Biology 9700/43 — May/June 2024

Background Concept

Antidiuretic hormone (ADH) is a small peptide hormone released from the posterior pituitary gland. Because ADH is too large and hydrophilic to cross the phospholipid bilayer, it binds to a specific receptor protein embedded in the cell surface membrane of its target cells (collecting-duct cells in the kidney, and similar receptors exist on hepatocytes for glucagon).

Binding of the hormone activates an associated G protein (a GTP-binding protein that acts as a molecular switch on the inner leaflet of the membrane). The activated G protein then stimulates the enzyme adenylyl cyclase, which catalyses the conversion of ATP into a small intracellular second messenger — cyclic AMP (cAMP). The rise in cAMP concentration triggers an enzyme cascade that ultimately causes vesicles containing aquaporin water-channel proteins to fuse with the cell surface membrane, increasing the permeability of the collecting-duct cell to water.

The structure C is therefore an aquaporin — a transmembrane protein that forms a pore allowing water to move down its water-potential gradient across the membrane.

Understanding the Question

The question provides Fig. 1.1, a labelled diagram of a collecting-duct cell showing the ADH cell-signalling pathway, and asks the candidate to name three structures: A, B and C. A is a structure inside the membrane associated with the receptor; B is produced from ATP; C is the channel through which water crosses the membrane on the lumen side.

Approach

Match each letter on the diagram to its position in the standard G-protein/adenylyl-cyclase/cAMP pathway. A sits on the inner face of the membrane next to the receptor — a G protein. B is generated from ATP by adenylyl cyclase — cyclic AMP. C is a transmembrane channel protein on the lumen-facing membrane through which water moves — an aquaporin.

Step-by-Step Reasoning

• A is shown attached to the inner side of the cell surface membrane, next to the ADH receptor. When ADH binds, this protein is activated — it is the G protein.
• B is produced by adenylyl cyclase from ATP. It is the second messenger that diffuses into the cytoplasm and triggers the enzyme cascade. This is cyclic AMP (cAMP).
• C is the water-permeable channel embedded in the cell surface membrane on the lumen side. The protein that forms this selective water pore is an aquaporin.

Key Takeaways

• The ADH signalling pathway uses a G protein, adenylyl cyclase, and cAMP as a second messenger — the same cascade used by glucagon in liver cells.
• Aquaporins are the water-selective channels inserted into the collecting-duct cell membrane in response to ADH.
• cAMP is generated from ATP; "cyclic" refers to the ring structure formed when the phosphate group loops back onto the ribose sugar.

Common Mistakes

• Writing "G-protein" is fine; avoid writing "G enzyme" or "GTP" for structure A.
• For B, students sometimes write "AMP" or "ATP" instead of "cyclic AMP / cAMP". The word cyclic (or the abbreviation cAMP) is the key marking point.
• For C, the mark scheme rejects "water channel" — the required term is the specific protein name aquaporin.

Things to Be Careful About

• The question stem notes that ADH's mechanism is similar to that of glucagon on liver cells, so identifying cAMP as the second messenger links this question to the glucagon/cAMP cascade students also learn.
• "Aquaporin" must be spelled correctly; "aquaporin-2" is the specific isoform inserted in the collecting duct but the name "aquaporin" alone is what the mark scheme accepts.

Background Concept

Water potential ($\Psi$) is a measure of the tendency of water to move by osmosis; pure water has $\Psi = 0\ \text{kPa}$ and adding solutes makes $\Psi$ more negative. The water potential of the blood depends on the balance between the water content of the plasma and the concentration of dissolved solutes (mainly Na$^+$, Cl$^-$, glucose, urea and plasma proteins). Anything that either removes water from the plasma or adds solute to it will lower (make more negative) the blood water potential. This fall is detected by osmoreceptors in the hypothalamus, which stimulate ADH release from the posterior pituitary.

Understanding the Question

The question gives a simple context: "ADH is secreted by the posterior pituitary gland when the water potential of the blood decreases." It then asks the candidate to suggest reasons why the water potential of the blood might decrease. The command word suggest means the answer does not have to be limited to textbook examples — any reasonable factor that would lower blood $\Psi$ earns credit, up to the mark allowance of three.

Approach

There are two routes by which $\Psi$ falls: (1) loss of water from the plasma, or (2) gain of solute by the plasma. Generate examples of each:

• Water loss — not drinking, sweating, vomiting, diarrhoea, evaporation, exercise.
• Solute gain — eating salty food, drinking sea water, high blood glucose (e.g. uncontrolled diabetes mellitus), high blood amino acids after a high-protein meal.

Any three distinct, sensible points gain the three marks.

Step-by-Step Reasoning

1. Not drinking / dehydration reduces the water component of plasma, concentrating the remaining solutes — lower $\Psi$.
2. Ingesting salty food introduces a large NaCl load into the blood; the additional solute lowers $\Psi$ (and osmoreceptors respond before the kidney can excrete the excess).
3. More ions, glucose or amino acids in the blood — any of these increases plasma osmolarity and so lowers $\Psi$; this is why uncontrolled diabetes mellitus (high blood glucose) causes osmotic diuresis and thirst.
4. Sweating / perspiration is evaporative water loss from the skin; the water leaves without solute, raising plasma concentration and lowering $\Psi$.

Key Takeaways

• Blood water potential is set by the ratio of water to solutes in plasma — either variable can change it.
• The hypothalamus detects small changes in plasma osmolarity (≈1%) and triggers ADH release and thirst — this is the afferent limb of an osmoregulatory negative-feedback loop.
• The question tests the application of water potential, not just rote learning.

Common Mistakes

• Saying "ADH is low" or "ADH is not secreted" — the question asks for causes of decreased water potential, not consequences.
• Listing only one type of cause (only water loss, or only solute gain) — the mark scheme rewards variety.
• Vague answers such as "hot weather" without explaining the mechanism (evaporative water loss → lower plasma $\Psi$) — the link must be made.

Things to Be Careful About

• "Dehydration" is acceptable and worth a mark on its own; further detail (no water intake) is optional.
• "Salty food" is a single cause — no need to also list "more ions in the blood" as a separate point, as the mark scheme treats them as alternatives (separated by ; not \n).
• Any creditable alternative (e.g. haemorrhage, severe burns, osmotic diuresis in diabetes mellitus) is accepted under the AVP line.

Background Concept

ADH increases the permeability of the collecting duct (and distal tubule) to water by triggering the insertion of aquaporin-2 channels into the luminal cell surface membrane. When ADH is low, the collecting duct remains relatively impermeable to water, so the filtrate that entered as plasma at the glomerulus passes through with little water reabsorption. The result is a large volume of dilute urine, and the body loses water it would normally have conserved. The drop in body water content raises plasma osmolarity and lowers blood volume (and pressure), which is sensed by osmoreceptors (thirst) and baroreceptors.

Understanding the Question

The stem tells the candidate that a pituitary tumour causes decreased ADH secretion (this is the central / neurogenic form of the condition). The candidate must therefore predict the symptoms: if ADH is low, water cannot be reabsorbed efficiently, so the body loses water and becomes dehydrated and thirsty.

Approach

Work forward from "less ADH" → "fewer aquaporins in collecting-duct membrane" → "less water reabsorbed" → "large volumes of dilute urine" + "net water loss from body" → "thirst, dehydration, fatigue".

Step-by-Step Reasoning

1. With little ADH, the collecting-duct cells do not insert aquaporins into their luminal membrane; water is not reabsorbed and remains in the filtrate → large volume of dilute urine and frequent urination.
2. The body therefore loses water faster than it is taken in → plasma becomes more concentrated and total body water falls → thirst / dry mouth (osmoreceptor response) and signs of dehydration.
3. Loss of fluid from the circulation reduces blood volume and pressure; dehydration and electrolyte disturbance cause fatigue / lethargy.

Key Takeaways

• The triad of diabetes insipidus is polyuria (large urine volume), polydipsia (excess thirst) and dilute urine — these are the symptoms any candidate should aim to recall.
• This is mechanistically distinct from diabetes mellitus, where polyuria is caused by glucose-driven osmotic diuresis rather than ADH deficiency.
• The question illustrates how a single hormone deficiency produces a coherent set of symptoms through one disrupted pathway.

Common Mistakes

• Writing "less urine" — the opposite is true; without ADH the kidney cannot concentrate urine and produces a large volume of dilute urine.
• Saying "no aquaporins" — there is always a small baseline permeability; the right wording is fewer aquaporins or reduced water reabsorption.
• Confusing diabetes insipidus with diabetes mellitus and listing "high blood glucose" — the question specifies a pituitary tumour causing low ADH.

Things to Be Careful About

• The command word is suggest, so the marks reward any plausible symptom consistent with the mechanism. Aim for two clear, mechanism-based points.
• "Dilute urine" is one of the most distinctive features of diabetes insipidus and is worth a mark on its own.

Background Concept

The cell surface membrane of a collecting-duct cell is normally only sparingly permeable to water; its water permeability is regulated by the number of aquaporin-2 channels in the membrane. These channels are stored in vesicles just beneath the membrane and are inserted by exocytosis only when the cAMP cascade triggered by ADH is active. When ADH is removed, the channels are retrieved by endocytosis, and the membrane reverts to low water permeability.

In the figure, membrane P is the cell surface membrane on the lumen side of the collecting-duct cell — exactly where aquaporins are inserted to allow water to leave the tubular fluid and enter the cell (and then the blood).

Understanding the Question

The question sets up a different form of diabetes insipidus — neurogenic diabetes insipidus (NDI) — in which ADH molecules are present in the blood but cannot bind to the receptor proteins on the collecting-duct cells. The candidate must use Fig. 1.1 to explain the effect on membrane P.

Approach

Because the pathway is blocked at the very first step (no ADH–receptor binding), nothing downstream happens. Walk through Fig. 1.1 step by step and identify what fails to occur: no G-protein activation, no adenylyl cyclase activation, no cAMP produced, no enzyme cascade, and therefore no aquaporin vesicles fuse with membrane P. Then state the consequences: membrane P remains of low water permeability and little water crosses from the lumen.

Step-by-Step Reasoning

1. ADH cannot bind the receptor → the G protein (A) is not activated → adenylyl cyclase is not stimulated → cAMP (B) is not produced from ATP → the enzyme cascade is not triggered.
2. Without the cascade, vesicles carrying aquaporins (C) do not fuse with the cell surface membrane P, so few or no aquaporins are added to P.
3. The permeability of membrane P to water therefore remains low, and less water moves through P from the lumen into the cell — which is exactly why NDI patients pass large volumes of dilute urine (linked to part (c)).

Key Takeaways

• The water permeability of the collecting-duct membrane is hormonally regulated, not fixed. It is a dynamic property set by the number of aquaporins in the membrane at any moment.
• Blocking any step of the cascade has the same end result as having no ADH: low membrane water permeability and dilute urine.
• The ecf note in the mark scheme is important: if the candidate called structure C a "water channel" in part (a), they can still score in (d) by saying "few/no water channels added to P".

Common Mistakes

• Saying the membrane becomes "impermeable to water" — the mark scheme rejects "impermeable"; the correct wording is reduced / low permeability.
• Describing only the biochemical consequences (no cAMP, no cascade) without linking them to the physical state of membrane P — the question specifically asks about the effect on P, so the answer must say something about its aquaporin content and water permeability.
• Forgetting to state the consequence in terms of water movement, which is what would actually be observed.

Things to Be Careful About

• The question says "with reference to Fig. 1.1", so at least one mark is essentially guaranteed by mentioning membrane P and the aquaporin (structure C).
• This part tests the link between cell signalling and a measurable physiological outcome — keep the chain of reasoning tight.

Background Concept

Sex-linked genes in humans are carried on the X chromosome; the Y chromosome is much smaller and carries very few genes that are not also present on the X. A male has the sex chromosomes XY — his single X comes from his mother (who contributes only an X to a son) and his Y comes from his father (who contributes either an X or a Y to each child). For a recessive X-linked allele to be expressed in a male, he needs only one copy (because he has no second X to carry a dominant allele that could mask it). Such a male must therefore have received the recessive allele on the X chromosome he got from his mother; the X chromosome cannot have come from his father because his father gave him a Y.

Understanding the Question

The stem states: "NDI is caused by a recessive allele of the gene coding for the receptor protein. The gene is located on the X chromosome." The question asks the candidate to explain why a man with NDI could not have inherited the condition from his father. This is a sex-linkage inheritance question framed in a clinical context.

Approach

Use the sex chromosome contributions of each parent:

• A son's X chromosome comes only from his mother.
• A son's Y chromosome comes only from his father.
• The recessive allele for NDI is on the X chromosome, so it can only be transmitted on an X. The father does not pass an X to a son; he passes a Y. Therefore the allele could not have come from the father.

Step-by-Step Reasoning

1. The NDI allele is on the X chromosome; there is no corresponding allele on the Y chromosome.
2. To inherit an X-linked allele, a person must receive an X chromosome. A son receives his single X from his mother (and his Y from his father).
3. Therefore, if a man has the X-linked recessive allele, that X came from his mother. The Y he carries, which came from his father, does not carry the allele at all.
4. Conclusion: a man with NDI cannot have inherited the condition from his father; he must have inherited it from his mother (who would be a carrier, $\text{X}^H\text{X}^h$ if H is the dominant normal allele and h is the recessive NDI allele).

Key Takeaways

• A father's contribution to a son is the Y chromosome; the mother contributes the X.
• X-linked recessive conditions therefore appear in sons only via the maternal line; an affected father cannot pass an X-linked recessive allele to his son (he can only pass it to his daughters, who become carriers).
• Recognising which parent contributed which sex chromosome is the heart of any X-linked inheritance question.

Common Mistakes

• Saying "because the father is male he cannot pass the allele" — the father can carry and pass an X-linked allele, but only to his daughters (via his X); to a son he passes only the Y.
• Confusing X-linked with autosomal recessive inheritance and not invoking the sex chromosomes at all.
• Saying "the gene is only on the X chromosome, not on the Y" without connecting it to which parent gave which chromosome to the son.

Things to Be Careful About

• Use the words X chromosome and Y chromosome explicitly; vague statements like "from the mother because she's the one" will not earn both marks.
• A clear logical chain (X comes from mother, Y from father, allele is on X) is needed — the mark scheme awards one mark each for the X-from-mother and Y-from-father points.
• This question is a classic X-linkage reasoning item; practice with pedigree diagrams (aunts, uncles, grandfathers) reinforces the same logic.

Background Concept

Phenotypic variation among individuals in a population can be split into three underlying causes:

• $V_G$ — variation caused by genetic factors, i.e. differences in the alleles an individual carries. This variation is heritable.
• $V_E$ — variation caused by environmental factors acting on individuals (e.g. nutrition, light, temperature, disease, exercise). This variation is not heritable through the alleles involved, although the environment itself can sometimes affect future generations via epigenetic or maternal effects.
• $V_G + V_E$ — variation caused by a combination of both, where the phenotype is the product of an underlying genetic predisposition modified by the environment an individual experiences. Most continuously varying traits in outbreeding populations fall into this category.

A useful diagnostic: if the individuals are genetically identical (clones, identical twins, pure-breeding line under controlled conditions) but still differ in phenotype, the cause must be environmental. If the trait follows a simple Mendelian pattern unaffected by external conditions, it is genetic. If both genes and lifestyle/environment are known to influence the trait, it is $V_G + V_E$.

Understanding the Question

The question presents five real-world examples of phenotypic variation and asks the candidate to classify each as caused by $V_G$, $V_E$ or $V_G + V_E$. Three marks are available, awarded as 5/4-3/2-1 correct. The marks are therefore about consistent, defensible classification, not a single tricky decision.

Approach

Read each example and ask two diagnostic questions:

1. Is there an explicit mention of genes/alleles/heritable differences? If yes, $V_G$ is at least partly involved.
2. Is there an explicit mention of different environments, diets, or non-heritable influences? If yes, $V_E$ is at least partly involved.
3. If only one is present, the answer is that one alone; if both are present, the answer is $V_G + V_E$.

A special case to watch for: asexual reproduction produces genetically identical offspring. If those offspring still vary, the variation MUST be environmental, because no new genetic combinations are produced.

Step-by-Step Reasoning

Row 1 – Tomato yield, glasshouse vs outside, 17 genes identified:
Both genes (the 17 identified yield-associated genes) and environment (glasshouse vs outside growing conditions) are explicitly stated as influences. → $V_G + V_E$.

Row 2 – Strawberry plants from 'Sweet Ann' produced by asexual reproduction, but vary in size and fruit number:
Asexual reproduction (e.g. runners, cuttings) produces clones with identical genotypes. Therefore any difference between the new plants cannot be due to genetics — the new plants are genetically the same as the parent. The variation in size and fruit number must come from differing growth conditions (soil quality, light, water, nutrients, position in the row, competition). → $V_E$.

Row 3 – Domestic cat blood groups A, B and AB, determined by antigens on red blood cell membranes:
Blood group antigens are coded for by alleles at specific gene loci. They show discrete categories (a classic discontinuous variation), are inherited, and the question gives no environmental factors. → $V_G$.

Row 4 – Human weight gain: 50+ gene variants + risk factors including diet and exercise:
Both genetic variants AND environmental/lifestyle factors (diet, exercise) are stated. → $V_G + V_E$.

Row 5 – Human resting heart rate: influenced by biological sex and family history (genetic) and by smoking and medication (environmental):
Biological sex and family history both indicate heritable/genetic influences, while smoking and medication are environmental influences. → $V_G + V_E$.

Key Takeaways

• $V_G$ = variation between genotypes; heritable.
• $V_E$ = variation between individuals of the same genotype caused by their environment; not heritable through the alleles themselves.
• $V_G + V_E$ = the common situation where a trait has both a genetic component and an environmental component (most continuous traits in humans, including weight, height, blood pressure, heart rate).
• Asexually reproduced organisms (clones) that vary in phenotype demonstrate $V_E$ directly, because the genetic component is held constant.
• Blood group antigens are a textbook example of a discontinuous, purely genetic trait.

Common Mistakes

• Marking the asexually reproduced strawberries as $V_G + V_E$ because "genes are involved". This loses the point: the new plants are genetic copies, so any difference must be environmental.
• Marking the cat blood groups as $V_G + V_E$ because "the environment could matter". Blood group antigens are determined directly by alleles; the question gives no environmental influence, so it is $V_G$ only.
• Confusing $V_E$ with "non-genetic" in a general sense and missing the genetic component in the tomato/weight/heart-rate rows.

Things to Be Careful About

• The classification is about the cause of the observed variation, not whether genes exist somewhere in the organism. Even where a trait could be influenced by environment in principle, only assign $V_E$ (or $V_G + V_E$) if the question explicitly mentions an environmental factor.
• "Family history" and "biological sex" are proxies for genetic influence; "diet", "exercise", "smoking" and "medication" are proxies for environmental influence.
• Read each row as a stand-alone scenario — the marks are independent across rows.

Background Concept

Variation between individuals has two ultimate sources. One is the re-shuffling of existing alleles during meiosis and sexual reproduction (covered in part (c)). The other is the creation of new alleles — and that requires a change in the DNA sequence itself. Such a change is called a mutation: a spontaneous, random alteration in the nucleotide sequence of a gene or chromosome.

Mutations are described as:

• Spontaneous — they arise from random errors in DNA replication, damage from radiation, or chemical mutagens, without being directed by the organism.
• Random — they occur at any base in any gene, with no prediction of when, where, or what will change.
• Heritable only if they occur in the cells that give rise to gametes (germ-line mutations).

If a mutation occurs in a gene that influences a phenotype (e.g. the HBB sickle-cell allele, the TYR albino allele), the resulting new allele contributes to $V_G$.

Understanding the Question

This is a one-mark "name" question. The command word "name" requires a single biological term, not a sentence. The clue words in the question are "spontaneous" and "random event occurring in cells" — these are textbook descriptors of mutation.

Approach

Identify the only term in the syllabus that matches "spontaneous, random event occurring in cells" that introduces new phenotypic variation. That term is mutation.

Step-by-Step Reasoning

• Spontaneous + random = mutation (not crossing over, not independent assortment — those are programmed features of meiosis, not random events).
• "Occurring in cells" is satisfied because mutations happen during DNA replication in any dividing cell.
• The accepted one-word answer is mutation.

Key Takeaways

• Mutation is the original source of all new genetic variation; without it, evolution cannot occur.
• Mutations are random with respect to the needs of the organism — they are not produced in response to an environmental demand.
• Only germ-line mutations (in cells that form gametes) are inherited and contribute to $V_G$ in the next generation.

Common Mistakes

• Writing "gene" or "allele" — these are units that can be affected by mutation, but they are not the event itself.
• Writing "meiosis" — meiosis re-shuffles existing variation; it does not create new variation.
• Writing "crossing over" — crossing over is part of the answer to part (c) and is a programmed feature of meiosis, not a spontaneous random event.

Things to Be Careful About

• A single word is sufficient and is what the mark scheme expects.
• The marking note in the scheme lists "mutation" as the only acceptable answer for this part.

Background Concept

Sexual reproduction produces genetically unique offspring because the gametes that fuse are themselves genetically non-identical. This uniqueness is generated at two stages: during meiosis (when gametes are formed) and at fertilisation (when gametes fuse).

During meiosis three processes introduce variation:

1. Crossing over — in prophase I, non-sister chromatids of homologous chromosomes exchange segments of DNA at chiasmata. This physically recombines the maternal and paternal alleles on the same chromosome, producing chromatids that contain a mixture of alleles from both grandparents. After meiosis, gametes therefore contain chromosomes that are neither purely maternal nor purely paternal.

2. Independent (random) assortment — at metaphase I, the orientation of each pair of homologous chromosomes on the equator is independent of every other pair. For an organism with $n$ chromosome pairs this gives $2^n$ possible combinations of maternal and paternal chromosomes in the gametes. In humans ($n = 23$) this alone gives $2^{23} \approx 8.4 \times 10^6$ chromosome combinations per gamete.

3. Random fusion of gametes at fertilisation — any of the huge variety of male gametes can fuse with any of the huge variety of female gametes, so the allele combinations in the zygote are not predetermined.

In many populations, random mating (no preference for particular genotypes) is an additional feature, ensuring the gametes that meet are not biased by mate choice.

Understanding the Question

The question asks for features of sexual reproduction (other than the mutation named in (b)) that contribute to genetically different offspring. The command word is "describe", so each point should be a short statement of the feature, not a long essay. Three marks are available, so three distinct features are needed.

Approach

List the four classic mechanisms covered above and select any three. Each marking point requires the mechanism (e.g. "crossing over") and, where helpful, a brief reason it creates variation. The mark scheme accepts the bare name of each mechanism.

Step-by-Step Reasoning

• Crossing over: in prophase I of meiosis, non-sister chromatids of homologous chromosomes exchange DNA at chiasmata, producing new combinations of alleles on a single chromosome. This is the first source of variation inside meiosis.
• Independent (random) assortment: at metaphase I, the orientation of each bivalent is random, so homologous chromosomes (and therefore the alleles they carry) are distributed independently into daughter cells. This produces $2^n$ combinations per gamete.
• Random fusion of gametes at fertilisation: any sperm may fertilise any egg, so the combination of alleles in the zygote is left to chance. This multiplies the variation produced by meiosis.
• Random mating: in many populations, mate choice is not constrained by genotype, so the alleles entering a fertilisation event are not pre-selected.

Any three of the above four are credited.

Key Takeaways

• Sexual reproduction creates unique offspring because meiosis generates varied gametes and fertilisation combines them randomly.
• The three meiotic sources of variation are crossing over, independent assortment, and (random) gamete fusion at fertilisation.
• Random mating is an additional population-level feature, but only when mate choice is genuinely random.

Common Mistakes

• Confusing crossing over (which exchanges DNA between chromatids of homologous chromosomes) with the random fusion of gametes (which happens at fertilisation, not during meiosis).
• Calling the mechanism "segregation" — the CIE syllabus uses the term "independent/random assortment" for the separation of homologous chromosomes; "segregation" alone is the Mendelian term and is not the credited wording.
• Listing the same mechanism twice in different words (e.g. "random assortment of chromosomes" and "independent assortment of alleles") — only one mark can be awarded.
• Naming mutation again, even though the question explicitly excludes it ("other than the event named in (b)").

Things to Be Careful About

• The mark scheme explicitly accepts four alternative answers; any three earn the three marks.
• Use precise phrasing: "crossing over" rather than "DNA exchange"; "independent / random assortment" rather than "shuffling"; "random fusion of gametes / fertilisation" rather than "chance meeting".
• The four marks are awarded independently — credit any combination of three from the four listed.

Background Concept

The $F8$ gene codes for coagulation factor VIII, one of the plasma proteins in the blood-clotting cascade. The gene lies on the long arm of the X chromosome (Xq28). A loss-of-function mutation in $F8$ reduces or abolishes factor VIII activity, which in turn slows the activation of factor X (and ultimately thrombin), so fibrinogen is not converted to fibrin efficiently and a stable blood clot does not form. Because the gene is on the X chromosome, inheritance is sex-linked: a female has two X chromosomes and is usually unaffected unless both carry the mutation ($X^f X^f$); a male has only one X, so a single copy of the mutant allele ($X^f Y$) is enough to cause the disease. Different mutations in $F8$ produce factor VIII of differing activity, so the severity of haemophilia A ranges from mild to severe.

Understanding the Question

The question asks for an explanation of the relationship between three things: the $F8$ gene, the factor VIII protein, and the clinical condition haemophilia. The candidate must chain genotype to protein to phenotype.

Approach

Lay out the logic in order: gene → protein → physiological role → clinical effect. Be sure to include the sex-linkage aspect (it is haemophilia A) and the recessive nature of the allele.

Step-by-Step Reasoning

• The $F8$ gene is located on the X chromosome. Inheritance is therefore sex-linked.
• The allele responsible for haemophilia A is recessive to the wild-type allele.
• The recessive allele codes for a factor VIII protein that is non-functional, or is produced in much smaller amounts, or is absent altogether.
• Because factor VIII is required to activate factor X in the clotting cascade, its absence / reduction means the cascade runs too slowly. Thrombin is not activated efficiently, fibrinogen is not converted to fibrin, and the blood clot does not form properly.
• The phenotypic consequence is that bleeding (after injury, surgery, or spontaneously into joints) is excessive and prolonged.
• A male inheriting the recessive allele on his single X ($X^f Y$) is affected; a female needs two copies ($X^f X^f$) to be affected. Heterozygous females ($X^F X^f$) usually have enough factor VIII from their one functional allele to clot normally — which is why haemophilia A is much more common in males than in females.

Key Takeaways

• Sex-linked recessive inheritance: males are hemizygous and therefore express the recessive phenotype from a single copy.
• The molecular chain is: $F8$ allele → factor VIII activity → clotting-cascade speed → bleeding phenotype.
• Different $F8$ mutations give a spectrum of severity, not a single 'all-or-nothing' disease.
• Haemophilia A (factor VIII, $F8$) and haemophilia B (factor IX, $F9$, 'Christmas disease') are clinically similar but are caused by mutations in different genes; do not confuse them.

Common Mistakes

• Writing 'haemophilia is caused by a lack of red blood cells / platelets' — it is a clotting-factor deficiency, not a cellular deficiency.
• Calling the gene $F8$ but not mentioning its location on the X chromosome; the mark scheme requires 'X-linked' or equivalent.
• Saying 'the recessive allele causes the disease' without saying what the recessive allele does at the protein level.
• Writing $F^F$ and $F^f$ instead of $X^F$ / $X^f$ — the allele is on the X chromosome, not on an autosome.
• Mixing up haemophilia A (factor VIII, $F8$) with haemophilia B (factor IX, $F9$, 'Christmas disease').

Things to Be Careful About

• Sex-linkage must be explicit: write $X^F$ / $X^f$ to make clear the allele is on the X chromosome.
• Phenotype examples help: $X^F Y$ normal male, $X^f Y$ affected male, $X^F X^F$ normal female, $X^F X^f$ carrier female, $X^f X^f$ affected female.
• The 'AVP' mark in the mark scheme is for any valid additional point — e.g. that the severity varies with the mutation, or that the protein is required upstream of thrombin activation.

Background Concept

For a recombinant gene to be expressed in a host cell, three conditions must be met:

1. The gene must actually be present in the plasmid (i.e. the ligation step must have inserted the fragment, not just re-sealed the empty vector).
2. The gene must lie downstream of a promoter that the host's RNA polymerase (with its transcription factors) can recognise.
3. The gene must be inserted in the correct orientation — its coding strand must lie on the same strand that the promoter directs RNA polymerase to read, so that a sense mRNA is produced.

If the orientation is reversed, RNA polymerase will transcribe the opposite (template) strand and the resulting mRNA will be the antisense of the gene, which cannot be translated into a functional polypeptide. If the plasmid has simply re-circularised without picking up the gene, the bacterium receives an intact vector but no insulin gene, so again no insulin is made. In real industrial work, only a fraction of the plasmids recovered from a ligation reaction contain the insert in the correct orientation; the rest are either empty, recircularised without insert, or carry the gene in the wrong orientation.

Understanding the Question

The stem tells the student that only some plasmids taken up by host bacteria lead to expression of insulin. The figure shows three possible outcomes of the ligation step:

• Plasmid X — gene present, but inserted in the wrong direction (note the direction-of-transcription arrow on the gene points opposite to that of the promoter).
• Plasmid Y — gene present and in the correct orientation, with the transcription arrow matching that of the promoter.
• Plasmid Z — recircularised plasmid with no gene of interest inside (only the original marker gene and promoter).

The student is asked to comment and explain — so the answer must give the outcome (insulin / no insulin) and the reason (orientation or absence of gene).

Approach

For each plasmid, apply the three conditions above and decide:

1. Is the gene present? (look at the figure)
2. Is a promoter upstream of it? (look at the figure)
3. Is the gene the right way round? (compare the direction of transcription arrow on the gene with that of the promoter)

Step-by-Step Reasoning

• Plasmid X — the gene is present and a promoter is upstream, but the gene's direction-of-transcription arrow points opposite to the promoter's. RNA polymerase will read the wrong strand; no sense mRNA is made, so no insulin polypeptide is produced. The reason to give is therefore gene inserted the wrong way round / backwards.
• Plasmid Y — the gene is present, a promoter is upstream, and the direction-of-transcription arrows on the promoter and the gene point the same way. RNA polymerase will read the coding strand, produce a sense mRNA, and the host's ribosomes will translate it into the insulin polypeptide. The reason is therefore gene inserted the correct way round.
• Plasmid Z — there is no gene of interest inside the plasmid at all; the cut plasmid simply re-ligated to itself (its own sticky ends paired up). Although the marker gene and promoter are present, they have no gene to drive expression of, so no insulin polypeptide is produced. The reason is that the gene was not inserted (the plasmid re-circularised; it is a non-recombinant plasmid).

Key Takeaways

• The presence of a promoter is not enough — the gene must be inserted in the correct orientation.
• A recircularised plasmid (no insert) is a common, useless product of a ligation reaction; this is why the marker gene is also needed for screening.
• This question illustrates why expression efficiency is rarely 100 % in a transformation experiment, even when transformation is successful.

Common Mistakes

• Stating that plasmid Z "does not contain a promoter" — it does contain a promoter; the problem is the absence of the gene of interest.
• Saying plasmid X produces insulin because the gene is present — ignoring the orientation issue.
• Vague wording such as "the gene is not the right one" or "the gene is damaged"; the specific point is the orientation of insertion.

Things to Be Careful About

• The question is a comment with explanation — both elements are needed for full credit.
• Three independent marks: one per plasmid, each requiring a clear reason.
• The figure is the evidence: read the direction of the arrows on promoter and gene carefully; do not rely on what a "correct" plasmid usually looks like.

Background Concept

During prophase I of meiosis, homologous chromosomes pair up to form bivalents (tetrads). Programmed double-stranded breaks (DSBs) are deliberately introduced into the DNA of these chromosomes by the enzyme SPO11. These breaks are not damage in the harmful sense — they are a normal part of meiosis and initiate the physical exchange of DNA segments between non-sister chromatids of homologous chromosomes. The visible manifestation of this exchange is a chiasma (plural: chiasmata), which is the point where two non-sister chromatids of a homologous pair cross over one another.

Understanding the Question

The stem of the question establishes that BRCA2 is involved in repairing double-stranded DNA breaks, and Part (a) asks specifically what biological event in meiosis is initiated by these double-stranded breaks. The question is essentially asking you to identify the natural, programmed use of DSBs during gamete formation.

Approach

Recognise that DSBs in meiosis are not pathological here — they are a controlled, deliberate part of the mechanism that generates genetic variation. Recall the sequence: SPO11 cuts DNA → strands are resected → single-stranded overhangs invade the homologous chromatid → crossover resolution → chiasma formation. So the event initiated is crossing over.

Step-by-Step Reasoning

1. Double-stranded DNA breaks are made at chosen sites in prophase I.
2. The 3' single-stranded ends produced by resection invade the homologous (non-sister) chromatid.
3. Repair of the breaks using the homologous chromatid as a template produces recombinant chromatids.
4. The cytological point at which the exchange is visible is the chiasma.
5. The biological process that this represents is crossing over — a key source of genetic variation through recombination.

Key Takeaways

• Programmed DSBs in meiosis initiate crossing over / chiasma formation, not damage.
• This is the basis of genetic recombination and one of the major sources of genetic variation in sexually reproducing populations.
• BRCA2's role in repairing DSBs is essential because unrepaired breaks in somatic cells can lead to mutations and cancer, but in meiosis the same kind of break is exploited to generate diversity.

Common Mistakes

• Writing "replication" — DSBs do not initiate replication; they initiate recombination.
• Writing "mutation" — DSBs in meiosis are not random mutations; they are programmed events leading to legitimate crossing over.
• Writing "separation of homologues" or "independent assortment" — these occur later (anaphase I) and are consequences of alignment, not DSBs.

Things to Be Careful About

The mark scheme accepts "crossing over" or "chiasma formation" as alternatives. Either is a complete answer. Do not give a long description of the whole prophase I — the question asks for the single event initiated by the breaks.

Background Concept

Stomata are small pores, mainly on the underside of leaves, each flanked by a pair of bean-shaped guard cells. They are the main route by which CO2 enters the leaf for photosynthesis and by which water vapour leaves the leaf (transpiration). Because CO2 is needed for the Calvin cycle but water is constantly lost through the same pore, the plant faces a fundamental trade-off: it must keep stomata open enough to feed photosynthesis, but not so open that it loses more water than its roots can replace.

Guard cells respond to several environmental signals. Blue light (via the phototropin receptors) and low internal CO2 stimulate opening, while darkness, drought (via abscisic acid, ABA) and high CO2 inside the leaf stimulate closure. This is a classic homeostatic compromise: a single opening is regulated by multiple, sometimes opposing, environmental cues.

Understanding the Question

The command word is explain, so for each marking point you need to state WHY the response helps the plant — not just that the response happens. The question gives you 3 marks, so you need any three valid points from the mark scheme: one point about why opening is needed, and two about why and when closing is needed. Always link the stomatal behaviour to a plant process (photosynthesis, transpiration) and to an environmental trigger.

Approach

Start with the photosynthesis side (why open), then the water-balance side (why close), and finally give a specific example of an environmental cue that forces closure. Three short, distinct sentences — each making a clear cause-and-effect link — will score full marks.

Step-by-Step Reasoning

1. Photosynthesis requires CO2 — atmospheric CO2 (~0.04%) enters the leaf through the open stomata by diffusion down its concentration gradient. Without an open pore, the Calvin cycle in the chloroplasts would run out of CO2 and photosynthesis would stop.
2. Transpiration loses water — water evaporates from the wet cell walls of the mesophyll and diffuses out of the leaf through the stomata. If stomata stayed permanently wide open in hot, dry or windy conditions, the plant would lose water faster than the roots could replace it, leading to wilting and death.
3. Darkness removes the need to be open — at night there is no light, so the light-dependent reactions of photosynthesis cannot occur and CO2 fixation is not needed. Closing stomata in the dark conserves water without any cost to photosynthesis.
4. Stressful conditions force closure — high light intensity, high temperature, low humidity, high wind speed and drought all increase the rate of transpiration. The guard cells respond (often via abscisic acid in drought) by becoming flaccid and closing the pore, limiting water loss.

Any three of these four points earn the 3 marks.

Key Takeaways

• Stomata must balance two competing demands: CO2 uptake for photosynthesis and water retention.
• Closure is triggered by conditions that increase evaporative loss (heat, wind, drought) and by conditions where photosynthesis cannot occur (darkness).
• The opening/closing response is a homeostatic mechanism that maintains the plant's internal water status.

Common Mistakes

• Saying stomata open "to let air in" — vague; the mark requires the specific gas (carbon dioxide) and its role in photosynthesis / the Calvin cycle.
• Stating that stomata close "to save water" without naming the process (transpiration) — be specific.
• Forgetting to give a trigger — closing in response to high temperature is a different marking point from closing at night.

Things to Be Careful About

• "Explain" demands both the observation and the reason; "describe" only requires the observation.
• The mark scheme accepts "high light intensity / high temperatures / high wind speed / drought" as alternative stressful triggers — give any one to score the point.

Background Concept

Guard cells are specialised epidermal cells that change shape to open and close the stomatal pore. Their behaviour is driven by turgor — when they are turgid they bow apart and the pore opens; when they lose solutes and become flaccid the pore closes. The opening mechanism is a beautiful example of how plants use active transport to drive a passive osmotic response.

The plasma membrane of a guard cell contains an H$^+$-ATPase (proton pump) and voltage-gated K$^+$ channels. Because the inner wall (next to the pore) is thicker and reinforced with radial cellulose microfibrils, while the outer wall is thinner and more elastic, a uniform increase in volume forces the guard cells to bend away from each other. This asymmetry converts a turgor change into an aperture change.

Understanding the Question

The command word is describe, so you must set out the sequence of events clearly. Six marks = six distinct creditable points (or fewer if you also give some of the AVP points). The mark scheme explicitly warns that if you describe CLOSURE instead of opening you can earn at most 3 marks — so be sure your answer is about opening. The mechanism starts with light (or another trigger) and ends with the pore opening.

Approach

Work through the mechanism in strict order:

light / trigger → H$^+$ pumped out → K$^+$ moves in → solute concentration rises → water potential falls → water enters by osmosis → guard cells become turgid → differential wall stretching bends the cells apart → pore opens.

Each arrow is a marking point. Keep the chain tight so the examiner can credit every step.

Step-by-Step Reasoning

1. Trigger and pump activation — Blue light (absorbed by the phototropin receptor) activates the H$^+$-ATPase in the guard cell plasma membrane. This pump uses ATP to actively transport protons (H$^+$) out of the guard cell into the surrounding apoplast. The active transport step is essential; passive movement of H$^+$ would not produce the required gradient.
2. Establishing the gradient — Pumping H$^+$ out lowers the H$^+$ concentration inside the cell and hyperpolarises the membrane (the inside becomes more negative). This sets up both a chemical gradient and an electrical gradient that together drive the next step.
3. K$^+$ uptake — The hyperpolarisation opens voltage-gated inward-rectifying K$^+$ channels in the plasma membrane. K$^+$ ions flow into the guard cells down their electrochemical gradient. Chloride ions (Cl$^-$) and malate$^2-$ often accompany the K$^+$ to balance charge.
4. Solute accumulation lowers water potential — The high internal concentration of K$^+$ (and accompanying anions) makes the cytoplasm and vacuole hypertonic to the surrounding apoplast. The water potential ($\Psi$) of the guard cell therefore decreases (becomes more negative).
5. Osmotic water entry — Water moves into the guard cell by osmosis, down the water potential gradient. Aquaporins in the plasma membrane and tonoplast facilitate rapid water flow.
6. Turgor increases — As water enters, the central vacuole swells and the protoplast pushes against the cell wall. The guard cell becomes more turgid.
7. Differential wall stretching opens the pore — The inner wall (next to the stomatal pore) is thicker, more rigid and reinforced by radially arranged cellulose microfibrils, so it resists stretching. The outer wall is thinner and more elastic, so it stretches much more. The combined effect is that the guard cells bow outwards (like a pair of curved sausages), pulling the inner walls apart and opening the pore between them.

Any six of these points earn the marks; the seventh (asymmetric walls) is often the one students forget and is well worth including.

Key Takeaways

• Stomatal opening is driven by an active proton pump, but the bulk water movement is passive (osmosis).
• Solute accumulation (K$^+$ and accompanying anions) lowers water potential, drawing water in.
• The asymmetric wall structure converts isotropic turgor into a directional opening of the pore.
• The sequence is: active H$^+$ efflux → K$^+$ influx → osmotic water uptake → turgor → shape change → open pore.

Common Mistakes

• Describing closure (loss of K$^+$, ABA signalling, Ca$^{2+}$ channels) instead of opening — capped at 3 marks.
• Saying water "enters by active transport" — water only moves by osmosis.
• Saying K$^+$ is actively transported in — K$^+$ moves down its electrochemical gradient through channels; it is the H$^+$ pump that uses ATP.
• Omitting the wall asymmetry point and so losing a free mark — the bowed shape of the guard cells is the mechanical reason the stoma actually opens.
• Confusing the location of the thicker wall (it is the INNER wall, bordering the pore, that is thicker and less elastic — students often reverse this).

Things to Be Careful About

• Mark scheme accepts either "outer wall thinner, allows more stretching" OR "inner wall thicker, allows less stretching" — either phrasing is fine.
• "AVP" points (mark point 8) include: light stimulating ATP production, Cl$^-$ co-transport, and aquaporin-mediated water entry — include any of these for extra credit if you have covered the core points.
• The mechanism is initiated by blue light, not by photosynthesis directly; photosynthesis lowers internal CO2 which also contributes, but the standard CIE mechanism starts with H$^+$-ATPase activation by light.

Background Concept

The light-dependent reactions of photosynthesis take place on the thylakoid membranes inside the chloroplast. The thylakoid membrane is the site of two key pieces of molecular machinery: the photosystems (which capture light energy) and the electron transport chain with ATP synthase (which convert that energy into ATP). ATP is made indirectly, by building up a difference in $\text{H}^+$ concentration across the thylakoid membrane and then letting those protons flow back down their gradient through ATP synthase. The synthesis of ATP in this way, driven by a proton gradient, is called chemiosmosis — the same principle that operates in mitochondria during oxidative phosphorylation.

Understanding the Question

Part (b) asks the candidate to describe, in a clear sequence, how the absorption of a photon by chlorophyll leads to the synthesis of ATP. The mark scheme explicitly offers seven creditable points (any four of which are needed for full marks), and the candidate should arrange them in the correct causal order so that the mechanism reads as a continuous story rather than a list of disconnected facts.

Approach

Walk through the pathway in the order the events actually happen on the membrane:

1. Light absorption → photoactivation (an electron in chlorophyll is raised to a higher energy level).
2. The high-energy electron leaves chlorophyll and is passed along a series of electron carriers (the electron transport chain), losing energy at each step.
3. Some of that released energy is used by the carriers to actively transport $\text{H}^+$ across the membrane, from the stroma into the thylakoid lumen.
4. The accumulation of $\text{H}^+$ inside the lumen sets up a proton gradient (and an electrochemical gradient) across the thylakoid membrane.
5. $\text{H}^+$ flows back down its gradient through the channel part of ATP synthase, into the stroma.
6. The energy of this proton flow drives the catalytic part of ATP synthase to combine ADP and $\text{P}_i$ into ATP — this is chemiosmosis.

A complete answer should make all of these points, but the mark scheme only requires four of the seven listed.

Step-by-Step Reasoning

• Photoactivation. A photon of light is absorbed by a chlorophyll molecule in a photosystem. An electron in the chlorophyll is promoted to a higher energy level; this is photoactivation. The excited electron can leave the chlorophyll and enter the electron transport chain.
• Electron transport. The electron is passed from one carrier to the next along the chain of electron carriers embedded in the thylakoid membrane. At each transfer, a small amount of energy is released.
• Proton pumping. Some of the carriers use the released energy to actively pump $\text{H}^+$ from the stroma, across the membrane, into the thylakoid space (lumen). This raises the $\text{H}^+$ concentration inside the thylakoid space relative to the stroma.
• Proton gradient. The difference in $\text{H}^+$ concentration across the thylakoid membrane, together with the membrane potential that builds up, constitutes a proton motive force (an electrochemical gradient).
• ATP synthase. $\text{H}^+$ ions flow back from the thylakoid space into the stroma through a channel in the enzyme ATP synthase. The enzyme is shaped so that this flow rotates part of its structure, and the rotation drives the catalytic synthesis of ATP from ADP and inorganic phosphate ($\text{P}_i$) in the stroma.
• Chemiosmosis. The entire process — building a gradient by active transport and then harvesting it through ATP synthase — is called chemiosmosis, the same mechanism that operates in respiration.

Key Takeaways

• Photoactivation of chlorophyll is the start of the process; the energy of the excited electron is not used directly to make ATP, but is converted into a proton gradient.
• The proton gradient is set up across the thylakoid membrane, with $\text{H}^+$ being high in the thylakoid space and low in the stroma.
• ATP synthase is the only route for protons to flow back; the energy released as they flow drives the synthesis of ATP from ADP and $\text{P}_i$.
• The whole process is called chemiosmosis and is essentially the same idea as oxidative phosphorylation in mitochondria.

Common Mistakes

• Saying that ATP is made 'in the thylakoid space' — ATP is actually made in the stroma, where ADP and $\text{P}_i$ are available to ATP synthase.
• Confusing the direction of proton flow: protons are pumped into the thylakoid space, but they flow out of the thylakoid space (back into the stroma) through ATP synthase.
• Omitting the electron transport chain and jumping straight from 'excited electron' to 'ATP is made'.
• Forgetting to say where the energy for proton pumping comes from (it comes from the electron transport chain).
• Calling the thylakoid space the 'thylakoid' (it is the fluid-filled space inside the thylakoid, often called the lumen).

Things to Be Careful About

• Use the term 'chemiosmosis' at least once — it is a specific mark-scheme point.
• The mark scheme requires the candidate to state both that the protons are pumped AND that this creates a gradient. Writing only one of the two is a common half-credit mistake.
• The direction of proton flow through ATP synthase (from thylakoid space to stroma) must be stated; examiners penalise the reverse direction.

Background Concept

A mitochondrion is bounded by a smooth outer membrane and a highly folded inner membrane (the folds are called cristae). The two membranes enclose two compartments: the intermembrane space (between the membranes) and the matrix (inside the inner membrane). This double-membrane system is essential to aerobic respiration because the inner membrane houses the protein complexes of the electron transport chain (ETC) and ATP synthase. As electrons pass along the ETC, protons (H+) are pumped from the matrix into the intermembrane space, generating an electrochemical gradient. Protons flow back into the matrix through ATP synthase, driving the phosphorylation of ADP to ATP — this is chemiosmosis. For this gradient to be harnessed for ATP synthesis, the inner membrane must remain largely impermeable to H+, and the surface area available for ETCs and ATP synthases must be large — which is exactly what the cristae provide.

Understanding the Question

Part (a) carries a Fig. 8.1 of a mitochondrion with the outer membrane and the folded inner membrane labelled. The command word is "outline", which means a brief description of the main features. The question asks specifically about the roles of the membranes (not the matrix, not the enzymes) in respiration, so the answer should be membrane-centred. Four marks are available, and the mark scheme supplies eight possible points; the candidate only needs any four of them.

Approach

Think about what each membrane physically does: it is a barrier and a surface. As a barrier it (i) controls the movement of respiratory substrates and products (O2, pyruvate in; CO2 out), (ii) maintains the H+ gradient by being selectively impermeable, and (iii) creates compartments (matrix vs intermembrane space) so that the different stages of respiration are separated. As a surface it provides the location for the ETC and ATP synthase, and its folding (cristae) enlarges that surface.

Step-by-Step Reasoning

• Substrate and product exchange: small, non-polar molecules (O2, CO2) and pyruvate must cross the membranes to reach their sites of use. The outer membrane has large non-specific porins; the inner membrane is more selective, but specific carriers admit pyruvate. CO2 produced by the link reaction and Krebs cycle must diffuse out.
• Compartmentalisation: the matrix contains the enzymes of the Krebs cycle, the link reaction and (in prokaryotic-like relics) mitochondrial DNA and ribosomes. The intermembrane space is the destination of the H+ pumped by the ETC. Keeping these regions separate is essential to the chemiosmotic mechanism.
• Site of oxidative phosphorylation: the ETC complexes I–IV and ATP synthase are integral membrane proteins of the inner membrane. Pumping H+ across the membrane as electrons travel along the chain stores energy as a proton-motive force.
• Surface area: the inner membrane is thrown into cristae, which can occupy a large proportion of the volume of a mitochondrion in active cells (e.g. cardiac muscle). This means many ETCs and ATP synthases can be packed in, increasing the rate of ATP production.
• Impermeability to H+: the inner membrane does not allow protons to leak back into the matrix except through ATP synthase, so the proton gradient built up by the ETC is preserved and used for ATP synthesis rather than being dissipated.

Key Takeaways

• The mitochondrion's two membranes do three jobs: control what enters and leaves, separate the matrix from the intermembrane space, and provide the surface for oxidative phosphorylation.
• Cristae = increased surface area for ETCs and ATP synthases.
• An intact, H+-impermeable inner membrane is a prerequisite for chemiosmosis.

Common Mistakes

• Saying the Krebs cycle occurs on the inner membrane — it occurs in the matrix (the mark scheme explicitly ignores this).
• Describing the outer membrane as the site of oxidative phosphorylation — this is wrong; it is the inner membrane.
• Saying the membrane "produces ATP" without mentioning ATP synthase or chemiosmosis — too vague for a mark.
• Confusing the intermembrane space with the matrix when describing where protons accumulate.

Things to Be Careful About

• The command word "outline" does not require extensive detail — short, distinct points, one mark each, are sufficient.
• Link function explicitly to the membrane (e.g. "inner membrane holds ATP synthase", not just "ATP synthase is present").
• Remember that the cristae are infoldings of the inner membrane, not separate structures.

Background Concept

Mitochondria have a limited lifespan and are continually turned over. The average mitochondrion survives for only days to weeks depending on cell type. Old or damaged mitochondria are typically removed by mitophagy — they are engulfed by autophagosomes and fused with lysosomes, whose hydrolytic enzymes (proteases, lipases, nucleases) digest the organelle. The amino acids, fatty acids, sugars and nucleotides released are recycled by the cell and used, together with new mitochondrial proteins encoded by nuclear DNA, to assemble new mitochondria. This turnover maintains a healthy population of functional mitochondria and prevents a fall in the cell's capacity for oxidative phosphorylation.

Understanding the Question

This "suggest" question is worth 2 marks. The stem tells the candidate that old or damaged mitochondria reduce the cell's ability to respire and asks what happens to them so that the cell's overall rate of respiration and ATP production is kept constant. The candidate must therefore describe the removal and recycling of dysfunctional mitochondria, not their repair.

Approach

Think about (i) how the cell destroys an organelle, and (ii) what happens to the breakdown products. The two marks correspond to these two ideas: degradation by lysosomes, and recycling of the components to make new mitochondria.

Step-by-Step Reasoning

• Degradation: damaged mitochondria are tagged (e.g. by PINK1/Parkin in mammals), engulfed and fused with lysosomes, and the proteins, lipids and DNA of the mitochondrion are hydrolysed.
• Recycling: the resulting amino acids, fatty acids, sugars, nitrogenous bases and phosphate are re-used in cellular metabolism. Together with fresh proteins translated from nuclear mRNA, they are assembled into new mitochondria. Because new functional mitochondria replace the old non-functional ones, the total mitochondrial mass and the cell's overall capacity for oxidative phosphorylation are maintained, even though individual mitochondria are short-lived.

Key Takeaways

• Mitochondria are continually turned over by lysosomal degradation (mitophagy).
• The breakdown products are recycled to build new mitochondria.
• This turnover is essential for sustained cellular ATP production and for quality control of the mitochondrial population.

Common Mistakes

• Saying the mitochondria are "repaired" — the question explicitly states they are old or damaged, and the mark scheme expects degradation.
• Saying only that lysosomes digest them, without mentioning the recycling of components (this only earns one of the two marks).
• Confusing lysosomes with proteasomes — proteasomes degrade individual cytosolic proteins, not whole organelles.

Things to Be Careful About

• The question asks how the cell maintains the same overall rate of respiration — the answer must therefore show that new mitochondria are produced to compensate for the lost ones; the word "overall" is the key.
• Use precise language: "hydrolytic enzymes of lysosomes" or "lysosomal digestion" is the correct phrasing, not just "broken down".

Background Concept

Biodiversity — the variety of living organisms — has value to humans at several levels:

• Utilitarian value (instrumental): direct uses such as food, timber, fibres, medicines, and the breeding stock for agriculture.
• Ecosystem-service value: functioning ecosystems provide pollination, soil formation, nutrient cycling, climate regulation, flood control and water purification.
• Genetic-resource value: wild relatives of crops carry alleles for disease resistance, drought tolerance and other traits that breeders may need in the future.
• Intrinsic value: aesthetic, ethical, cultural and spiritual reasons — the argument that species have a right to exist independent of human utility.
• Option value: the future, unknown uses of species we have not yet discovered (e.g. a wild plant whose secondary metabolites may one day treat a new disease).

Cambridge syllabus explicitly expects students to be able to outline these categories and give at least one example for each.

Understanding the Question

This is a "outline" question worth four marks. The candidate must give four distinct reasons for maintaining plant biodiversity. The mark scheme accepts any four from a list of eight, and rewards a single supporting example where it is offered. There is no requirement to structure the answer in any particular way, but clearly separated points (bullet list, numbered list) help the examiner to award the marks.

The question is specifically about plant biodiversity, so examples and reasons should be drawn from the plant kingdom (e.g. timber, fibre, medicines derived from plants, wild relatives of crops).

Approach

1. Decide on four distinct categories. Avoid repeating the same idea in two different wordings.
2. For each category, give a short justification. A concrete example strengthens the point.
3. Order the points for readability (e.g. from utilitarian to intrinsic, or vice versa).

A strong four-point answer picks one each from: a direct economic use, a medical/option use, an ecosystem-service use, and an aesthetic/ethical/cultural reason. Candidates can also include genetic-resource value and soil stability if preferred.

Step-by-Step Reasoning

• Reason 1 — Future / option value. We do not yet know all the ways plants might be useful; some plants may contain as-yet-undiscovered medicines or industrial chemicals. Mark-scheme line 1.
• Reason 2 — Aesthetic / ethical / cultural. Plants have beauty (wildflowers, forests), and many people believe species have an intrinsic right to exist. Cultural value includes sacred groves, national emblems, and traditions tied to particular plants. Mark-scheme line 2.
• Reason 3 — Medical. Plants are a major source of pharmaceutical drugs: aspirin (from willow bark, Salix), quinine (from Cinchona), morphine (from opium poppy, Papaver somniferum), vincristine (from Catharanthus roseus), digitalis (from Digitalis). Mark-scheme line 3.
• Reason 4 — Ecological stability. Plant communities underpin food webs and provide ecosystem services (pollination, soil formation, water regulation, climate buffering). Loss of plant diversity destabilises these systems. Mark-scheme line 5.
• Reason 5 — Economic resources. Plants supply wood for building, fibres (cotton, linen, jute) for clothing, food directly (cereals, fruit, vegetables), and feedstock for agriculture. Mark-scheme line 6.
• Reason 6 — Genetic resource. Wild relatives of crops carry alleles that breeders can use to improve yield, disease resistance or environmental tolerance. Maintaining genetic diversity preserves this raw material. Mark-scheme line 7.
• Reason 7 — Soil stability. Plant roots bind soil; loss of vegetation (e.g. through deforestation or overgrazing) accelerates erosion. Mark-scheme line 8.
• Reason 8 — Ecotourism. Wild plant communities attract tourists, supporting local economies and giving those communities a financial incentive to conserve rather than exploit. Mark-scheme line 4.

Any four of these earns full marks.

Key Takeaways

• Biodiversity has multiple categories of value — utilitarian, ecological, genetic, aesthetic/ethical, option.
• Concrete examples are always rewarded; vague platitudes ("plants are important") are not.
• "Outline" questions reward breadth over depth — four crisp points beat one long paragraph.

Common Mistakes

• Repeating the same idea twice (e.g. "food" and "agriculture" as two separate points) — only one mark.
• Omitting an example when one would clearly earn extra credit.
• Confusing reasons for plant biodiversity with reasons for animal biodiversity — the question is specifically about plants, so animal-only examples (e.g. "pandas in zoos") are off-target.
• Forgetting the "plant" focus and listing general biodiversity reasons without plant-specific grounding.

Things to Be Careful About

• The mark scheme gives any four from eight, so the order in which reasons are presented does not matter.
• "Maintain stability of ecosystems" is a stronger answer than "keeps nature balanced" — the examiner rewards mechanistic understanding.
• "Gene pool / genetic diversity" should be tied to a use (e.g. crop breeding), otherwise it is just a phrase.

Background Concept

The resting potential of a neurone (about −70 mV inside relative to outside) is maintained by the sodium–potassium (Na⁺/K⁺) pump, an active-transport protein embedded in the cell surface membrane. For every one cycle it uses one molecule of ATP to pump 3 Na⁺ out of the axon and 2 K⁺ in, against their respective concentration gradients. This unequal pumping makes the inside of the axon negative relative to the outside and keeps the Na⁺ and K⁺ gradients steep — gradients that are then exploited during the action potential.

In a typical textbook diagram of the pump:

• Triangles (△) are used to represent sodium ions (Na⁺).
• Circles (○) are used to represent potassium ions (K⁺).
• The pump protein is the bulky structure spanning the membrane.
• The square (□) symbol represents the ATP molecule that binds to the pump on its cytoplasmic (axoplasm) side to provide energy for active transport.

The pump operates only on the axoplasm side of the membrane, because that is where ATP is generated (by mitochondria) and where the relevant ion-binding sites face.

Understanding the Question

Part (a) gives Fig. 10.1, a simplified diagram of the Na⁺/K⁺ pump. The candidate must add two label lines and the letters K and A to the correct features. The question provides the key:

• K – potassium ions
• A – ATP

The mark scheme awards one mark for each correct placement, so both must be correct to score 2.

Approach

Read the question first: it tells you exactly what each label refers to. Then look at the symbols in the figure. The trick is to notice that both potassium ions and the ATP symbol are located in the axoplasm (lower side), so both labels are added below the membrane.

Step-by-Step Reasoning

1. Look at the symbols. Triangles are in the tissue fluid (top); from the question and the standard convention these are Na⁺. Circles are in the axoplasm (bottom) — these are the K⁺ ions. The square is the ATP molecule that drives the pump.
2. Draw a label line from the letter K to any circle in the axoplasm. (1 mark)
3. Draw a label line from the letter A to the square in the axoplasm. (1 mark)
4. Common trap: do not point K to a triangle (that would be Na⁺), and do not point A to a circle (that would be a K⁺ ion). Both labels belong in the axoplasm because that is the side of the membrane where K⁺ is concentrated by the pump and where ATP binds to the pump protein.

Key Takeaways

• In standard CIE diagrams, triangles = Na⁺ and circles = K⁺.
• The Na⁺/K⁺ pump is an active-transport protein — it needs ATP, and the ATP-binding site is on the axoplasm (cytoplasmic) side.
• Always read the key supplied with the question before labelling, since symbol conventions can vary between papers.

Common Mistakes

• Pointing K to a triangle (Na⁺): the candidate has confused the ion shapes.
• Pointing A to a circle or triangle: ATP is the square symbol, not an ion.
• Drawing both labels above the membrane: both K⁺ and ATP are on the axoplasm side of the membrane, not in the tissue fluid.

Things to Be Careful About

• Use label lines (straight lines ending in a small touch, not arrows pointing into structures) and place the letter at the end of the line, as required by CIE drawing conventions.
• Do not write the answer in prose in the margin — the marks are for the labels on the figure itself.

Background Concept

An action potential is a rapid, transient reversal of the membrane potential of an excitable cell, propagating along the axon without losing amplitude. It is generated by the coordinated opening and closing of voltage-gated sodium (Na⁺) channels and voltage-gated potassium (K⁺) channels in the axon membrane. The Na⁺/K⁺ pump has not finished its work — it has already established the steep ion gradients (high Na⁺ outside, high K⁺ inside) during the resting potential; the action potential is the controlled, brief collapse and restoration of those gradients.

The key terms to use precisely are:

• Depolarisation – the inside of the membrane becoming less negative (more positive) than the resting potential.
• Repolarisation – the inside returning towards the negative resting potential.
• Hyperpolarisation – the inside becoming more negative than the resting potential (an 'undershoot').
• Threshold – the critical membrane voltage (about −55 mV) at which enough Na⁺ channels are open that the depolarisation becomes self-propagating.

Understanding the Question

This is a 'describe the sequence' question worth 4 marks. The mark scheme credits any 4 points from a list of 7 specific statements, plus an AVP. The candidate must give a clear ordered account using the correct technical vocabulary — marks are tied to specific terms such as 'depolarised', 'threshold', 'repolarised' and 'hyperpolarised'.

Approach

Mentally run through the action potential as it would appear on a voltage-versus-time trace: a flat baseline (resting), an upward stroke (depolarisation), a downward stroke (repolarisation), a dip below the baseline (hyperpolarisation), then a return to baseline. Match each phase to the ion channel responsible and to the precise vocabulary the mark scheme requires.

Step-by-Step Reasoning

1. Na⁺ channels open (voltage-gated). Because Na⁺ is more concentrated outside the axon and the inside is negative, Na⁺ rushes into the axon down its electrochemical gradient. (MP1)
2. The influx of positive charge makes the inside less negative — the membrane depolarises. (MP2)
3. If depolarisation reaches the threshold potential, an all-or-nothing action potential fires. Below threshold, the signal simply decays. (MP3)
4. At the peak of the action potential, Na⁺ channels inactivate and voltage-gated K⁺ channels open. K⁺ diffuses out of the axon, taking positive charge with it. (MP4)
5. Loss of positive charge from the axoplasm repolarises the membrane — the inside becomes negative again. (MP5)
6. K⁺ channels are slow to close, so K⁺ continues to leave briefly, driving the potential below resting level — this is hyperpolarisation. (MP6)
7. The Na⁺/K⁺ pump and leak channels restore the original ion distribution, and the membrane returns to resting potential. (MP7)
8. (AVP) A strong candidate may add that the entry of Na⁺ makes the local membrane potential more positive, which in turn opens more voltage-gated Na⁺ channels — the positive-feedback loop that produces the rapid upstroke of the action potential.

Any four of these marking points (1–7) score full marks; the AVP is an alternative route to the fourth mark.

Key Takeaways

• The action potential is a sequence of voltage-gated channel events; describe it in order.
• Memorise the exact vocabulary: depolarised, repolarised, hyperpolarised, threshold, resting potential. Vague terms such as 'the charge changes' or 'ions move' will not earn marks.
• The Na⁺/K⁺ pump does not generate the action potential — it sets up the gradients that make it possible. The action potential itself is generated by facilitated diffusion of Na⁺ in and K⁺ out through voltage-gated channels.

Common Mistakes

• Writing 'sodium ions move in and potassium ions move out' without naming the channels or the resulting change in membrane potential.
• Saying the membrane 'reverses polarity' instead of using depolarised / repolarised.
• Forgetting to mention threshold, or describing the action potential as graded rather than all-or-nothing.
• Omitting hyperpolarisation — it is a separate, named phase worth a mark.
• Confusing the role of the Na⁺/K⁺ pump (which restores gradients afterwards) with the channels that generate the action potential.

Things to Be Careful About

• 'Describe the sequence' demands an ordered account; marks can be lost for jumping backwards and forwards.
• The mark scheme credits 'Na⁺ enters' or 'sodium channels open' (and similarly for K⁺); the candidate may give either but it is safer to name the channel and the direction of ion movement.
• The word 'refractory period' is closely linked to this topic but is not required to answer this specific question — do not pad the answer with it unless it is the AVP you choose to use.