missing the "out" / "in" direction is a common way to lose it.
• Hyperpolarisation is a separate mark point from depolarisation; do not combine them into a single confused statement.
• "Voltage-gated" is not strictly required at Y/Z by the mark scheme, but using it shows precise knowledge and is a safe wording.

Background Concept

An action potential is not "pushed" down the axon like electricity in a wire — it is regenerated at every successive patch of membrane. The driving force for this regeneration is the difference in potential between the currently depolarised patch (inside positive) and the neighbouring resting patch (inside negative). Ions flow between these two regions through the axoplasm and the extracellular fluid, and this movement of charge is called a local circuit current (or local current).

Where the local current crosses the membrane of the next resting patch, it depolarises that membrane. If the depolarisation reaches threshold (~ –55 mV), voltage-gated Na⁺ channels in that next patch open and an action potential is fired there. The previous patch is in its refractory period, so the impulse can only travel in one direction.

In an unmyelinated axon there are no nodes of Ranvier; voltage-gated Na⁺ channels line the entire length of the membrane, and the action potential is regenerated at every adjacent patch in turn — a relatively slow, continuous process.

Understanding the Question

Part (b) follows directly from part (a): once one section of the membrane has fired an action potential, how is the next section triggered? The question specifically says "unmyelinated neurone" and "adjacent section of the axon", so the answer must be about local circuits between neighbouring patches of membrane, not saltatory conduction (which belongs to part (c)).

The command word is suggest — examiners want a concise, mechanistically correct explanation, not a list of all possible ideas.

Approach

Two marking points are available:

1. Name the mechanism: local circuit / local current.
2. State the direction of current flow: from the depolarised (positive) region to the resting (negative) region.

The second point is what makes the next patch reach threshold. State both, in that order, for full marks.

Step-by-Step Reasoning

1. After point Y, the membrane at that location has its pd reversed (inside ≈ +40 mV, outside negative). The membrane either side of it is still at resting potential (≈ –70 mV, outside positive, inside negative).
2. Positive charge inside the axon flows away from the depolarised patch along the axoplasm toward the next resting patch, while in the extracellular fluid positive charge flows in the opposite direction toward the depolarised patch. This is the local circuit current.
3. The current crossing the membrane of the next patch depolarises it. If the depolarisation is big enough to reach threshold, voltage-gated Na⁺ channels in that next patch open and a full action potential is generated there.
4. Meanwhile, the previous patch is in its refractory period (Na⁺ channels inactivated, K⁺ channels still open), so the impulse cannot travel backwards — it propagates in one direction only.

Key Takeaways

• Action potentials are propagated, not conducted, along an axon — each new section of membrane regenerates the impulse.
• The trigger is a local circuit current from the depolarised region to the adjacent resting region.
• The previous section's refractory period enforces one-way travel.

Common Mistakes

• Saying "the electrical impulse travels along the axon" without mentioning local circuits or regeneration at the next patch — too vague for the mark.
• Confusing the direction: saying the current flows from the resting area to the depolarised area.
• Describing saltatory conduction / nodes of Ranvier here — the question is about an unmyelinated axon, so this earns no credit.
• Saying "ions diffuse along the axon" — diffusion of ions across the membrane is not the same as the local circuit current flowing in the axoplasm.

Things to Be Careful About

• "Local circuit" is the technical term; "local current" is accepted, but "electrical signal" or "nerve impulse moves" is too vague.
• Specify the polarity at each end of the current path: "from the depolarised (positive) area to the resting (negative) area".
• The question says "adjacent section of the axon" — make it clear the new action potential is generated in the next patch of membrane, not by the original one.

Background Concept

In a myelinated axon, specialised glial cells — Schwann cells in the peripheral nervous system (oligodendrocytes in the central nervous system) — wrap their plasma membranes around the axon in many tight layers, forming the myelin sheath. The sheath is a very good electrical insulator because the lipid-rich membrane layers prevent ion flow across the wrapped regions of the axolemma.

The sheath is not continuous: there are small gaps every 1–2 mm called the nodes of Ranvier, where the axolemma is exposed to the extracellular fluid and where the voltage-gated Na⁺ (and K⁺) channels are clustered at very high density (orders of magnitude higher than in unmyelinated axons). Because ion exchange can only occur at these nodes, the action potential cannot be regenerated along the myelinated stretches. Instead, the depolarisation at one node sets up a local circuit that travels passively (electrotonically) through the insulated axoplasm to the next node, where it depolarises that membrane to threshold and triggers a new action potential.

This node-to-node regeneration is called saltatory conduction (from Latin saltare, "to jump"). Because each node is depolarised almost instantaneously by the incoming local current — without having to regenerate the impulse at every micrometre of membrane — the overall speed of conduction is much greater than in an unmyelinated axon (up to ~120 m s⁻¹ vs ~1–2 m s⁻¹). A faster impulse means a faster reaction time and a faster muscle contraction, which is biologically advantageous for an animal responding to its environment.

Understanding the Question

The question stem reminds you that vertebrate motor neurones are myelinated and asks you to explain why this is an advantage — i.e. how does the myelin sheath improve the functioning of the motor neurone? The command word "explain" requires you to give a reason and a consequence, not just a label. Three marks are available, so two or three linked points are needed.

Approach

Build the answer as a causal chain:

1. Myelin insulates the axon (structure).
2. Therefore the action potential can only be regenerated at the nodes of Ranvier (consequence 1).
3. The impulse effectively jumps from node to node — saltatory conduction (consequence 2, named phenomenon).
4. This makes impulse transmission much faster than in an unmyelinated axon (consequence 3 — the advantage).
5. A faster impulse gives a faster response / muscle contraction (biological benefit for a motor neurone).

The mark scheme credits any three of these. For full marks, name saltatory conduction explicitly.

Step-by-Step Reasoning

1. Insulation: the lipid-rich, multilamellar myelin sheath prevents ion movement across the membrane in the wrapped regions, so no action potential can be generated under the sheath. The membrane behaves like the insulation around an electrical cable.
2. Nodes of Ranvier: at the small gaps between adjacent Schwann cells, the axolemma is exposed and densely packed with voltage-gated Na⁺ channels. These are the only sites where the action potential can be regenerated.
3. Local circuits between nodes: when one node fires, the depolarisation drives a local current through the axoplasm (and back through the extracellular fluid) to the next node. The high density of Na⁺ channels at the next node means that even a small depolarising current quickly brings the membrane to threshold.
4. Saltatory conduction: because each node fires in turn, and the depolarisation "jumps" the insulated stretches between them, the action potential appears to leap along the axon. This is far faster than the continuous regeneration seen in unmyelinated axons.
5. Functional advantage: for a motor neurone, faster impulse conduction means a shorter reaction time and more rapid, precisely timed muscle contraction. For a predator or prey animal this can be the difference between catching a meal and being caught; in any vertebrate it allows swift, coordinated movement.

Key Takeaways

• Myelin is an insulator; voltage-gated Na⁺ channels sit at the nodes of Ranvier.
• Action potentials "jump" between nodes — this is saltatory conduction.
• Saltatory conduction increases the speed of nerve impulses compared with unmyelinated axons.
• A faster impulse gives a faster biological response (e.g. quicker muscle contraction).

Common Mistakes

• Saying the myelin sheath "speeds up the impulse" without explaining how (insulation + nodes + jumping) — this is the consequence without the mechanism, and only earns one of the three marks.
• Omitting the word "saltatory conduction" — it is the named phenomenon the mark scheme explicitly rewards.
• Saying the impulse "travels along the myelin sheath" or "in the myelin" — the impulse is in the axon; the myelin is outside it.
• Confusing the function of myelin with the function of the synapse — myelin is about speed along the axon, not about transmitting the signal to the next neurone.
• Saying the myelin sheath "prevents the action potential" — it does not; it only prevents regeneration of the action potential under the myelinated regions, while allowing it at the nodes.

Things to Be Careful About

• The question specifies a motor neurone — credit goes to answers that connect the faster impulse to a faster muscle contraction or faster response, not just "faster thinking".
• "Insulator" is the precise word the mark scheme rewards; "protects" or "covers" is too vague.
• "Nodes of Ranvier" should be named explicitly, not just "gaps".
• "Saltatory conduction" is a technical term worth using in full — it is a discrete marking point (mark 4 in the scheme) and a single word that shows understanding.