Biology 9700/22 — October/November 2024

Background Concept

Water's exceptional solvent properties come from its polarity. The oxygen atom is much more electronegative than hydrogen, so the shared electrons in each O–H bond spend more time near the oxygen. Combined with water's bent geometry, this gives the molecule a permanent dipole: a partial negative charge ($\delta^-$) on the oxygen and partial positive charges ($\delta^+$) on each hydrogen. The resulting partial charges allow water to form hydrogen bonds with other polar molecules and ions, surrounding them and pulling them into solution. Non-polar molecules (e.g. the long hydrocarbon chains of triglycerides) lack partial charges; placing them in water would disrupt the water–water hydrogen-bond network without any compensating water–solute interaction — energetically unfavourable. This is the basis of the hydrophobic effect.

Understanding the Question

The question sets up a contrast: glycerol (a small alcohol with three –OH groups) dissolves in water; triglycerides (esters of glycerol with three long fatty acids) do not. The candidate must explain this contrast at the molecular level.

Approach

1. State water's polarity.
2. Explain why polarity lets water dissolve polar molecules (glycerol).
3. Explain why the same polarity leaves non-polar molecules (triglycerides) insoluble.

Step-by-Step Reasoning

1. Water is polar / dipolar — partial charges on O ($\delta^-$) and H ($\delta^+$) mean water can attract other polar or charged species.
2. Glycerol is polar — it has three hydroxyl (–OH) groups that can form hydrogen bonds with water. Water molecules cluster around glycerol molecules via hydrogen bonding, surrounding and dispersing them — i.e. dissolving them.
3. Triglycerides are non-polar — the long hydrocarbon chains have only C–C and C–H bonds, which are essentially non-polar. There are no partial charges for water to interact with, and inserting triglyceride molecules into water would force water to reorient around them, breaking water–water hydrogen bonds without forming water–triglyceride hydrogen bonds. The triglycerides therefore remain as a separate phase — insoluble.

Key Takeaways

• Water dissolves polar / ionic substances because it can form hydrogen bonds or ion–dipole interactions with them.
• Water does not dissolve non-polar substances — this is the hydrophobic effect.
• The polarity of the solute determines whether water can interact with it.

Common Mistakes

• Saying "water dissolves everything" or "oil and water don't mix because oil is thick" — neither engages with polarity.
• Referring to "the fatty acid tails" without explaining why they are non-polar.
• Confusing "insoluble" with "denser than water" — triglycerides do float, but the underlying reason is non-polarity, not density.
• Citing only one of the two contrast points (e.g. only why glycerol dissolves, not why triglycerides don't).

Things to Be Careful About

• Use the precise terms polar / dipolar and hydrogen bond — these are the mark-bearing words.
• The "ora" (or reverse argument) is acceptable: "water cannot hydrogen-bond with non-polar substances" also scores.

Background Concept

Phloem is the plant vascular tissue responsible for translocation — the transport of organic solutes (mainly sucrose, but also amino acids, fatty acids and other assimilates) from sources (e.g. photosynthesising leaves, storage tissues) to sinks (e.g. roots, fruits, growing tips). Unlike xylem, phloem is a living tissue at functional maturity. It is composed mainly of two cell types:

• Sieve tube elements — the conducting cells. They are joined end-to-end to form long sieve tubes, with perforated end walls called sieve plates between adjacent elements. At maturity a sieve tube element has lost its nucleus, ribosomes, vacuole and most other organelles — leaving a clear cytoplasmic pathway for the flow of sap.
• Companion cells — small cells closely associated with each sieve tube element (derived from the same parent cell by division). They retain their nucleus and normal organelles and provide metabolic support to the enucleate sieve tube element.

(The tissue also contains phloem fibres for support and phloem parenchyma for storage.)

Understanding the Question

The question asks for the cell type that forms the transport vessels of phloem tissue. "Transport vessels" points directly to the sieve tubes; the individual cells forming them are sieve tube elements.

Approach

Recall the cellular composition of phloem and identify the cell type whose job is to form the conducting tube.

Step-by-Step Reasoning

• The conducting tubes of phloem are sieve tubes.
• Each sieve tube is made up of many sieve tube elements joined end-to-end.
• "Sieve element" is an accepted alternative term.

Key Takeaways

• Phloem transport is performed by sieve tube elements (not companion cells, which support them).
• Sieve tube elements are unusual cells: alive at functional maturity but lacking a nucleus and most organelles, depending on their companion cell for metabolism.

Common Mistakes

• Writing "xylem vessel" — wrong tissue (xylem transports water and mineral ions).
• Writing "phloem cell" — too vague; the specific term "sieve tube element" is required.
• Writing "companion cell" — companion cells support the sieve tube elements; they do not form the conducting tube themselves.

Things to Be Careful About

• "Sieve tube element" (the individual cell) and "sieve tube" (the conducting tube made of many elements) are not the same thing. The question asks for the cell type — sieve tube element.

Background Concept

HIV is transmitted through the exchange of certain body fluids — blood, semen, vaginal secretions, rectal secretions and breast milk. The main transmission routes are unprotected sexual contact, blood-to-blood contact (contaminated needles in injecting-drug use, unscreened transfusions, unsafe medical or cosmetic procedures) and mother-to-child transmission (during pregnancy, birth or breastfeeding). Effective prevention therefore targets one or more of these routes with a specific intervention.

Understanding the Question

The stem tells you ART can both treat infected individuals and reduce onward transmission, but the question asks for two OTHER control methods. 'Control' here means population-level measures that reduce transmission in a community; ART alone is excluded.

Approach

For each transmission route, recall the specific public-health or behavioural intervention that interrupts it, then choose any two that together cover distinct routes.

Step-by-Step Reasoning

• Sexual transmission → practising safe sex with barrier contraceptives (condoms); reducing the number of partners.
• Blood-borne (transfusion) → screening donated blood, treating it to inactivate any virus (heat / chemical / solvent-detergent), refusing donations from high-risk donors.
• Blood-borne (drugs) → not sharing needles / syringes; needle-exchange and safe-injection schemes; using only new / sterile equipment.
• Mother-to-child → mothers with HIV avoiding breastfeeding (using formula milk); antiretroviral prophylaxis during pregnancy (this overlaps with ART but is sometimes credited under specific contexts).
• Iatrogenic (surgery, dental, tattooing, piercing) → using sterile equipment.
• Pre- / post-exposure prophylaxis (PrEP / PEP) for individuals at high risk of exposure.

The mark scheme rejects vague answers such as 'use contraceptives' without specifying barrier, and explicitly does NOT credit 'education', 'contact tracing', 'protective gear', 'early diagnosis' or 'testing' as standalone answers for this item.

Key Takeaways

Control of a sexually and blood-borne pathogen = identify each transmission route and apply the intervention that breaks it.

Common Mistakes

• Writing 'use contraceptives' without specifying barrier / condom — too vague to earn the mark.
• Writing 'screen blood' without saying 'donated' or 'before transfusion' — incomplete.
• Offering 'education' or 'contact tracing' alone — these are explicitly ignored by the mark scheme here.
• Confusing PrEP / PEP with ART — ART is excluded by the question; PrEP / PEP for uninfected individuals is acceptable.

Things to Be Careful About

Mention the specific element that earns the mark: barrier, screening OF DONATED blood, needle-exchange, PrEP / PEP, sterile equipment, etc. Vague answers are penalised.

Background Concept

Magnification describes how much larger (or smaller) an image is compared with the actual object. It is a ratio — a pure number with no units. The formula is:

A magnification greater than 1 means the image is enlarged; less than 1 means it is reduced.

Understanding the Question

This is a recall-of-definition item. You must give the verbal definition that matches the formula, not the formula itself (although the formula is accepted as a supplement).

Approach

State the definition in words, anchoring it to a comparison between image size and actual size.

Step-by-Step Reasoning

A correct one-line answer is 'the number of times larger the image is than the actual (object/specimen)'. The mark scheme accepts a formula as an alternative or addition, but a verbal definition is more reliable because it is unambiguous and free of mathematical ambiguity.

Key Takeaways

Magnification is a dimensionless ratio — it depends on BOTH the image size and the specimen size, not on the microscope alone.

Common Mistakes

• Saying simply 'makes it bigger' — too vague; doesn't anchor to a comparison with the actual object.
• Writing only a formula without any words.
• Confusing magnification with resolution (the ability to distinguish two close points as separate).

Things to Be Careful About

Anchor the definition to 'image' vs 'actual (specimen / object)'. Acceptable phrasings include 'the number of times larger the image is than the actual' or 'image size ÷ actual size'.

Background Concept

Enzyme inhibitors fall into two broad classes. Competitive inhibitors bind to the active site and block substrate binding reversibly; their effect can be overcome by raising substrate concentration. Non-competitive inhibitors bind to a different site (an allosteric site) and change the shape of the enzyme so the active site can no longer accommodate the substrate; raising substrate concentration does NOT overcome the inhibition. The drug caspofungin is the latter type.

Fungal cell walls are built from β-glucans (especially 1,3-β-D-glucan) and chitin. If you stop β-glucan synthesis, the wall cannot form properly. A cell without an intact wall is vulnerable to osmotic lysis — water rushes in down the water-potential gradient, the membrane cannot withstand the pressure, and the cell bursts.

Understanding the Question

You are told caspofungin is a non-competitive inhibitor of 1,3-β-D-glucan synthase, the enzyme that makes a key wall component. You must explain, with reference to the mechanism, how this drug could treat P. jirovecii pneumonia.

Approach

Walk through the standard non-competitive-inhibition chain, then chain it on to fungal cell-wall synthesis, osmotic lysis, and the reduction in pathogen load. That's exactly five linked points — one chain per mark.

Step-by-Step Reasoning

1. Caspofungin binds to a site on 1,3-β-D-glucan synthase OTHER than the active site (the allosteric site).
2. This changes the shape of the active site (the tertiary structure of the enzyme changes).
3. Substrate can no longer bind to the active site, so enzyme–substrate complexes do not form (or form in much smaller numbers).
4. The rate of 1,3-β-D-glucan synthesis falls; the fungal cell wall is weakened / not formed properly.
5. The cell cannot resist osmotic influx of water (water enters down the water-potential gradient) and undergoes osmotic lysis.
6. The P. jirovecii population no longer grows / falls; fewer cells are available to colonise the alveoli, easing the infection and giving the body's immune system a better chance of clearing the remaining fungus.

Key Takeaways

Non-competitive inhibitors work by altering enzyme shape at a remote site; the chain drug → enzyme inhibition → no wall → lysis → fewer pathogens illustrates how a single molecular event can have a population-level therapeutic effect. This is the same logic that underpins many antimicrobial drugs that target cell-wall synthesis (β-lactam antibiotics, azole antifungals, etc.).

Common Mistakes

• Saying caspofungin 'binds to the active site' — that would make it competitive, not non-competitive.
• Saying only 'substrate cannot bind' without explaining WHY (shape change at the allosteric site).
• Stating 'the fungus dies' without naming the mechanism (osmotic lysis) or the reason (cell-wall weakening).
• Stopping at 'fewer cells' without linking to the immune system or to the symptoms easing.

Things to Be Careful About

Use the term 'allosteric site' (or 'site other than the active site'). State explicitly that the active site changes SHAPE so that substrate can no longer bind. Mention osmotic lysis for the final link, and ideally that water enters DOWN the water-potential gradient.

Background Concept

A nucleotide is the monomer of a nucleic acid. It has three parts: a phosphate group, a pentose sugar (deoxyribose in DNA, ribose in RNA), and a nitrogenous base. Bases are of two structural kinds: purines (double ring — adenine, guanine) and pyrimidines (single ring — cytosine, thymine, uracil).

Nucleotides are joined into a polynucleotide by phosphodiester bonds. A phosphodiester bond forms between the 3′ carbon of the sugar of one nucleotide and the 5′ phosphate of the next nucleotide on the same strand. This is the bond that builds the sugar–phosphate backbone of DNA or RNA. The backbone runs 5′ → 3′.

Across the two strands, bases pair by hydrogen bonds (two between A and T/U, three between G and C). Hydrogen bonds are weak and reversible (they must be broken to open the strands for replication or transcription); phosphodiester bonds are strong covalent bonds and are not broken during these processes.

Understanding the Question

The question gives a molecular diagram of a single base pair between a DNA nucleotide and an RNA nucleotide, with three hydrogen bonds shown as dashed lines. Part (a) asks you to reason about a bond that is absent from the figure: the phosphodiester bond. The instruction is to explain, so you must give a reason — not just say "there is no phosphodiester bond".

Approach

Think about what a phosphodiester bond needs in order to exist: two adjacent nucleotides, on the same strand, with a phosphate linking the 3′ of one sugar to the 5′ of the next. Then look at the figure and count: how many nucleotides are on each strand? Is the phosphate attached to one sugar or to two? Use that count to give a clear reason.

Step-by-Step Reasoning

1. A phosphodiester bond is formed during polynucleotide synthesis, between the phosphate of one nucleotide and the 3′ carbon of the pentose sugar of the adjacent nucleotide on the same strand.
2. In Fig. 3.1 only one nucleotide is shown on each strand (one DNA nucleotide, one RNA nucleotide). There is no second nucleotide alongside either of them.
3. The phosphate group (the circle on the left of the DNA nucleotide) is therefore bonded to only one pentose sugar. A phosphodiester bond requires each phosphate to be bonded to two sugars.
4. The same logic applies to the right-hand nucleotide. Hence, the structural pre-conditions for a phosphodiester bond are not met in this figure.

Either of these ideas, clearly expressed, earns the mark. The strongest answers combine the two.

Key Takeaways

• Phosphodiester bonds join adjacent nucleotides on the same strand; hydrogen bonds join bases across the two strands.
• A phosphodiester bond cannot exist unless a phosphate is shared between two sugars, i.e. between two adjacent nucleotides.
• A figure showing a single base pair therefore has hydrogen bonds but no phosphodiester bonds.

Common Mistakes

• Saying "there is no phosphodiester bond because it is a single base pair" without explaining what is required for a phosphodiester bond to form.
• Confusing phosphodiester bonds with hydrogen bonds (hydrogen bonds are present in the figure; phosphodiester bonds are not).
• Saying "only one nucleotide is shown" but failing to add that a phosphodiester bond needs an adjacent nucleotide on the same strand.

Things to Be Careful About

• The mark scheme rewards the idea of two adjacent nucleotides on the same strand. Wording such as "the phosphodiester bond joins nucleotides in a chain" or "joins the sugar of one nucleotide to the phosphate of the next" is acceptable.
• An alternative acceptable point is that the phosphate shown is attached to only one sugar, when a phosphodiester linkage requires attachment to two.
• Use the term nucleotide, not base, when referring to the building blocks of a polynucleotide.

Background Concept

Every nucleotide has three components: a phosphate, a pentose sugar (deoxyribose in DNA, ribose in RNA), and a nitrogenous base. The five bases fall into two structural groups:

• Purines — double ring of carbon and nitrogen atoms: adenine (A) and guanine (G).
• Pyrimidines — single ring: cytosine (C), thymine (T) (DNA only) and uracil (U) (RNA only).

In a double-stranded nucleic acid, a purine on one strand always pairs with a pyrimidine on the other, so the helix has a constant width. The base-pairing rules are:

• A pairs with T (DNA) or U (RNA) via two hydrogen bonds.
• G pairs with C via three hydrogen bonds.

During transcription, the DNA double helix is unwound and one DNA strand acts as the template for synthesising a complementary mRNA strand. The pairing is therefore DNA–RNA: A (DNA) with U (RNA), T (DNA) with A (RNA), G (DNA) with C (RNA), and C (DNA) with G (RNA). Hydrogen bonds are shown in molecular diagrams as dashed lines.

Understanding the Question

Part (b) asks you to identify the specific DNA–RNA pair in Fig. 3.1 and to describe the features that allow you to recognise it. The figure shows a single base pair in molecular detail, with the RNA sugar already labelled. You are expected to use the visual evidence (number of rings, number of hydrogen bonds, the labelled sugar, the presence of a phosphate) together with the base-pairing rules to name the bases correctly and to label the diagram.

Approach

Use the visual clues systematically:

1. Count the rings on each base — purine (double) vs pyrimidine (single).
2. Count the hydrogen bonds (dashed lines) — 2 or 3 — to fix which base pair this is.
3. Use the labelled sugar on the right (ribose) to assign that nucleotide to RNA, and therefore the other to DNA.
4. Use the chemical groups drawn on the rings (e.g. C=O vs NH₂ positions) to distinguish guanine from adenine, and cytosine from uracil/thymine, if needed.
5. Identify the other labelled/visibly distinct features: the circle is a phosphate, the left pentagon is deoxyribose, the dashed lines are hydrogen bonds.

Step-by-Step Reasoning

Identifying the bases

• The right-hand base is on a ribose sugar (labelled), so this is the RNA nucleotide.
• The right-hand base has a single ring, so it is a pyrimidine — either C, T or U. Uracil is RNA-specific; cytosine is found in both DNA and RNA; thymine is DNA-specific and so cannot appear in an RNA nucleotide.
• The left-hand base has a double ring, so it is a purine — either A or G.
• There are three dashed lines between the two bases. Three hydrogen bonds = G–C. (Two hydrogen bonds = A–T or A–U.)
• Therefore the left base is guanine (the DNA base) and the right base is cytosine (the RNA base). The cross-strand base pair is G(DNA) · C(RNA).

Confirming from the chemical groups drawn

• The left base has a C=O at the top of the ring and an NH₂ on the side, which is the signature of guanine. (Adenine would have an NH₂ at the top and no C=O on the six-membered ring.)
• The right base has an NH₂ group and a C=O on its single ring, which matches cytosine. (Uracil has two C=O groups and no NH₂; thymine has two C=O groups and a methyl group.)

Describing the other features

• The three dashed lines are hydrogen bonds — weak, non-covalent interactions that hold the two strands together and can be broken to separate the strands for transcription.
• The circle bonded to the deoxyribose on the left is the phosphate group, the third component of a nucleotide. (The RNA nucleotide on the right also has a phosphate — it is the small circle drawn below the ribose pentagon.)
• The pentagon on the left is the pentose sugar; because the figure is a DNA nucleotide this is deoxyribose. (Ribose, on the right, has one more oxygen than deoxyribose — the 2′ OH — but at A-level this distinction is usually shown only schematically, as in Fig. 3.1, where the right sugar is explicitly labelled.)
• The left base is a purine (double ring) and the right base is a pyrimidine (single ring); this 1-purine-with-1-pyrimidine pairing is what keeps the double helix a constant width.

Optional annotation (AVP)
Other annotations that would earn credit include: labelling the solid lines as covalent bonds, drawing a box around one complete nucleotide, or indicating the 5′ and 3′ ends to show that the strands run antiparallel.

Key Takeaways

• Three hydrogen bonds = G–C pair; two hydrogen bonds = A–T or A–U pair.
• A purine (double ring) always pairs with a pyrimidine (single ring) — visible directly from the figure.
• The sugar identifies the nucleic acid: deoxyribose → DNA, ribose → RNA. (The question tells you this is a DNA–RNA pair, so the ribose label fixes which nucleotide is which.)
• A complete nucleotide has three parts: phosphate + pentose + base. The diagram shows all three on the left; on the right the phosphate (small circle below the ribose) and the ribose (pentagon, labelled) are both present.
• During transcription the DNA strand acts as the template; the RNA strand is built complementary and antiparallel to it.

Common Mistakes

• Writing "cytosine and guanine" without making clear which is on the DNA strand and which is on the RNA strand. The mark scheme ignores reversed orderings unless further qualified.
• Calling the bases "DNA" and "RNA" (a base is not a nucleic acid) instead of naming the specific base.
• Saying "there are hydrogen bonds" without specifying that there are three of them — the count is what identifies the G–C pair.
• Confusing the sugar and the base — calling the pentagon a "base" or the ring a "sugar".
• Saying "thymine pairs with adenine" — this is a DNA–DNA rule; the question is about a DNA–RNA pair during transcription, so the RNA pyrimidine here is cytosine, not thymine or uracil.
• Saying "the strands are parallel". The two strands of a nucleic acid (and of the DNA–RNA hybrid formed during transcription) are antiparallel.

Things to Be Careful About

• The mark scheme accepts "pentose" for deoxyribose if the DNA context is clear, but "deoxyribose" is the precise term and is safer.
• "Purine" / "pyrimidine" earns the mark even if the base name is not given, but the base identification mark is separate — both are needed to be safe.
• Hydrogen bonds between the bases are weak and allow the strands to be separated during transcription; phosphodiester bonds in the backbone are covalent and are not broken.
• Where the figure shows a small circle attached to a sugar, that is the phosphate group — not a hydroxyl or some other functional group.

Background Concept

Mitosis is a type of nuclear division that produces two genetically identical daughter nuclei, and is followed by cytokinesis to form two daughter cells. In a multicellular animal, mitosis is used for three main purposes: growth (increasing cell number during development), asexual reproduction (in some species), and repair (replacing cells lost through damage or wear).

Repair is a continuous process in many tissues because cells have a finite lifespan. Skin cells, intestinal epithelial cells, and red blood cell precursors, for example, are constantly being replaced. Adult stem cells supply the new cells by dividing mitotically; the daughter cells then either remain as stem cells (self-renewal) or go on to differentiate into the functional cells of that tissue.

Understanding the Question

Part (a) gives a short cue ('Mitosis is important for the repair of tissues') and asks the candidate to explain what 'repair of tissues' actually means. The command word is 'explain', but only one mark is awarded, so a single concise point is sufficient. The stem context (adult stem cells, identical daughter cells, differentiation) tells us that 'repair' is being used in the cellular sense.

Approach

The mark scheme rewards the idea that repair involves replacing cells that have been lost through damage, destruction, wear, or age. Pair this with the verb 'replacing' to make the link to mitosis explicit.

Step-by-Step Reasoning

1. The candidate should identify the key action: replacing cells.
2. The reason replacement is needed: because the original cells have become damaged, destroyed, worn out or old.
3. The new cells come from mitosis (the parent stem makes this link).

Key Takeaways

'Repair' in a cellular context means restoring cell numbers by mitosis after loss. This is distinct from 'regeneration' of whole structures.

Common Mistakes

• Saying 'to heal wounds' — too vague and does not mention cells.
• Saying 'to make new cells' — partially right but misses the WHY (replacing damaged/old ones).
• Confusing repair with growth.

Things to Be Careful About

Use precise phrasing: 'replacing cells that are damaged / destroyed / worn out / old' matches the mark scheme wording. Any one of those four descriptors is sufficient on its own.

Background Concept

A tumour is a mass of cells that has escaped from the normal controls on cell division. Normal differentiated cells only divide when they receive the correct growth signals and stop when they contact neighbouring cells (contact inhibition), when DNA damage is detected (G1, G2 and M checkpoints), or when they are programmed to die (apoptosis). A cancer cell has typically accumulated mutations that override several of these controls.

Two key classes of gene are commonly mutated: proto-oncogenes (which when mutated become oncogenes and drive uncontrolled division) and tumour suppressor genes (which normally restrain division; when both copies are inactivated the brake is lost). Telomerase is also re-activated in many cancer cells, allowing them to keep dividing beyond the Hayflick limit.

Understanding the Question

The stem already tells us one feature — 'uncontrolled cell division' — and asks for OTHER features. The command word is 'describe', so concise factual statements are enough; two marks are available, so two clear points should be made.

Approach

Read the question carefully: do not repeat 'uncontrolled division' as one of the answers. Pick two distinct, well-known features from the mark scheme list. Aim for breadth: one cause and one consequence, or one feature of division and one of behaviour.

Step-by-Step Reasoning

Possible strong combinations:

• Cause: it is the result of a mutation (e.g. proto-oncogene → oncogene).
• Division behaviour: short / fast / continuous cell cycles, with no contact inhibition.
• Consequence: loss of the original differentiated function.
• Survival: no apoptosis.
• Telomeres: telomerase reactivated so telomeres do not shorten as fast.
• Spread: angiogenesis, metastasis.

The marking scheme accepts many alternatives, so any two well-articulated features earn full marks.

Key Takeaways

Tumours are defined by multiple features, not just rapid division: mutation origin, loss of checkpoints, loss of apoptosis, ability to form new blood vessels, ability to invade and spread.

Common Mistakes

• Restating 'uncontrolled cell division' as one of the two answers (this is already given).
• Saying simply 'mutation' without context — better to link to oncogenes or tumour suppressors.
• 'Cells grow larger' — this is hypertrophy, not tumour formation.
• Vague references to 'cancer spreading' without naming metastasis or blood/lymph.

Things to Be Careful About

The mark scheme explicitly IGNORES 'mutation' as a stand-alone term and REQUIRES either the gene context (proto-oncogene / tumour suppressor) or 'result of a mutation'. Avoid the bare word 'mutation'.

Background Concept

Telomeres are repetitive, non-coding nucleotide sequences (TTAGGG in humans) at the ends of linear chromosomes. They protect chromosome ends from deterioration and from being recognised as DNA double-strand breaks (which would otherwise be joined together by repair enzymes).

Because DNA polymerase can only add nucleotides to an existing 3′-OH group, the very end of the lagging strand cannot be fully replicated — this is the 'end-replication problem'. Each round of DNA replication therefore removes a small stretch of nucleotides from the 5′ end, and telomeres get a little shorter every cell cycle. Eventually telomeres become critically short, the chromosome ends become unstable, and the cell enters senescence or dies. The enzyme telomerase adds sequence back onto telomeres and is active in stem cells and germ cells but not in most somatic cells.

Understanding the Question

The stem establishes that telomeres 'prevent loss of genes' and that adult stem cells have long telomeres. The candidate must explain why a long telomere is an ADVANTAGE to a cell that undergoes many cell cycles.

Approach

Connect two ideas: (1) telomeres shorten every division; (2) what would happen if they ran out (genes lost / chromosome ends damaged). Long telomeres delay or prevent this.

Step-by-Step Reasoning

1. Each cell cycle shortens telomeres because of the end-replication problem on the lagging strand.
2. Long starting telomeres therefore allow many more rounds of replication before telomere length becomes critically short.
3. Telomeres are non-coding / do not contain genes, so when they shorten no genetic information is lost — they act as a disposable buffer.
4. While telomeres are long enough, they continue to protect chromosome ends from being damaged or fused.
5. The net effect: the cell can keep dividing (long life span, many mitoses) while its coding DNA stays intact.

Key Takeaways

Telomeres are disposable, protective caps that shorten every division. Long telomeres in stem cells are the molecular reason these cells can divide many times without losing genes.

Common Mistakes

• Saying 'long telomeres prevent ageing' — too general and not a marking point.
• Confusing telomeres with centromeres.
• Saying long telomeres 'replicate faster' — telomeres are the part lost during replication, not the part replicated faster.
• Saying 'genes' are lost when telomeres shorten — telomeres are non-coding, so this is the opposite of the truth.

Things to Be Careful About

The mark scheme is generous (any two of many options) but rewards PRECISION: name the end-replication problem, name non-coding status, or link to protection of chromosome ends. Avoid vague 'they stop DNA being lost' — say WHAT is lost and WHY (because of how DNA polymerase works on the lagging strand).

Background Concept

A stem cell is defined by two cardinal properties:

1. Self-renewal — it can divide to produce at least one daughter cell that remains an identical, undifferentiated stem cell.
2. Potency — it can give rise to multiple differentiated cell types.

A haematopoietic stem cell (HSC) is a multipotent adult stem cell in the bone marrow that can produce every type of blood cell shown in Fig. 4.1: red blood cells, platelets, neutrophils, monocytes, B-lymphocytes and T-lymphocytes.

A progenitor cell (also called a precursor or transit-amplifying cell) sits one step downstream. It has lost some potency (it can only give rise to a limited range of cell types) and it cannot self-renew — when it divides, both daughters differentiate further. Progenitor cells are shaded grey in Fig. 4.1.

Understanding the Question

The stem explains that the first stage of HSC differentiation is the production of progenitor cells, and that these CAN divide by mitosis but are NOT stem cells. Part (d) asks the candidate to justify why GMP cells — a specific type of progenitor cell shown in Fig. 4.1 — fail the definition of a stem cell.

Approach

Use Fig. 4.1 to compare GMP cells with HSCs. Check the three stem-cell criteria in turn:

• Are they undifferentiated? No — they are already along the differentiation pathway (shaded progenitor box).
• Can they self-renew? No — Fig. 4.1 shows GMP cells dividing only into more differentiated daughters; there is no loop back to HSCs.
• Are they multipotent? No — they give rise only to neutrophils and monocytes.

Step-by-Step Reasoning

1. GMP cells lie on the CMP → GMP branch in Fig. 4.1; they are shaded as progenitor cells, meaning differentiation has already started.
2. The only self-renewal arrow in the whole diagram is the loop on HSCs. GMP cells have no such loop; each division simply produces more differentiated daughters, so there is no self-renewal.
3. Tracing GMP cells in the figure shows they lead only to immature neutrophils and immature monocytes (and then macrophages, platelets etc. are NOT on this branch). So GMP cells are committed to the granulocyte/monocyte lineage, not multipotent.
4. A stem cell, by definition, must have all three properties — potency AND self-renewal AND undifferentiated status. GMP cells satisfy none of these fully.

Key Takeaways

Three criteria define a stem cell: undifferentiated, self-renewing, multipotent. Loss of ANY ONE of them means the cell is a progenitor, not a stem cell.

Common Mistakes

• Saying 'they are not found in the bone marrow' — irrelevant; GMP cells ARE in the bone marrow.
• Saying 'they cannot divide' — wrong; the stem tells us progenitor cells CAN divide by mitosis.
• Repeating just one criterion (e.g. 'no self-renewal') without linking it to the figure.
• Confusing GMP with CLP — the figure shows GMPs give neutrophils and monocytes; CLPs give lymphocytes.

Things to Be Careful About

Refer EXPLICITLY to Fig. 4.1 — the mark scheme's 'AW' allows alternative wording but the answer must be grounded in the figure: trace which boxes GMP connects to and notice that the only self-renewal loop is on HSCs.

Background Concept

Monocytes circulate in the blood. When they migrate into tissues they mature and enlarge into macrophages. Macrophages are large phagocytic cells that engulf and digest pathogens, dead cells and cell debris by phagocytosis — the same broad function as neutrophils.

The monocyte → macrophage transition is the classical way in which the body converts short-lived blood phagocytes into long-lived tissue-resident phagocytes. Macrophages are also the cells that present pathogen antigens on their surface to T-helper cells, helping to activate the specific immune response.

Understanding the Question

The stem of part (e) tells us X is the cell a monocyte differentiates into, and that it has a similar function to neutrophils (i.e. it is phagocytic). The mark to be earned is naming X.

Approach

Recall the standard monocyte maturation pathway: blood monocyte → tissue macrophage. The 'similar function to neutrophils' cue (phagocytosis) confirms this.

Step-by-Step Reasoning

1. Fig. 4.1 shows: immature monocyte → monocyte → X.
2. Mature blood monocytes leave the bloodstream and mature into macrophages in the tissues.
3. Macrophages, like neutrophils, are phagocytes — they engulf and digest pathogens and cellular debris.
4. Therefore X = macrophage.

Key Takeaways

Monocytes are blood-borne phagocyte precursors; macrophages are their tissue-resident mature form.

Common Mistakes

• 'Mature monocytes' — the question already names monocytes; X must be the NEXT differentiated stage.
• 'Phagocyte' — too generic; the mark scheme explicitly rejects this.
• 'Granulocyte' — wrong lineage (granulocytes are neutrophils, eosinophils and basophils).

Things to Be Careful About

Do not write 'mature monocyte' — that is the box immediately above X in Fig. 4.1. The answer must be the differentiated cell type downstream of the monocyte.

Background Concept

B-lymphocytes mature in the bone marrow. On first exposure to a specific antigen (with help from T-helper cells and antigen-presenting cells), a B-lymphocyte is activated, proliferates and differentiates into:

• plasma cells, which secrete antibodies (also called immunoglobulins);
• memory B-cells, which remain dormant until re-exposure.

Antibodies are Y-shaped glycoproteins with two antigen-binding sites at the tips of the Y. They neutralise pathogens by binding to antigens, marking them for destruction (opsonisation), agglutinating them, or activating complement.

Understanding the Question

Part (f) tells us cell type Y releases molecules with antigen-binding sites. Fig. 4.1 shows Y at the bottom of the mature B-lymphocyte branch.

Approach

Cells that secrete molecules with antigen-binding sites are plasma cells, and the molecules they secrete are antibodies.

Step-by-Step Reasoning

1. Fig. 4.1 shows the B-lymphocyte branch ending at cell type Y, with 'immune response' labelled next to the mature B-lymphocyte box.
2. The molecules released by (differentiated) B-lineage cells that have antigen-binding sites are antibodies / immunoglobulins.
3. Therefore Y releases antibodies.

Key Takeaways

Plasma cells (differentiated B-lymphocytes) secrete antibodies; these antibodies bind specifically to antigens via their antigen-binding sites.

Common Mistakes

• 'Antigens' — antigens are what antibodies bind TO, not what B-cells release.
• 'Memory cells' — these are cells, not molecules.
• 'Cytokines' — released by T-helper cells, not B-cells.

Things to Be Careful About

'Antibody' and 'immunoglobulin' are both accepted and are interchangeable terms.

Background Concept

T-lymphocyte precursors migrate from the bone marrow to the thymus, where they mature. During maturation, each developing T-cell randomly generates a T-cell receptor (TCR) with a specific binding site. Because this generation is random, some TCRs will, by chance, bind to 'self' antigens — molecules normally displayed on the surface of the body's own cells (presented on MHC molecules).

The thymus tests each T-cell for self-reactivity. T-cells that bind self antigens too strongly are destroyed by apoptosis — this is positive and negative selection, and it establishes self-tolerance. T-cells that bind self antigens weakly (or not at all) are released into the circulation; only these become mature T-helper and T-killer cells that respond to foreign antigens.

Understanding the Question

Part (g) gives two pieces of context:

• T-lymphocyte differentiation starts in the bone marrow and is completed in the thymus.
• T-lymphocytes that bind self antigens are destroyed IN the thymus.
  The candidate must explain WHY this destruction is necessary.

Approach

Reason about what would happen if self-reactive T-cells were NOT destroyed:

1. They would leave the thymus and enter the general circulation.
2. There they would meet cells of the body, which display self antigens on MHC molecules.
3. They would bind these self antigens and trigger an immune response.
4. The consequences are: damage/destruction of the body's own cells; creation of memory T-cells against self antigens (so the attack persists on re-exposure) — i.e. autoimmune disease.

Step-by-Step Reasoning

1. The thymus is the last checkpoint before T-cells enter the body. Once they leave, they are exposed to all body tissues.
2. Self antigens are normal surface markers on the body's own cells. A T-lymphocyte whose receptor binds a self antigen would, if released, attack that body cell.
3. T-killer cells would directly destroy the cell; T-helper cells would release cytokines, amplifying the response and recruiting phagocytes and B-cells against the body's own tissues.
4. Some of these self-reactive lymphocytes would become memory cells, so the attack would be repeated and amplified on every subsequent exposure to those self antigens — this is autoimmunity.
5. Destroying self-reactive T-cells in the thymus prevents all of the above, ensuring the immune system only attacks foreign material.

Key Takeaways

Self-tolerance is essential. Negative selection in the thymus removes T-cells that would attack the body's own tissues; without it, autoimmune disease results. This is one of the central mechanisms preventing self-destruction by the immune system.

Common Mistakes

• Saying 'so the T-cells don't attack themselves' — T-cells don't attack themselves; they attack OTHER cells displaying self antigens.
• Saying 'so the T-cells don't get attacked' — confusion of direction.
• Forgetting to mention memory cells and autoimmunity — these give the long-term consequence and are explicit mark-scheme points.
• Saying 'because self antigens are bad' — self antigens are not bad; the problem is that an immune response AGAINST them is bad.

Things to Be Careful About

The mark scheme specifically notes that points 2 and 3 require the candidate to make clear that self antigens are on the body's OWN cells. A bare statement like 'they would cause an immune response' does not earn these marks; add 'against body cells'.

Background Concept

Malaria in humans is caused by protoctist parasites belonging to the genus Plasmodium. These are single-celled eukaryotes (protoctists) transmitted to humans through the bite of an infected female Anopheles mosquito. Five species of Plasmodium are known to cause malaria in humans, and they differ in their geographical distribution, severity and the clinical picture they produce.

The five species that infect humans are:

• Plasmodium falciparum — responsible for the majority of deaths worldwide, especially in sub-Saharan Africa; causes the most severe form (cerebral malaria).
• Plasmodium vivax — the most widespread geographically; causes relapsing malaria because it can lie dormant in the liver as hypnozoites.
• Plasmodium ovale — similar to P. vivax in causing relapses; found mainly in West Africa and the western Pacific.
• Plasmodium malariae — causes a milder, chronic form with a characteristic 72-hour fever cycle; widespread but less common.
• Plasmodium knowlesi — originally a parasite of macaques; now recognised as a significant cause of zoonotic malaria in South-East Asia.

Understanding the Question

This is a simple recall question. The stem has already told you that P. vivax and P. falciparum are the two species responsible for most malaria deaths. You simply have to name one of the remaining three species. The answer should be the binomial name written in the same italicised form, e.g. Plasmodium ovale.

Approach

Read the question carefully — it says "name one other species". Any one of the three remaining species will earn the mark. Spell the species name correctly; CIE mark schemes reject misspelt binomials.

Step-by-Step Reasoning

• Plasmodium ovale — the most common alternative answer, especially for CIE questions which list ovale and malariae as the standard "other two".
• Plasmodium malariae — also credited; the same spelling rules apply.
• Plasmodium knowlesi — credited, but the mark scheme warns that it "must be spelled correctly".

If you give two answers, both must be correct and correctly spelled, or the mark is lost.

Key Takeaways

• Five species of Plasmodium cause malaria in humans: falciparum, vivax, ovale, malariae, and (more recently recognised) knowlesi.
• P. falciparum and P. vivax are responsible for the great majority of malaria deaths and cases worldwide.
• Binomial names must be italicised with the genus capitalised and species in lower case.

Common Mistakes

• Writing just "ovale" or "malariae" without the genus Plasmodium — the binomial must be complete.
• Misspelling "knowlesi" — the mark scheme rejects incorrect spellings.
• Giving P. falciparum or P. vivax — these are explicitly excluded by the stem.

Things to Be Careful About

• Write the full binomial: Plasmodium in italics followed by the species epithet in lower case.
• If you write more than one answer, ensure each is correctly spelled — partial credit is not given.

Background Concept

The red blood cell (erythrocyte) is one of the most highly specialised cells in the body. Its only function is to transport oxygen (and to a lesser extent carbon dioxide) between the gas-exchange surface and the respiring tissues. Its structure is therefore pared down to the minimum needed for that single role.

A mature mammalian red blood cell:

• has a biconcave disc shape (thinner in the middle than at the edges);
• is about 6–8 µm in diameter;
• has no nucleus, no mitochondria, no ribosomes, no endoplasmic reticulum and no other organelles;
• consists essentially of a cell surface membrane enclosing a cytoplasm packed with haemoglobin;
• is flexible because its membrane is supported by a meshwork of proteins (spectrin) just beneath the bilayer.

In addition, the question stem excludes the most obvious feature ("a very large number of haemoglobin molecules") and asks you to discuss the other features.

Understanding the Question

The command word is "describe" — give an account of the size and structure and link each feature to its function. The marks reward paired structure–function statements. You are told to ignore the high haemoglobin content (otherwise it would be the obvious answer). You should also ignore the colour (red) and any reference to lifespan; only size and structure are required.

Approach

Work through the features systematically, pairing each with a functional consequence:

• size (6–8 µm);
• shape (biconcave);
• absence of nucleus;
• absence of mitochondria;
• flexibility.

For each, state the feature and then explicitly link it to the function. The mark scheme awards a mark for the feature and (often) a separate mark for the linked functional explanation, so both halves of the pair must be present.

Step-by-Step Reasoning

Size (6–8 µm diameter)

• Feature: the diameter is 6–8 µm, which is approximately the same as the lumen of a capillary. Mark point 1.
• Function: the cell just fits inside a capillary, so it travels in single file, one cell at a time. This slows the flow of blood past the capillary wall and gives maximum time for oxygen to diffuse in (and CO₂ to diffuse out). Mark point 2.
• The mark scheme does not credit "small size" on its own — you must give a number (or compare to a capillary) and then link to the single-file / slowed flow consequence.

Absence of nucleus

• Feature: the mature red blood cell has no nucleus. Mark point 3.
• Function: this leaves more space inside the cell for haemoglobin. Mark point 4. Note that mark point 4 is dependent on mark point 3; the function must be linked to the structural feature.

Absence of mitochondria

• Function: with no mitochondria, the cell cannot respire aerobically, so it does not consume any of the oxygen it carries — that oxygen is all available to be delivered to respiring tissues. Mark point 5.

Flexibility

• Feature: the cell is flexible and can deform, so it can squeeze through capillaries that are narrower than its own diameter. Mark point 6. The mark scheme ignores "squeeze between" unless it is clearly stated that the cell is inside a capillary.

Biconcave shape

• Feature: the cell is biconcave. Mark point 7.
• Function 1 (SA:V): compared with a sphere of the same volume, a biconcave disc has a larger surface area to volume ratio, so diffusion of oxygen across the cell surface membrane is faster. Mark point 8.
• Function 2 (diffusion distance): the thin centre of the disc means the diffusion distance from the cell surface membrane to the haemoglobin molecules in the middle of the cell is small, so oxygen reaches the haemoglobin quickly. Mark point 9.

Any four of these mark points earn the four marks. The cleanest single answer combines the four paired structure–function statements above.

Key Takeaways

• Red blood cells are exquisitely specialised for oxygen transport; their structure is shaped entirely by that function.
• The four most important structure–function links are: small diameter (capillary fit), biconcave shape (SA:V and short diffusion distance), absence of nucleus (more space for haemoglobin), and absence of mitochondria (oxygen is not used by the cell itself).
• A biconcave disc has a higher SA:V than a sphere of the same volume and a shorter maximum diffusion distance from the surface to the centre.
• "Describe" questions on cell structure reward paired structure and function statements, not the structure alone.

Common Mistakes

• Stating only the structural feature without the linked function (e.g. just "no nucleus"). You need the explanation that the absence of a nucleus gives more space for haemoglobin.
• Saying "small size" without quantifying it or linking it to single-file flow in a capillary. The mark scheme ignores "small size" on its own.
• Including the high haemoglobin content — explicitly excluded by the stem.
• Stating that the cell is "round" or "spherical" — it is biconcave; the shape is the whole point of the question.
• Confusing red blood cells with other cell types (white blood cells, prokaryotes) and listing irrelevant features.

Things to Be Careful About

• Each mark point typically has two halves: a structural feature and its functional consequence. The functional half is often the harder one to articulate and is required for the mark.
• Mark point 4 (more space for haemoglobin) is conditional on mark point 3 (no nucleus); if the structure is not stated, the function cannot be credited.
• A flexible cell that "squeezes between cells" is ignored unless the candidate makes it clear that the red blood cell is inside a capillary.
• Use the term "biconcave" (not just "disc-shaped" or "round").
• Note that the mark scheme describes the SA:V point as "compared to a sphere" — the comparison is with what a sphere of the same volume would give, not with another cell.

Background Concept

Water is taken up by a plant root from the soil and travels across the root to the xylem, from where it is pulled up to the leaves and lost by transpiration. The soil solution is relatively dilute and has a high (less negative) water potential, while the inside of the xylem and the air in the leaf have progressively lower water potentials. This continuous water potential gradient from soil → root → xylem → leaf → atmosphere is what drives the bulk movement of water into and through the plant.

Two parallel routes carry water across the cortex of the root:

• Apoplast pathway — the non-living network consisting of cell walls and the spaces between cells (intercellular spaces). Water (and dissolved mineral ions) can move freely through this continuous matrix of cellulose without ever crossing a cell surface membrane.
• Symplast pathway — the living route, in which water passes from cell to cell through the cytoplasm via plasmodesmata, having crossed a cell surface membrane to enter the first cell.

At the innermost layer of the cortex, the endodermis, the apoplast pathway is blocked by the Casparian strip — a ring of suberin (a waxy, hydrophobic, impermeable material) impregnating the radial and transverse walls of the endodermal cells. Because suberin is waterproof, water and dissolved ions cannot continue to flow freely through the cell walls; they must cross the cell surface membrane of an endodermal cell, enter its cytoplasm, and continue via the symplast. This forces water and solutes to pass through a living, selectively permeable barrier before reaching the xylem, which is essential for controlling which substances enter the transpiration stream.

Understanding the Question

This part has a compound command word: describe the apoplast pathway from soil to endodermis, and explain why it cannot continue at the endodermis. The stem tells us that transport of water from soil to xylem uses the apoplast and symplast pathways, so candidates are expected to know the difference between them and the special role of the endodermis. Marks are awarded for naming the cell-wall/intercellular-space route, naming the cells that the apoplast passes through, identifying the Casparian strip, describing its suberin composition, and linking overall movement to a water potential gradient.

Approach

• First describe the route: cell walls and intercellular spaces of epidermal / root hair cells and cortical cells, without crossing membranes.
• Then identify the barrier: the Casparian strip in the walls of endodermal cells, made of suberin (waxy / waterproof / impermeable).
• Finally add context: the overall driving force is a water potential gradient from soil to xylem, and that beyond the endodermis water must enter the symplast.

Step-by-Step Reasoning

1. Cell walls form the apoplast pathway. Apoplast literally means "outside the cell" and refers to the network of cell walls and intercellular spaces that permeate plant tissue. Because cellulose cell walls are fully permeable to water and small solutes, water in the apoplast can travel through them without ever entering a cell. Mark scheme point 1.

2. Water also moves through intercellular spaces. Where adjacent cells do not meet tightly, water can flow through the gaps between cells. Mark scheme point 2.

3. Name the cells the apoplast passes through on the way to the endodermis. Going from outside in, water in the apoplast passes through the walls of root hair / epidermal cells and then through the walls of the cortical (parenchyma) cells. Mark scheme point 3.

4. The pathway stops at the Casparian strip. At the endodermis, the radial and transverse cell walls are impregnated with suberin, forming a continuous waterproof ring. Because suberin is hydrophobic, water cannot pass along the cell walls or intercellular spaces past this point. Mark scheme point 4.

5. Suberin is waxy / waterproof / impermeable. This detail is required to earn the explanation mark — it is not enough to say "stops at the Casparian strip"; the candidate must describe the material of the strip and its physical property. Mark scheme point 5.

6. Overall movement is down a water potential gradient. Water moves passively from a region of higher (less negative) water potential in the soil to a region of lower (more negative) water potential in the xylem and leaf. The mark scheme explicitly rejects references to "osmosis" or "active transport" here — this is a passive process driven by the gradient. Mark scheme point 6.

7. AVP — apoplast is non-living and does not cross membranes. This is a useful framing statement that the mark scheme accepts as an extra point. It is also the underlying reason that water must cross a membrane at the endodermis, which is what allows the plant to control what enters the xylem. Mark scheme point 7.

Key Takeaways

• The apoplast pathway is the non-living route through cell walls and intercellular spaces; it does not cross cell surface membranes.
• It is blocked at the endodermis by the Casparian strip, a ring of suberin in the radial and transverse walls of endodermal cells.
• Beyond the endodermis, water must enter the symplast (cytoplasm of endodermal cells) before passing into the xylem.
• The Casparian strip is a control point: by forcing water and dissolved ions to cross a living cell membrane, the plant can selectively regulate what enters the transpiration stream.
• Overall water uptake is driven by a water potential gradient from soil to xylem (a passive process).

Common Mistakes

• Saying water moves by osmosis / active transport along the apoplast. The mark scheme explicitly rejects "osmosis" and "active transport" in the context of the apoplast — apoplastic water movement is bulk flow through wet cell walls, driven by a pressure / water potential gradient, not osmosis or active transport.
• Vagueness about which cells the apoplast passes through. Saying "through the root" is not enough; the cells (epidermal / root hair cells, cortical cells) must be named and the context "cell walls" stated.
• Naming only the Casparian strip but not the suberin. The mark scheme requires both the structure AND its composition/property (suberin / waxy / waterproof / impermeable).
• Confusing the Casparian strip with the cell surface membrane. The strip is in the cell walls, not in the membrane.
• Mixing up apoplast and symplast. Apoplast = cell walls + intercellular spaces; symplast = cytoplasm connected by plasmodesmata.

Things to Be Careful About

• Use the exact term Casparian strip (not "Casparian band" or "suberin ring" alone).
• Use the exact term suberin for the waxy material; "waxy" or "waterproof" alone is acceptable but suberin is the precise term.
• The endodermis is a single cell layer; do not describe it as the "inner cortex" or "pericycle".
• "Movement down a water potential gradient" must not be replaced by "osmosis" or "active transport".
• Keep the description tied to the apoplast specifically — the mark scheme rejects "passing through vacuoles" or "passing through cytoplasm" in the context of movement to the endodermis.

Background Concept

The cell surface membrane is a phospholipid bilayer with embedded proteins. Small, non-polar molecules (e.g. O₂, CO₂) can dissolve in the bilayer and cross it directly, but ions (charged particles such as K⁺, Na⁺, Ca²⁺) cannot pass through the hydrophobic core of the bilayer on their own — they are repelled by the non-polar fatty acid tails. To cross the membrane, ions must use a membrane transport protein (either a channel protein or a carrier protein).

There are two ways ions can move through transport proteins:

• Facilitated diffusion — ions move down their concentration gradient (or electrochemical gradient) through a channel or carrier protein. No ATP is used because movement is passive, down the gradient. The protein simply provides a hydrophilic route through the bilayer.
• Active transport — ions move against their concentration gradient using a carrier protein. Energy from ATP is required because ions are being pumped "uphill". The carrier protein undergoes a conformational change powered by ATP hydrolysis.

So three diagnostic features identify the mechanism:

1. Direction of movement (with or against the concentration gradient)
2. Whether a membrane protein is needed (always yes for ions)
3. Whether ATP is used (yes for active transport, no for facilitated diffusion)

Understanding the Question

Part (b) presents Table 6.1, which has two pre-filled rows describing two scenarios for K⁺ uptake into root epidermal cells. The first row describes K⁺ moving against its concentration gradient; the second row describes K⁺ moving down its concentration gradient. The candidate must fill in three remaining columns: whether a membrane protein is needed, whether ATP is used, and the name of the transport mechanism.

The stem reminds us that mineral ions (including K⁺) are transported dissolved in water, so we are reasoning about ion transport across cell surface membranes — the standard membrane-transport scenarios.

Approach

For each row:

1. Decide whether the ion can cross the phospholipid bilayer on its own. K⁺ is a charged ion → yes, a membrane protein is needed in both rows.
2. Decide whether ATP is required. Movement against the concentration gradient requires energy from ATP. Movement down the concentration gradient does not.
3. Name the mechanism: against gradient + ATP = active transport; down gradient + no ATP = facilitated diffusion.

Step-by-Step Reasoning

Row 1: against the concentration gradient

• K⁺ is a charged ion and cannot cross the phospholipid bilayer directly → a membrane protein (carrier) is needed.
• Moving against the concentration gradient requires energy → ATP is used.
• The transport of a substance against its concentration gradient using a membrane protein and ATP is, by definition, active transport.

Row 2: down the concentration gradient

• K⁺ still cannot cross the phospholipid bilayer on its own (it is charged) → a membrane protein (channel or carrier) is still needed. This is the key distinction from simple diffusion, which is restricted to small non-polar molecules.
• Movement is down the gradient, so it is spontaneous and does not require energy → ATP is not used.
• Passive movement of ions (or polar molecules) through a membrane protein, down the concentration gradient and without ATP, is facilitated diffusion.

Key Takeaways

• The direction of movement relative to the concentration gradient is the single most important diagnostic for choosing between active transport and facilitated diffusion.
• Charged ions always require a membrane protein to cross the phospholipid bilayer, whether they are moving with or against the gradient. This is what distinguishes facilitated diffusion from simple diffusion.
• ATP is required only for active transport — i.e. when moving against the concentration gradient.
• A useful mnemonic: "uphill = active (uses ATP)"; "downhill = passive (no ATP)".
• Active transport allows root cells to accumulate K⁺ to concentrations far higher than in the soil solution, which is essential because K⁺ is needed for many cytoplasmic functions (e.g. as a cofactor for enzymes, in stomatal opening).

Common Mistakes

• Writing "no" for the membrane protein column in the second row. Candidates sometimes confuse simple diffusion (small non-polar molecules, no protein) with facilitated diffusion (ions / polar molecules, protein required). The question is about K⁺, an ion — a protein is required in BOTH rows.
• Writing "no" for ATP use in the first row. Movement against a concentration gradient is energetically unfavourable and can only happen with an energy supply. ATP must be used.
• Writing "diffusion" instead of "facilitated diffusion" for the second row. Plain diffusion cannot move ions through the bilayer; the protein-mediated route is called facilitated diffusion.
• Swapping the two mechanisms. Always check: against gradient = active; down gradient = facilitated.
• Writing "osmosis" anywhere. Osmosis is specifically the diffusion of water molecules across a partially permeable membrane, not ion transport.

Things to Be Careful About

• The candidate receives one mark per correctly completed column (both rows right in that column), with a fallback mark available if one entire row is correctly filled — so it is worth filling every cell thoughtfully even if uncertain about one.
• "Yes / no" answers must be unambiguous — "y", "✓", or blank are not accepted; only "yes" or "no" earn the mark.
• The transport mechanism name should be the standard CIE term: active transport (not "active uptake" or "pumping") and facilitated diffusion (not "passive transport" or "channel diffusion").
• The concentration gradient refers to the gradient of K⁺ across the cell surface membrane of the root epidermal cell — not across the whole root or the Casparian strip.