Biology 9700/41 — October/November 2025

Background Concept

When two genes are on separate (non-homologous) chromosomes, they obey Mendel's law of independent assortment: during meiosis I, the bivalents orient randomly at the metaphase plate, so each gamete receives one allele from each locus independently. A dihybrid (heterozygous at two loci) therefore produces $2 \times 2 = 4$ genetically distinct gametes in equal proportions. A cross between two dihybrids produces a characteristic $9:3:3:1$ phenotypic ratio in the F2, provided both genes show complete dominance and are unlinked.

Understanding the Question

The P generation is built from two homozygous parents with different dominant phenotypes. The black long-fur parent must be aaBB (long because recessive aa, black because dominant BB). The chocolate short-fur parent must be AAbb (short because dominant AA, chocolate because recessive bb). Crossed, every F1 receives A from one parent and a from the other, and B from one parent and b from the other — all F1 are therefore AaBb and phenotypically short, black. The question asks the student to:

1. Show the F1 × F1 gametes.
2. Complete the Punnett square to give all 16 F2 genotypes.
3. State the F2 phenotype ratio with a key linking each phenotype to its genotypes.

Approach

• Treat the F1 × F1 cross as a standard dihybrid cross.
• Use independent assortment to list the four gametes per parent.
• Fill the 4 × 4 grid.
• Group the 16 F2 cells into the four phenotype classes by inspection of the genotype.
• Express the phenotype counts as a ratio and attach a key.

Step-by-Step Reasoning

Step 1 — Gametes. Each F1 (AaBb) makes four gametes:

• AB — one from each dominant allele
• Ab — dominant length, recessive colour
• aB — recessive length, dominant colour
• ab — both recessive

All four are equally likely because the loci are on separate chromosomes.

Step 2 — Punnett square. Combining each pair of gametes (4 × 4 = 16 F2 genotypes) gives the standard dihybrid grid: one AABB, two AABb, two AaBB, four AaBb, one AAbb, two Aabb, one aaBB, two aaBb, one aabb.

Step 3 — Phenotype classes. Reading off the grid:

• Short, black: any cell with at least one A and at least one B → AABB, AABb, AaBB, AaBb = 9 cells.
• Short, chocolate: at least one A and bb → AAbb, Aabb = 3 cells.
• Long, black: aa with at least one B → aaBB, aaBb = 3 cells.
• Long, chocolate: aabb only = 1 cell.

Step 4 — Ratio. $9 : 3 : 3 : 1$, with a key linking each phenotype to the genotype(s) (e.g. A_B_ for short black, A_bb for short chocolate, aaB_ for long black, aabb for long chocolate).

Key Takeaways

• Two unlinked dihybrids produce a $9:3:3:1$ F2 ratio, and only a $9:3:3:1$ ratio (or its variants under codominance / epistasis) — deviation suggests linkage, a lethal allele, or a counting error.
• Showing the gametes is essential, because independent assortment (not random allele pairing) is what generates the four gamete types.
• The phenotype key is required so the ratio is unambiguous.

Common Mistakes

• Showing only one or two gamete types per parent — the grid then collapses and the ratio is wrong.
• Filling cells with phenotype abbreviations instead of genotypes.
• Quoting the ratio as $1:1:1:1$ or $3:1$ — those are monohybrid, not dihybrid, outcomes.
• Forgetting to attach a key linking each phenotype to its underlying genotype(s).

Things to Be Careful About

• Both F1 parents are heterozygous at both loci, so each makes four gametes, not two.
• Independent assortment is conditional on the two loci being on separate chromosomes — the stem states this explicitly.
• Genotypes must use paired allele symbols (AaBb, not AB/ab).
• The phenotype ratio must be presented in the same order in both the numbers and the key (short black first, long chocolate last).

Background Concept

Not every gene codes for a protein that does a metabolic or structural job in the cell. Some genes code for proteins whose job is to regulate the expression of other genes. These two categories are called structural genes and regulatory genes.

• A structural gene is transcribed and translated into a polypeptide that performs a direct role in the cell — for example, an enzyme (such as rubisco or tyrosinase), haemoglobin, collagen, or an ion-channel protein.
• A regulatory gene is transcribed and translated into a protein (typically a transcription factor or a repressor) whose role is to bind to DNA near other genes and switch their transcription on or off, or modulate it. In prokaryotes, regulatory genes encode repressors such as the lac repressor; in eukaryotes, they encode transcription factors that bind promoters or enhancers.

Understanding the Question

The question asks for the difference between a structural gene and a regulatory gene. The mark scheme rewards two points: one for what each codes for, and one for the regulatory gene's role in controlling other genes' expression. Both halves of the contrast must be present for full marks.

Approach

Identify the two gene types and state, for each:

1. The product it codes for.
2. The role of that product in the cell.

Step-by-Step Reasoning

• A structural gene codes for a protein that has a direct function in the cell — most often an enzyme, but also structural proteins (collagen, keratin), transport proteins (haemoglobin, membrane pumps), or signalling molecules. Such proteins are the molecular workhorses of metabolism and structure.
• A regulatory gene codes for a protein that does not carry out a metabolic role itself; instead it regulates the expression of other (usually structural) genes. In prokaryotes this is the role of the lacI gene, which produces the lac repressor; in eukaryotes, regulatory genes encode transcription factors that bind DNA sequences to switch transcription on or off.

Key Takeaways

• The contrast is product and role: structural genes produce functional/structural proteins; regulatory genes produce regulators of gene expression.
• "Regulatory" describes what the protein does (control other genes), not a separate class of DNA sequence.

Common Mistakes

• Vague answers such as "a structural gene makes proteins" without naming structural/functional, and "a regulatory gene controls things" without naming transcription factors or other genes.
• Saying that regulatory genes "control cell activities" — too broad; they specifically control gene expression.
• Confusing the two (e.g. saying a structural gene controls other genes).

Things to Be Careful About

• Both halves of the description must be present: the product and the function.
• "Controls expression of other genes" is the precise wording; "controls traits" or "controls metabolism" is not credited.

Background Concept

Each photosynthetic pigment absorbs light most strongly at particular wavelengths, and this is shown by its absorption spectrum (percentage absorption plotted against wavelength). The shape of the spectrum is determined by the pigment's molecular structure:

• Chlorophylls (a and b) have a porphyrin head with a central Mg²⁺ and absorb strongly in the blue (~430 nm) and red (~660 nm) regions, with a gap in the green.
• Carotenoids (β-carotene and xanthophyll) are long isoprenoid chains with conjugated double bonds; they absorb mainly in the blue region (400–500 nm) and appear yellow / orange because they transmit the longer wavelengths.

The combined absorption of all leaf pigments covers most of the visible spectrum, which is why leaves look green (the wavelengths they do not absorb are reflected / transmitted).

Understanding the Question

Fig. 3.2 shows the absorption spectra of chlorophyll a (solid line), chlorophyll b (dash-dot line) and carotenoids (dotted line). This question asks you to compare just chlorophyll a with the carotenoids — similarities and differences. The mark scheme allows up to two similarity points and up to three difference points, for a total of four marks.

Approach

Read each curve carefully:

• Note the wavelengths at which both curves are above zero.
• Note the position(s) of the peak(s) of each curve.
• Note any wavelength ranges where one curve is above the other.

Then organise the answer as 'Similarities' followed by 'Differences', picking the strongest, most specific points and quoting wavelength ranges or peak positions.

Step-by-Step Reasoning

1. Both curves are well above zero between roughly 400 nm and 500 nm — the blue region. → Similarity 1.
2. Both curves fall close to zero between about 525 nm and 600 nm — the green / yellow region — so both pigments absorb very little green / yellow light. → Similarity 2.
3. Chlorophyll a has a second prominent peak near 660 nm (red), where the carotenoid curve is essentially zero. → Difference 1.
4. Chlorophyll a's strongest blue peak is at about 430 nm; the carotenoids' strongest peak is at about 460–465 nm — they peak at different wavelengths in the blue region. → Difference 2.
5. Chlorophyll a has two distinct peaks (blue and red); carotenoids have three peaks, all in the blue / blue-green region. → Difference 3.
6. The range of wavelengths absorbed is wider for chlorophyll a than for the carotenoids. → Additional valid difference.

Any four of these clearly stated points — mixing similarities and differences — earn the four marks.

Key Takeaways

• Compare questions need both similarities and differences.
• Always quote the wavelength range (e.g. '400–475 nm') when describing a region of the spectrum.
• The absorption spectrum of a pigment is a fingerprint of its molecular structure.
• Different pigments complement each other: chlorophyll a absorbs where carotenoids do not (red) and vice versa (the 445–525 nm range where carotenoids still absorb but chlorophyll a does not).

Common Mistakes

• Saying 'chlorophyll absorbs more light than carotenoids' without specifying a wavelength range — the answer depends on wavelength.
• Ignoring the red peak of chlorophyll a.
• Confusing absorption spectra with action spectra (action spectra plot the rate of photosynthesis against wavelength, not absorption).
• Using only colour names ('blue', 'red') without the corresponding wavelength range.

Things to Be Careful About

• For each difference, frame it as a comparison (chlorophyll a X / carotenoids Y), not as two unrelated statements.
• Use 'between 400 and 475 nm' rather than just 'blue'.
• For chlorophyll a's red peak, write 'between 600 and 680 nm' or 'red region'; carotenoids essentially do not absorb above about 530 nm.
• Three differences can be drawn from the mark scheme; choosing any three is fine.

Background Concept

Gibberellins (GAs) are a class of plant hormones with multiple roles: stem elongation, breaking dormancy, and (importantly here) mobilising food reserves in germinating cereal seeds. In a barley- or wheat-type seed, the embryo is small and the bulk of the seed is starchy endosperm. The embryo cannot use starch directly, so it must first be hydrolysed to sugars by the enzyme α-amylase.

α-amylase is not stored pre-formed in the aleurone layer; it is synthesised after gibberellin reaches the aleurone. The molecular pathway is now well characterised:

• The gibberellin receptor GID1 sits inside aleurone cells.
• In the absence of gibberellin, DELLA proteins act as repressors, holding back transcription factors called PIFs (phytochrome-interacting factors).
• When gibberellin binds GID1, the DELLA protein is recruited, ubiquitinated and broken down by the proteasome.
• PIFs are freed to enter the nucleus and switch on target genes — including the gene for α-amylase.
• Newly synthesised α-amylase is secreted into the endosperm where it catalyses starch → maltose.
• Maltose → glucose, and the embryo uses this sugar for ATP production (respiration) and as a carbon skeleton for new cells (growth).

Understanding the Question

The stem says that in some plant species light absorption by the seed stimulates gibberellin production in the embryo. You are asked to describe the role of this gibberellin in germination — what gibberellin does after it has been produced. Four marks are available; the mark scheme accepts any four of seven creditable points, so an answer that lists the pathway in order is the safest.

Approach

Outline the pathway in logical order, starting from gibberellin moving out of the embryo and ending with the embryo receiving glucose. Aim to include the modern molecular detail (GID1, DELLA, PIF) as well as the classical end-point (α-amylase, starch → maltose).

Step-by-Step Reasoning

1. Gibberellin synthesised in the embryo diffuses to the aleurone layer (a thin layer of living cells just inside the seed coat, surrounding the starchy endosperm).
2. Inside aleurone cells, gibberellin binds to its receptor, GID1.
3. The gibberellin–GID1 complex recruits DELLA repressor proteins and triggers their breakdown (by the ubiquitin–proteasome system).
4. With DELLA removed, PIF transcription factors are no longer held back; they enter the nucleus and bind to promoter regions of target genes.
5. One target gene encodes α-amylase. Transcription is switched on, the mRNA is translated, and the enzyme is synthesised.
6. α-amylase is secreted from aleurone cells into the starchy endosperm where it hydrolyses starch to maltose.
7. Maltose is further broken down to glucose, which is taken up by the embryo and used in respiration (to make ATP) and as building blocks for new cells (growth).

Key Takeaways

• Gibberellin's role in germination is to mobilise stored food: it triggers the synthesis of α-amylase in the aleurone layer.
• The signalling pathway uses the GID1 receptor and the DELLA / PIF repressor system — a classical example of how a hormone can switch on specific genes.
• The embryo does not directly digest the endosperm; the aleurone layer is the intermediary tissue that produces the digestive enzyme.
• The sugars released by this process fuel both respiration and growth of the embryo, allowing the seedling to establish before it has any photosynthetic capacity.

Common Mistakes

• Saying 'gibberellin breaks down starch' — gibberellin does not digest starch directly; it triggers the synthesis of α-amylase, which then digests starch.
• Confusing gibberellin with auxin (auxin promotes elongation growth in shoots; gibberellin promotes germination and stem elongation).
• Omitting the aleurone layer — the answer should specify where gibberellin acts.
• Stating only one or two points when four marks are available — list the steps in the pathway.
• Saying 'gibberellin turns on the amylase gene' with no mention of DELLA / PIF — modern mark schemes expect the molecular detail.

Things to Be Careful About

• Use precise terminology: GID1 (the receptor), DELLA (the repressor), PIF (the transcription factor), α-amylase (the enzyme), aleurone (the tissue).
• Make the sequence of events clear: gibberellin → receptor → DELLA breakdown → PIF release → gene activation → α-amylase → starch → maltose → embryo.
• Spelling: 'aleurone' (not 'alerone'), 'gibberellin' (not 'giberelin'), 'DELLA' (all capitals — it is a protein-motif name), 'amylase' (not 'amalase').
• For four marks, the mark scheme accepts any four creditable points; including more than four shows strong understanding.

Background Concept

The CPD103 locus in Canis lupus is introduced with the information that each wolf inherits two copies — one from each parent. Three core genetics terms must be defined to set up the rest of the question.

• Diploid — a cell or organism whose cells contain two complete sets of chromosomes, and therefore two copies (alleles) of every autosomal gene. Most somatic cells of mammals are diploid (2n).
• Allele — one of two or more alternative nucleotide sequences (forms) of a gene that occupies the same locus on homologous chromosomes. The CPD103 gene has at least two alleles in the population: one associated with a black coat and one with a grey coat.
• Dominant allele — an allele that is expressed in the phenotype even when only one copy is present (i.e. in the heterozygote). The stem tells us that a single copy of the black allele is enough to produce a black coat, so the black allele is dominant over the grey allele.

Understanding the Question

This is a "state the term" item — three blanks, one mark each. The student simply has to supply the correct vocabulary word that matches the definition printed in the question. No biological reasoning is needed; the definitions are given in the question itself.

Approach

Read each definition, identify the matching term from the standard CIE genetics vocabulary, and write it in. The order of definitions mirrors the order in which the concepts are introduced in any genetics course: ploidy first, then allele, then dominance relationship.

Step-by-Step Reasoning

1. "An organism that has two copies of each gene" — this describes ploidy. Because there are two complete sets of chromosomes carrying two copies of every gene, the organism is diploid (2n). A gamete, with one set, is haploid (n).
2. "A form of a gene" — the technical term for a variant version of a gene at a particular locus is an allele. Both the black and grey versions of CPD103 are alleles of the same gene.
3. "A form of a gene that gives a phenotypic effect in a heterozygote" — the allele whose effect is visible in a heterozygote (one copy) is the dominant allele. A recessive allele is masked in a heterozygote. Because the question explicitly states that one copy of the black allele produces a black coat, the black allele is dominant.

Key Takeaways

• diploid = two sets of chromosomes (two copies of each gene)
• allele = one variant form of a gene
• dominant = expressed in the heterozygote; recessive = only expressed in the homozygote
• These three terms together allow a phenotype to be predicted from a genotype.

Common Mistakes

• Writing "gene" instead of "allele" for the second blank — a gene is the whole locus; an allele is a particular version of it.
• Writing "recessive" for the third blank — the question describes an allele that is expressed in a heterozygote, which is the definition of dominant.
• Writing "phenotype" or "genotype" for the first blank — these are not terms for the number of gene copies.

Things to Be Careful About

• "Dominant" alone is sufficient; "dominant allele" is also accepted but the bare word scores the mark.
• Do not write "heterozygous" or "homozygous" — these describe the genotype of the individual, not the dominance status of the allele.

Background Concept

Natural selection is the differential survival and reproduction of individuals in a population because of differences in their inherited characteristics. The four logical steps are:

1. Variation exists in the population — here, wolves have two coat-colour alleles at CPD103 producing grey or black phenotypes.
2. A selection pressure acts on that variation — Canine distemper virus (CDV), a lung pathogen spread from domestic dogs, is the selection pressure.
3. Individuals with the advantageous phenotype survive better — black wolves, whose CPD103 protein also defends against lung infection, are more likely to survive CDV.
4. Survivors reproduce and pass their alleles to the next generation, so the frequency of the advantageous allele increases over generations.

The stem also gives a crucial pleiotropic detail: the same protein that produces black coat colour also defends the lung against CDV. This is a key example of pleiotropy — one gene, multiple phenotypic effects — and it links coat colour directly to disease resistance.

Understanding the Question

The question supplies the observation: black wolves are proportionally more common in the south of the range (where domestic dogs are more numerous and CDV is more common) than in the north. It asks the candidate to explain, by natural selection, why this geographic trend has arisen. The command word "explain" requires both the mechanism and the link to the specific context (north–south gradient, domestic-dog density).

Approach

Write a connected explanation that follows the selection cycle, naming each element in turn: the selection pressure (CDV), the advantageous allele (black), the differential survival in the southern context, the differential reproduction, and the change in allele frequency. The mark scheme insists on the south context being made explicit at least once, and on pairing each survival/reproduction statement with the ORA for grey.

Step-by-Step Reasoning

1. Identify the selection pressure. CDV is a virus that causes serious lung disease and is transmissible from domestic dogs to wolves. It kills some wolves and prevents others from reproducing. This is the agent of natural selection.
2. Identify the advantageous phenotype. Because the CPD103 protein that produces black coat colour also defends the lung against CDV, black wolves (which carry at least one black allele) have greater resistance to CDV than grey wolves. The black allele therefore confers a selective advantage where CDV is common.
3. Establish the spatial context. Domestic dogs are more numerous in the south, so CDV transmission to wolves is higher in the south and lower in the north. The selective pressure is therefore stronger in the south.
4. Differential survival. In the south, where CDV is common, black wolves are more likely to survive infection than grey wolves. Grey wolves, lacking the protective CPD103 variant, are more likely to die. (This is the ORA pair the mark scheme requires.)
5. Differential reproduction. The surviving black wolves are more likely to reach reproductive age and produce offspring than the grey wolves that died. (Again, ORA.)
6. Change in allele frequency. The black wolves pass the black CPD103 allele to their offspring, so the frequency of the black allele — and hence the proportion of black wolves — increases in the southern population. Over generations this produces the observed north-to-south gradient.

The mark scheme asks for any four of these points; an excellent answer covers all six in a logical chain.

Key Takeaways

• Natural selection has four ingredients: variation, selection pressure, differential survival, differential reproduction.
• Pleiotropy can couple an apparently cosmetic trait (coat colour) to a survival trait (disease resistance), making the visible trait a marker of the underlying selective process.
• The intensity of selection varies with environment: stronger selection in CDV-rich south, weaker in CDV-poor north.

Common Mistakes

• Saying that wolves "chose" to be black or that black wolves "adapted" to CDV during their lifetime — natural selection acts on pre-existing heritable variation; individuals do not change their alleles.
• Talking about black wolves being better camouflaged or warmer — the question gives a different mechanism (the same protein defends against CDV); camouflage/temperature arguments are not credited.
• Omitting the south context — the mark scheme says "context must be established at least once"; a generic answer about selection without placing it in the southern range loses marks.
• Confusing survival with reproduction. Selection requires both differential survival AND differential reproductive success to change allele frequency.

Things to Be Careful About

• Use the term selection pressure explicitly when naming CDV.
• For the differential-survival and differential-reproduction points, the mark scheme allows the ORA form ("grey wolves more likely to die / fail to breed").
• "Allele" is the precise word; "gene", "DNA" or "characteristic" will not always be credited at the same point — match the mark scheme wording where it matters.

Background Concept

Antibodies as a record of past infection. When a vertebrate recovers from a viral infection, memory B-lymphocytes continue to produce specific antibodies against that virus for months or years. The presence of anti-CDV antibodies in a wolf's blood is therefore evidence that the wolf was once infected with — and survived — CDV. The proportion of a population carrying those antibodies is a measure of how much CDV the population has been exposed to over recent history.

Pleiotropy revisited. As established in part (b), the same CPD103 protein that produces black coat colour also defends the lung against CDV. So the visible trait (black coat) is a marker for the disease-resistance genotype, and differential mortality from CDV reshapes the phenotypic composition of the population.

Correlation and causation. A positive correlation between two variables does not prove that one causes the other, but in this case a clear biological mechanism (differential survival of black vs. grey wolves in CDV-rich populations) links the two. The line of best fit summarises the trend across several populations.

Understanding the Question

The question has two parts. First, the candidate must state the relationship shown by the line of best fit in Fig. 4.1 — its direction and form. Second, the candidate must suggest reasons for that relationship, using biological reasoning grounded in the information given earlier in the question. There are four marks: one for the relationship and three for the reasons.

Approach

• Read the graph. Identify the axes, the direction of the line, and the form (linear, positive). Describe the trend in plain language.
• Bridge to biology. Connect antibody prevalence → past CDV exposure → differential mortality of grey wolves → shift in the proportion of black wolves that survive.
• Use the pleiotropy from part (b). The same protein that gives black coat colour defends against CDV; this is the proximate cause of the differential mortality.

Step-by-Step Reasoning

1. State the relationship from the graph. The line of best fit on Fig. 4.1 rises from the origin towards the top-right corner: as the percentage of wolves with anti-CDV antibodies increases, the percentage of black wolves in the population also increases. The relationship is a positive (linear) correlation. Award 1 mark.
2. Antibodies indicate past CDV exposure. Wolves that have been infected with CDV and survived will have anti-CDV antibodies in their blood. A high percentage of the population with these antibodies therefore signals that the population has experienced a high level of CDV infection. Award 1 mark.
3. Differential mortality. In populations where CDV has been common, the grey wolves (which lack the protective black variant of CPD103) are more likely to have died from the disease. The black wolves, with the pleiotropic defence, are more likely to have survived. The surviving population therefore contains a higher proportion of black wolves. Award 1 mark.
4. Link to coat-colour gene. The CPD103 allele that gives the black coat produces a protein that also defends against CDV. Black wolves are therefore more likely to survive CDV infection and to be present in the population when antibody prevalence is measured. Award 1 mark.
5. (Optional AVP) The data are population averages across many populations; on a finer scale other factors (e.g. migration) could modify the trend, but the dominant explanation is differential survival under CDV pressure.

Key Takeaways

• A positive correlation between two variables can be explained by a shared underlying cause (here, CDV pressure) acting on one of the variables (mortality of grey wolves), with the other (antibody prevalence) being a marker of the cause's intensity.
• Antibody prevalence is a retrospective measure of pathogen exposure in a population.
• The same line of reasoning that explained the spatial gradient in part (b) — pleiotropic defence, differential survival, change in allele/phenotype frequency — also explains this graph.

Common Mistakes

• Writing only "positive correlation" with no biological explanation — that scores only the first mark; the other three marks require the reasons.
• Saying antibodies cause black coat colour, or that black coat colour causes antibody production. The two variables are correlated because each is independently linked to CDV exposure/survival.
• Saying the relationship is "directly proportional" without checking the line — the line on the figure does not pass through a defined constant of proportionality, so "positive correlation" is safer than "directly proportional".
• Forgetting to use the pleiotropy link from part (b). The candidate must explicitly state that the black-coat protein also defends against CDV; "black wolves are better" on its own is not enough.

Things to Be Careful About

• The mark scheme requires the candidate to name the positive correlation in their own words (e.g. "as % with antibodies increases, % of black wolves increases") rather than just saying "positive".
• Reasons should each be a complete, biologically explicit point; single-word answers ("resistance") are too vague.
• Read the axes carefully: x = % with anti-CDV antibodies, y = % of black wolves. Reversing the variables in the description will lose the relationship mark.

Background Concept

The endocrine system uses chemical messengers called hormones, released into the blood by glandular tissue, to coordinate activity in target cells elsewhere in the body. The three hormones tested here — ADH, glucagon and insulin — are all small proteins or polypeptides. Because they are water-soluble they cannot diffuse across the phospholipid bilayer of a cell-surface membrane; instead they bind to receptors on the outer surface of their target cells, triggering an intracellular second-messenger cascade (e.g. the cAMP pathway for glucagon).

• ADH (antidiuretic hormone, also called vasopressin) is synthesised in the hypothalamus and released from the posterior pituitary. It increases the permeability of the distal convoluted tubule and collecting duct to water by promoting the insertion of aquaporins into the apical membrane, so more water is reabsorbed from the filtrate back into the blood. This dilutes the blood and raises its water potential back towards normal.

• Glucagon is secreted by the α cells of the islets of Langerhans in the pancreas in response to low blood glucose. It stimulates glycogenolysis (breakdown of glycogen to glucose) and gluconeogenesis (synthesis of new glucose from non-carbohydrate sources) in liver cells; the glucose produced is released into the blood, raising blood glucose concentration.

• Insulin is secreted by the β cells of the islets of Langerhans in response to high blood glucose. It stimulates the uptake of glucose from the blood by muscle and adipose cells (via GLUT4 transporters) and the conversion of glucose to glycogen (glycogenesis) and to fat (lipogenesis). Insulin therefore moves glucose out of the blood and into cells — it does not move molecules out of cells into the blood.

The osmoreceptors in the hypothalamus detect a fall in blood water potential (an increase in osmolality); only ADH is released in response to their stimulation. Glucagon and insulin are released in response to changes in blood glucose concentration detected by the α and β cells of the pancreatic islets themselves.

Understanding the Question

This part presents a three-row feature table covering the three hormones and asks the candidate to put a tick or cross in every box. The mark scheme awards one mark per correctly completed row. The question explicitly says "put a tick or cross in every box" — leaving a box blank loses the mark for that row even if the other entries are correct.

Approach

For each row, decide what the feature really means in the context of the three molecules:

• "binds to receptors on cell surface membranes" — applies to any water-soluble hormone that cannot cross the membrane. All three are peptide hormones, so all three tick.

• "results in molecules moving from cells into the blood" — apply to each hormone's action. ADH makes the body reabsorb more water (water moves from filtrate into blood); glucagon makes liver cells release glucose (glucose moves out of cells into blood). Insulin does the OPPOSITE — it makes cells take up glucose from the blood — so insulin does not result in molecules leaving cells for the blood.

• "is secreted as a result of detection by osmoreceptors" — only ADH is released in response to osmoreceptor stimulation. Glucagon and insulin respond to blood glucose concentration.

Step-by-Step Reasoning

Row 1 — cell-surface receptors: ADH, glucagon and insulin are all peptide hormones. Because they are hydrophilic they cannot cross the plasma membrane directly, so they bind to receptors on the OUTER surface of the cell-surface membrane of their target cells. All three boxes get a tick (✓, ✓, ✓).

Row 2 — molecules moving from cells into blood:

• ADH increases water reabsorption in the collecting duct, so water leaves the cells/tissue and enters the blood. ✓
• Glucagon triggers glycogenolysis and gluconeogenesis in liver cells, releasing glucose from the cells into the blood. ✓
• Insulin has the OPPOSITE effect — it promotes glucose uptake by cells and glycogenesis, so molecules are moving from blood into cells, not the other way. ✗

Row 3 — secretion triggered by osmoreceptors:

• Osmoreceptors in the hypothalamus detect raised blood osmolality (low water potential); they signal the posterior pituitary to release ADH. ✓
• Glucagon secretion is triggered by low blood glucose detected by α cells in the islets of Langerhans — not by osmoreceptors. ✗
• Insulin secretion is triggered by high blood glucose detected by β cells in the islets of Langerhans — not by osmoreceptors. ✗

Key Takeaways

• All three hormones (ADH, glucagon, insulin) are peptide/protein hormones and therefore bind to cell-surface receptors.
• The chemical nature of a hormone (hydrophilic vs hydrophobic) determines whether it acts at a cell-surface receptor (peptide hormones) or an intracellular receptor (steroid/thyroid hormones).
• The actions of ADH and glucagon cause substances to leave cells/filtrate and enter the blood, while insulin causes substances to leave the blood and enter cells.
• The stimulus for secretion is specific to each hormone's homeostatic role: ADH → osmoreceptors; insulin/glucagon → blood glucose concentration.

Common Mistakes

• Ticking insulin in row 2 because students confuse the direction of glucose movement. Insulin promotes glucose UPTAKE from blood into cells — it does not move molecules out of cells into the blood.
• Ticking glucagon in row 3 because students forget that glucagon responds to glucose, not osmolality.
• Crossing out ADH in row 1 because ADH is sometimes thought of as "going into the cell" — in fact ADH binds to V2 receptors on the basolateral membrane of collecting-duct cells and triggers an intracellular cAMP cascade; it never enters the cell.

Things to Be Careful About

• Use the EXACT chemical nature of each hormone (peptide vs steroid) to deduce whether it acts at a cell-surface or intracellular receptor — this is a transferrable skill that recurs across the syllabus.
• The "osmoreceptor" stimulus is unique to ADH (and to thirst) in the endocrine system tested here; do not confuse it with the glucose-detection role of the pancreatic islets.
• The question states "put a tick or cross in EVERY box" — leaving a box blank will lose the mark for that row even if the other entries in the row are correct.

Background Concept

Mammals coordinate their internal activities and their responses to the environment using two principal communication systems: the nervous system and the endocrine system. Both are ultimately under the control of the hypothalamus, which acts as the bridge between them. The two systems differ fundamentally in:

• Nature of the signal — the nervous system uses rapid, transient changes in the membrane potential of neurones (action potentials) and releases neurotransmitter into synapses. The endocrine system releases hormones into the blood, where they persist until broken down or taken up by target cells.

• Pathway of transmission — nervous signals travel along specific anatomical routes (axons), allowing very precise targeting. Hormones are released into the general circulation and can reach almost every cell in the body, but only those with the appropriate receptor respond.

• Duration of the effect — neurotransmitter is rapidly broken down or taken back up, so nervous effects are brief (milliseconds to seconds). Hormones persist in the blood for longer (seconds to hours or days), giving sustained responses.

• Specificity of the target — the wiring of neurones means that the effect is highly localised (a single neuromuscular junction, a single gland). Hormones can act on many cells simultaneously (e.g. insulin acts on every body cell), giving a more diffuse but coordinated response.

Understanding the Question

The question stem has already told the candidate that the endocrine system has a slower transmission speed than the nervous system. The candidate is asked to describe other ways in which the two systems differ — i.e. other than the speed of transmission. The mark scheme accepts any four valid contrasting pairs from a list of six.

Approach

For each potential difference, state the feature of the nervous system and the corresponding feature of the endocrine system, using comparative phrasing such as "X, whereas Y" or "X in the nervous system but Y in the endocrine system". The mark scheme is a "matching pair" list — each pair counts as one mark, and a mark requires BOTH halves of the pair.

Step-by-Step Reasoning

1. Nature of the signal — Nervous system: electrical impulses (action potentials) along the axon, with chemical neurotransmitters across the synapse. Endocrine system: chemical hormones released into the blood. (Mark scheme points 1 and 2.)

2. Pathway — Nervous system: along neurones. Endocrine system: in the blood. (Mark scheme point 3.)

3. Specificity of target — Nervous system: specific / localised — the signal only reaches the cells wired to the active neurone. Endocrine system: can be widespread — a hormone released into the blood can act on any cell with the appropriate receptor anywhere in the body. (Mark scheme point 4.)

4. Duration of effect — Nervous system: short-lived (neurotransmitter is broken down or reabsorbed within milliseconds). Endocrine system: long-lasting (hormones persist in the blood until metabolised by the liver or kidneys). (Mark scheme point 6.)

5. Speed of response — Nervous system: fast (already given in the question, so do NOT use this if asked for "other" differences; mark scheme point 5 is also covered by the question stem).

Key Takeaways

• The two systems differ in the nature of the signal, the pathway, the specificity of the target, the duration of the effect, and the speed — the last is given in the question.
• Always present differences as a MATCHED PAIR with both halves stated; giving only one half of a pair does not earn the mark.
• The nervous system is best suited to rapid, precise, short-lived actions (e.g. withdrawal reflex, posture); the endocrine system is best suited to slower, sustained, body-wide regulation (e.g. glucose homeostasis, water balance, growth, reproduction).

Common Mistakes

• Stating only ONE half of a comparison (e.g. "the nervous system is electrical") without the corresponding half for the endocrine system — does not earn the mark.
• Repeating the speed difference that has been given in the question stem — this is what the question explicitly says NOT to do ("other ways").
• Confusing the ENDOCRINE system with the AUTONOMIC nervous system (which is part of the nervous system) — many autonomic effects are slow, diffuse and long-lasting, but they are still nervous, not endocrine.
• Vague statements like "they use different chemicals" or "they are located in different places" — these do not earn marks because they are not specific contrasts.

Things to Be Careful About

• Use precise CIE terminology: "action potential" or "nerve impulse", not just "signal"; "hormone", not just "chemical".
• Make the comparison explicit: the nervous system DOES X, while the endocrine system DOES Y.
• The mark scheme accepts any four valid pairs — pick the four you can express most clearly and accurately, not the four that are listed first.

Background Concept

ATP (adenosine triphosphate) is a nucleotide derivative consisting of the nitrogenous base adenine, the pentose sugar ribose, and a chain of three phosphate groups. The bonds between the terminal phosphate groups are often described as 'high-energy' (more accurately, they are bonds whose hydrolysis releases a large amount of free energy, $\Delta G^\circ \approx -30.5\ \text{kJ mol}^{-1}$). Removing the terminal phosphate by hydrolysis yields ADP (adenosine diphosphate) and inorganic phosphate ($\text{P}_i$), and this reaction powers almost every energy-requiring process in the cell — from active transport and biosynthesis to cell movement and maintaining body temperature.

For a molecule to act as a universal energy currency it must satisfy several criteria. It must release an intermediate amount of energy (not so much that it is destructive, not so little that it cannot drive reactions), be regenerated easily, and be transportable within the cell.

Understanding the Question

Part (a) is a 3-mark 'outline' question. The command word 'outline' requires a brief description of the relevant features — it does not require deep explanation. The question is restricted to the features of the ATP molecule itself (its structure, chemistry and properties), not to where the energy comes from or how ATP is used.

Approach

The mark scheme offers five possible creditable points. Pick the three that together give the most complete picture: one for the hydrolysis reaction itself, one for its reversibility/regeneration, and one for the physical mobility of the molecule. Always pair the property with its consequence (e.g. 'small/soluble' AND 'so can diffuse') — the mark scheme credits the reasoning, not just the label.

Step-by-Step Reasoning

• Point 1 — Hydrolysis and energy release. ATP is described as an 'energy currency' because the hydrolysis of its terminal phosphoanhydride bond, $\text{ATP} \rightarrow \text{ADP} + \text{P}_i$, releases energy. The energy released is sufficient to drive otherwise unfavourable (endergonic) reactions, often by phosphorylating the substrate.
• Point 2 — Reversibility / regeneration. The same reaction is reversible: ADP and $\text{P}_i$ can be re-phosphorylated back to ATP using energy from catabolic processes (respiration, photosynthesis). This 'recyclability' gives ATP a high turnover and means a cell only needs a small pool of ATP at any one moment.
• Point 3 — Small and soluble / mobility. The ATP molecule is small and water-soluble, so it can diffuse through the cytosol and reach all the locations in the cell where energy is needed, including organelles such as mitochondria and chloroplasts (although the inner mitochondrial membrane requires the ATP/ADP translocase).

Other valid points that could earn a mark include: only a small amount of energy is released per hydrolysis (so it is not destructive / wasteful); the adenine portion allows it to be recognised by many different enzymes; it can be made from ADP + $\text{P}_i$, so a constant supply of $\text{P}_i$ is not a limiting factor.

Key Takeaways

• ATP is the immediate energy source for most cellular work; the energy is released by hydrolysis of a terminal phosphate bond.
• The hydrolysis reaction is readily reversible, allowing ATP to be continuously recycled.
• Its small, soluble structure allows free movement in the cytoplasm.

Common Mistakes

• Saying only that ATP 'contains energy' — the mark is for stating that hydrolysis releases energy.
• Writing 'ATP has high-energy bonds' without mentioning that these bonds are broken to release energy.
• Describing how ATP is made (respiration) instead of features of the molecule that suit it as a currency.
• Forgetting to link each property to a consequence — e.g. 'small' alone is incomplete; it must be tied to mobility/diffusion.

Things to Be Careful About

• 'Outline' = brief description, not exhaustive explanation. Three well-stated points are sufficient for full marks.
• Acceptable alternative wording (e.g. 'phosphorylated' / 'converted to ADP and Pi' / 'donates a phosphate group') is credited, but the term 'energy currency' is not a substitute for the actual points.
• A common confusion: ATP does not store energy long-term — that role belongs to carbohydrates, lipids and proteins. ATP is an immediate, short-term carrier.

Background Concept

Oxidative phosphorylation is the process by which the energy of electron transport is used to phosphorylate ADP to ATP. In the inner mitochondrial membrane are four large protein complexes (I, II, III, IV) plus mobile electron carriers (ubiquinone and cytochrome c). Together these form the electron transport chain (ETC).

The flow of electrons through the chain, from higher-energy carriers to lower-energy ones, releases energy. Complexes I, III and IV use this energy to pump protons ($\text{H}^+$) from the matrix into the intermembrane space, creating an electrochemical (proton) gradient. Protons then flow back down this gradient through ATP synthase (Complex V), and the energy of that flow drives the synthesis of ATP from ADP and $\text{P}_i$ — this is chemiosmosis.

At the end of the chain, in Complex IV, electrons are donated to oxygen (the terminal electron acceptor), which combines with $\text{H}^+$ to form water:

This is why water is produced as a by-product of aerobic respiration. The electrons entering the chain come from reduced coenzymes (NADH and FADH$_2$) generated in glycolysis, the link reaction and the Krebs cycle. Re-oxidation of these coenzymes (NADH $\rightarrow$ $\text{NAD}^+$, FADH$_2$ $\rightarrow$ FAD) is essential because the cell has only a small pool of each, and they must be recycled for the earlier stages of respiration to continue.

Understanding the Question

The question gives a piece of information: rotenone disrupts the first carrier in the ETC so that it cannot pass electrons on. From this single fact you must reason through every consequence.

• The command word is suggest and explain — a two-step demand. You must state what happens AND why. Each marking point on the scheme is essentially a 'what' followed by a 'why'.
• Two products are explicitly named in the question: ATP and water. The answer should address both, plus the inevitable knock-on effect on the substrate-level stages through the lack of $\text{NAD}^+$ regeneration (this is an additional high-yield point the mark scheme rewards).
• The question is worth 6 marks, so the mark scheme expects around six clearly reasoned points.

Approach

Trace the consequences of the blockage in a logical cascade:

1. Block at the first carrier → no electron flow through the ETC. (Direct, mechanical consequence.)
2. No electron flow → no energy released to pump $\text{H}^+$. (Cause of the next effect.)
3. No $\text{H}^+$ pumping → no / smaller proton gradient. (The gradient is what drives ATP synthesis.)
4. No proton gradient → $\text{H}^+$ cannot flow through ATP synthase → no ATP made by oxidative phosphorylation. (Answers the 'ATP' part of the question.)
5. No electrons reach the final carrier (Complex IV) → no electrons to reduce $\text{O}_2$ → no / less water formed. (Answers the 'water' part of the question.)
6. Reduced NAD cannot be re-oxidised → shortage of $\text{NAD}^+$ → earlier stages (glycolysis, link reaction, Krebs cycle) slow / stop → less ATP from substrate-level phosphorylation too. (Indirect, downstream consequence.)

Aim to deliver these six points in a clear order so the examiner can tick each one.

Step-by-Step Reasoning

Point 1 — Electron flow stops.
Rotenone binds to Complex I (NADH dehydrogenase), the first carrier. Electrons from NADH cannot be transferred on to ubiquinone, so the entire electron-transport chain 'downstream' of this point receives no electrons. The chain is effectively 'broken'.

Point 2 — No energy released.
The energy used to pump protons is released as electrons pass from carrier to carrier and fall to lower energy levels. With no electron flow there is no fall in energy, so no usable energy is released to do the work of pumping.

Point 3 — No / reduced $\text{H}^+$ pumping.
Complexes I, III and IV normally pump $\text{H}^+$ from the matrix into the intermembrane space. With Complex I inactive, and the rest of the chain not receiving electrons, none of these complexes can pump protons. The $\text{H}^+$ concentration in the intermembrane space therefore falls.

Point 4 — No / smaller proton gradient.
A proton gradient requires a high $\text{H}^+$ concentration in the intermembrane space relative to the matrix. Without active pumping, the gradient dissipates (any gradient that exists is quickly used up by ATP synthase or by leakage). The proton-motive force that normally drives ATP synthesis is therefore lost.

Point 5 — ATP synthase cannot make ATP (oxidative phosphorylation halts).
$\text{H}^+$ normally flows back into the matrix through the channel in ATP synthase, and the energy released by this exergonic flow rotates the enzyme, catalysing the synthesis of ATP from ADP and $\text{P}_i$. With no gradient, there is no driving force, so ATP synthase cannot function. The cell's main source of ATP (most ATP comes from oxidative phosphorylation) is cut off.

Point 6 — Electrons never reach the final carrier.
Because rotenone blocks the very first step, electrons cannot reach Complex IV, where they would normally be passed to oxygen. The final reduction reaction ($\text{O}_2 + 4\text{H}^+ + 4\text{e}^- \rightarrow 2\text{H}_2\text{O}$) therefore cannot take place. The cell produces much less / no water as a by-product of oxidative phosphorylation.

Point 7 — Reduced NAD builds up; $\text{NAD}^+$ runs out.
The electrons blocked at Complex I are still attached to NADH. Because the chain cannot accept more electrons, NADH cannot be re-oxidised to $\text{NAD}^+$. The pool of $\text{NAD}^+$ in the mitochondrion therefore falls.

Point 8 — Earlier stages of respiration slow down.
Glycolysis, the link reaction and the Krebs cycle all need $\text{NAD}^+$ (and FAD) as electron acceptors. Without $\text{NAD}^+$ these reactions stall (or, in the case of glycolysis, switch to lactate fermentation in some tissues). Less pyruvate enters the link reaction, less acetyl-CoA enters the Krebs cycle, and the small amounts of ATP made by substrate-level phosphorylation in glycolysis and the Krebs cycle also fall. (This is an indirect but very important consequence — it is why rotenone is highly toxic: oxidative phosphorylation itself produces most ATP, but the cell can still be damaged by the secondary block on glycolysis and the Krebs cycle.)

Key Takeaways

• Blocking any one step in the ETC has knock-on effects throughout aerobic respiration.
• ATP production by oxidative phosphorylation depends on three things: electron flow, a proton gradient, and functional ATP synthase. Remove any one and ATP synthesis halts.
• Water is a by-product of electron transport (the reduction of $\text{O}_2$ at Complex IV), not a by-product of ATP synthase.
• Re-oxidation of NADH is essential to keep the earlier stages of respiration running — it is the link between the substrate-level and oxidative stages.

Common Mistakes

• Saying that rotenone 'stops respiration' without explaining the chain of events — the mark scheme rewards the explanation (no electron flow → no proton gradient → no ATP), not the conclusion.
• Confusing the direction of proton flow: protons are pumped out of the matrix into the intermembrane space, and they diffuse back in through ATP synthase.
• Saying that water is made 'by respiration' without specifying that it is the electrons reaching $\text{O}_2$ at the end of the chain that combine with $\text{H}^+$ to form water.
• Stating that glycolysis and the Krebs cycle are unaffected — they are heavily dependent on $\text{NAD}^+$ regeneration by the ETC.
• Confusing the roles of oxygen: oxygen is the terminal electron acceptor (combines with electrons and $\text{H}^+$ to form water), not the thing that is 'used up' directly in ATP synthesis.

Things to Be Careful About

• 'Suggest and explain' requires both elements for every point: state the effect, then state why it occurs.
• The question is asking for the overall effect on the cell, so a good answer integrates the ATP and water consequences with the knock-on effects on $\text{NAD}^+$ and the earlier stages.
• Stick to the named products (ATP and water) and the linked concept of $\text{NAD}^+$ regeneration — do not drift into unrelated detail (e.g. the specific name of the first carrier, the number of ATP per NADH, etc.) unless it earns a mark.
• Use precise terms: 'electron transport chain' (not 'respiratory chain'), 'proton gradient' (not 'concentration gradient'), 'ATP synthase' (not 'ATPase', although the mark scheme may accept either).
• Significant quantities matter: 'no / less / fewer' is the wording the mark scheme uses; be consistent and do not over-state absolute 'zero' unless you are confident.

Background Concept

LibertyLink® soybean carries the pat gene, which was isolated from the bacterium Streptomyces viridochromogenes. The gene encodes phosphinothricin N-acetyltransferase (PAT), an enzyme that inactivates glufosinate — the active ingredient of the Liberty® herbicide. Glufosinate normally inhibits glutamine synthetase, causing a lethal build-up of ammonia in plant cells. When PAT is present, glufosinate is acetylated and no longer toxic, so the soybean survives while surrounding weeds die.

The wider context is the use of herbicide-tolerant GM crops as part of modern, large-scale agriculture. Weeds compete with the crop for the same limited resources: light (the energy source for photosynthesis), water (the solvent and reactant in photosynthesis), and soil minerals such as nitrates and phosphates (needed to make proteins, nucleic acids and ATP). If weeds are removed, the crop can grow faster and produce more biomass and grain.

Understanding the Question

The command word is "explain", so the answer must link the genetic modification (resistance to glufosinate) to the agricultural benefit (more food produced). The mark scheme rewards a chain of reasoning: resistance → spray kills weeds → less competition → higher yield. The candidate does not need to describe how the pat gene was inserted, only what it lets the farmer do.

Approach

Identify the gene product (PAT) and what it does (detoxifies glufosinate). Then move from the cellular effect to the field-scale effect and finally to the harvest outcome. Use the named herbicide and a named resource for the mark.

Step-by-Step Reasoning

• Mark point 1: state that the LibertyLink® soybean is not harmed / killed by glufosinate because PAT inactivates the herbicide.
• Mark point 2: state that the farmer can therefore spray glufosinate over the field to kill weeds / unwanted plants.
• Mark point 3: link weed removal to reduced competition between the crop and weeds.
• Mark point 4: name a resource the weeds would otherwise take — e.g. sunlight, water, or a soil mineral such as nitrate or magnesium.
• Mark point 5: conclude that the crop produces a higher yield, helping to feed more people.

Key Takeaways

• Herbicide-resistance genes let a farmer spray an entire field with a broad-spectrum herbicide; only the weeds die.
• The biological benefit flows up: molecular resistance → field-scale weed control → reduced competition for resources → higher yield.
• GM herbicide-tolerant crops are a real, commercial example used worldwide.

Common Mistakes

• Saying only "it is resistant to herbicide" without going on to explain what that lets the farmer do (no link to weeds or yield).
• Naming "pesticide" instead of "herbicide". Glufosinate targets plants, not insects.
• Saying the soybean itself kills the weeds — it does not; the herbicide does, and only because the crop is unaffected.
• Stopping at "more food is produced" without mentioning reduced competition for a resource.

Things to Be Careful About

• A full "explain" answer needs a reason for each claim, not just a list of effects.
• Always name at least one resource (light, water, or a mineral) when discussing competition — "nutrients" on its own is too vague.
• Stay on the pat / glufosinate system described in Table 8.1; do not import Bt or insect-resistance ideas from other GM crops.

Background Concept

LibertyLink® soybean was first commercialised in 1996, and since then its use in food products has spread faster than its cultivation. The 21 countries that use it as a food ingredient include nations that are net importers of soy (e.g. for animal feed, oil, processed protein), while the 6 countries that grow it tend to be large agricultural producers with permissive regulation. The mismatch between "used in" and "grown in" is therefore a question of why a country would import a product without producing it.

The reasons cluster into five families: biophysical (climate and soil), regulatory (GM laws and herbicide bans), socio-political (public/activist opposition), economic (land availability and cost) and trade (some countries buy cheap soy from elsewhere rather than growing it).

Understanding the Question

The command word is "suggest", so the candidate is expected to come up with plausible reasons, not recall a single fact. The mark scheme accepts any three distinct, sensible points from the list it gives, and an AVP (any valid point) is also credited.

The structure of the answer is: a country is on the "used in" list but not the "grown in" list. Why? Because something about the country prevents cultivation but not import.

Approach

Think of the 21 countries as a mixed group: some have climates or soils that simply cannot grow soybean profitably; others can grow soybean but have chosen not to grow this GM variety. Distinguish between reasons that apply even to non-GM soybean (climate, land) and reasons that apply specifically to LibertyLink® (regulation, public opposition, glufosinate ban).

Step-by-Step Reasoning

• Mark point 1: biophysical unsuitability. Soybean needs warm, temperate climates with adequate rainfall. Many of the 21 user countries are too cold, too dry, or have the wrong soil type, so they import soy rather than grow it.
• Mark point 2: regulatory. The EU, for example, has had a slow and politically cautious approval process for GM cultivation. Many countries will accept imported GM food that has been approved as safe but will not allow domestic farmers to plant it.
• Mark point 3: public/activist opposition. Even where regulation is permissive, consumer or activist pressure can deter cultivation (e.g. opposition to "frankencrops").
• Mark point 4: land. Some countries are small, heavily urbanised or have most of their land under other crops; they have no spare agricultural land for soybean.
• Mark point 5: herbicide ban. If glufosinate itself is banned in a country, growing LibertyLink® soybean loses its point, so the country simply imports conventional soy or non-GM soy.

Key Takeaways

• Adoption of a GM crop is governed by more than just agronomic performance — regulation, public opinion and trade policy all matter.
• A country can consume a GM product while still opposing its cultivation. "Use" and "grow" are decoupled.
• This is a recurring theme in A2 Biology: socio-economic context shapes the real-world uptake of biotechnology.

Common Mistakes

• Saying "people don't like GM food" — this would explain why countries are on neither list. The question is specifically about countries that import but do not grow.
• Repeating the same idea twice in different words (e.g. "the government doesn't allow it" and "it's illegal").
• Conflating bans on the crop with bans on the herbicide — these are separate regulatory decisions.

Things to Be Careful About

• The question says "LibertyLink® soybean" specifically, so where possible refer to glufosinate (the matching herbicide) rather than generic weedkillers.
• "Suggest" requires plausible, reasoned points, not single-word answers. A short explanation is needed for full credit on each mark.
• Three marks = three distinct ideas, so do not pad one point into three sentences.

Background Concept

The human kidney has a characteristic gross structure visible in a longitudinal (sagittal) section. From outside in, the regions are: an outer cortex (paler, granular appearance — it contains the renal corpuscles and convoluted tubules), an inner medulla (darker, organised into cone-shaped pyramids whose striated appearance comes from the parallel loops of Henle and collecting ducts), and a central renal pelvis (the funnel-shaped cavity that collects urine and funnels it into the ureter). Between the pyramids of the medulla and the cortex, finger-like extensions of cortex called renal columns project inward. The renal pelvis is continuous with the ureter at the hilum, where the renal artery, renal vein and ureter enter and leave the kidney.

Understanding the Question

The candidate is given a photograph of a bisected kidney with two labels, X and Y, and must name the regions pointed to. X is in the central cavity of the kidney; Y is in the striated inner zone between the cortex and the pelvis. The question uses the command word "Name" — the candidate simply has to give the correct anatomical term, no description is needed.

Approach

Recall the layered organisation of the kidney from outside to inside: cortex → medulla → pelvis. Match each label's position on the photograph to the correct term.

Step-by-Step Reasoning

• X is in the centre of the kidney, where the branching white cavity is visible. This is the renal pelvis, the chamber that collects urine from the calyces before passing it into the ureter. (1 mark)
• Y is in the middle zone of the kidney, between the pale outer cortex and the central pelvis. The darker, striated appearance comes from the parallel arrangement of collecting ducts and loops of Henle. This is the medulla. (1 mark)
• The outermost rim — not labelled here — would be the cortex.

Key Takeaways

• The kidney is organised into three main regions: cortex (outer), medulla (middle, pyramidal), and pelvis (central cavity).
• Recognising a longitudinal section of a kidney is a routine Paper 2 skill.

Common Mistakes

• Confusing pelvis with ureter — the pelvis is the internal collecting chamber, while the ureter is the tube leading from the pelvis to the bladder.
• Confusing medulla with cortex — students often label the lighter outer band as the medulla; the medulla is the darker, striated, inner zone.
• Writing "renal pyramid" instead of "medulla" — the pyramids are subdivisions of the medulla and are not the same as the medulla as a whole.

Things to Be Careful About

• Always read the position of the arrow on the photograph carefully, not the line on the page — Y points to the middle band, not the outer band.
• Use the correct anatomical name; "pelvis" on its own is acceptable, but "renal pelvis" is more precise and is credited.

Background Concept

A biosensor combines a biological recognition element (an enzyme, antibody, or whole cell) with a physical transducer that converts the biological event into a measurable signal — most often an electrical one. The classic glucose biosensor used in clinical dipsticks and handheld meters uses the enzyme glucose oxidase (GOD). Glucose oxidase catalyses the oxidation of β-D-glucose by molecular oxygen, producing gluconolactone (which hydrolyses to gluconic acid) and hydrogen peroxide (H₂O₂):

The H₂O₂ produced can be detected electrochemically because it is readily oxidised at a platinum electrode, releasing electrons. The resulting current is proportional to the rate of H₂O₂ production, which in turn is proportional to the glucose concentration in the sample. The current is then converted to a digital reading (in mmol L⁻¹ or mg dL⁻¹) by the meter's electronics.

Understanding the Question

The question asks the candidate to outline how a biosensor measures glucose in urine. "Outline" means to give the main points in a logical order — a short structured description covering the enzyme, the reaction, the signal, and the read-out. Four marks are available, so four clear points are needed.

Approach

Follow the chain: enzyme on the strip → reaction producing H₂O₂ → electrochemical detection of H₂O₂ → proportional current → digital display. The mark scheme explicitly looks for these stages and the proportionality between current and concentration.

Step-by-Step Reasoning

1. Recognition element: Glucose oxidase is immobilised on the test strip / electrode of the biosensor. Immobilisation lets the enzyme be reused and keeps it in contact with the sample. (1 mark)
2. Specific reaction: Glucose in the urine is oxidised by O₂ (with glucose oxidase as catalyst) to produce H₂O₂. (1 mark)
3. Transduction: The H₂O₂ diffuses to the electrode, where it is oxidised, releasing electrons (or producing O₂ + H⁺ + e⁻). (1 mark)
4. Quantification: The flow of electrons produces a small current / voltage whose magnitude is proportional to the rate of H₂O₂ formation — and therefore to the glucose concentration in the original sample. (1 mark)
5. Read-out (AVP worth a mark): The current is converted to a digital / numerical display giving the glucose concentration.

Any four of the five creditable points score the four marks.

Key Takeaways

• A biosensor = biological recognition element + transducer + display.
• Glucose biosensors use glucose oxidase to generate H₂O₂, which is detected electrochemically.
• The signal magnitude is proportional to the analyte concentration, allowing quantitative measurement.
• Biosensors are specific (glucose oxidase only acts on glucose) and fast, which is why they have replaced older chemical strip methods in clinical use.

Common Mistakes

• Stating only that "the enzyme reacts with glucose" without naming glucose oxidase — the name of the enzyme is the specific marking point.
• Omitting the proportionality between current and concentration — without this, the sensor would be qualitative, not quantitative.
• Saying the biosensor "measures glucose directly" — it measures the current produced by H₂O₂ breakdown, and only infers glucose from that.
• Confusing glucose oxidase with glucose dehydrogenase (a different enzyme used in some modern meters; not the one credited on CIE mark schemes).
• Writing about colour change — that is how a chemical test strip (e.g. Clinistix) works, not a biosensor. The biosensor's read-out is a digital/electrical value, not a colour comparison.

Things to Be Careful About

• Use the exact term glucose oxidase (not just "enzyme").
• Mention the immobilised nature of the enzyme, since this is a defining feature of a biosensor (as opposed to a free-enzyme assay).
• Make the link glucose → H₂O₂ → current explicit, and state the proportionality between current and concentration.
• Give the digital / numerical nature of the read-out — this is what distinguishes a biosensor from a chemical colour strip.