Biology 9700/23 — May/June 2025

Background Concept

In a transverse section (TS) of a young dicot root, the vascular cylinder (stele) sits in the centre of the root, surrounded by the cortex and the outer epidermis. From the outside in, the stele is bounded by a single layer of tightly packed cells called the endodermis, with a Casparian strip running through the anticlinal walls that regulates the apoplastic movement of water and minerals into the stele. Inside the endodermis lies the pericycle, then the vascular tissues themselves.

In Ranunculus, the xylem occupies the centre of the stele as a star-shaped (or central) mass of large, thick-walled, lignified xylem vessel elements — these are dead at maturity and appear as the biggest, darkest-walled, empty lumens in the centre of the section. Between the arms of the xylem sit small groups of phloem sieve tube elements (smaller, thinner-walled living cells), and between the xylem and phloem lie the small, densely cytoplasmic cells of the procambium.

Understanding the Question

The stem shows a TS photomicrograph of a Ranunculus root. Cell A is labelled on the single ring of cells that encloses the vascular cylinder, and cell B is labelled on one of the large, thick-walled empty cells in the centre of the stele. The candidate simply has to name each cell type. This is a "name" command word question: one correct biological term per cell.

Approach

Recognise that the ring of cells forming the outer boundary of the stele is the endodermis (A), and the large, lignified, empty cell in the centre of the vascular cylinder is a xylem vessel element (B). Use the precise CIE-accepted names from the mark scheme.

Step-by-Step Reasoning

• A: The line points to a single cell forming the innermost layer of the cortex, immediately outside the procambium. This is the endodermis (a single endodermal cell). The mark scheme accepts "endodermal (cell)" as the named answer; bare "endodermis" is also credited as referring to the cell/tissue.
• B: The line points to one of the very large, thick-walled, hollow cells in the centre of the stele. These are lignified, dead at maturity, and conduct water and mineral ions. This is a xylem vessel element. The mark scheme accepts "xylem vessel element", "vessel element" or "xylem element"; bare "xylem" is ignored.

Key Takeaways

• The endodermis is the boundary layer of the stele; it controls which substances enter the xylem.
• The large, thick-walled, empty cells in the centre of a young dicot root TS are xylem vessel elements (lignified, dead at maturity).
• Always use the full tissue-level name, not just the organ name ("xylem vessel element", not "xylem").

Common Mistakes

• Calling B a "tracheid" or a "xylem fibre": Ranunculus is a dicot and its xylem contains vessel elements, not just tracheids.
• Calling A "cortex" (A is the innermost layer of cells, forming a discrete ring — the endodermis, not the broader cortex).
• Writing just "xylem" for B: the mark scheme explicitly ignores this and requires "xylem vessel element" (or equivalent).

Things to Be Careful About

• The endodermis and the pericycle are easy to confuse: pericycle sits just inside the endodermis and gives rise to lateral roots; endodermis sits at the very outer edge of the stele.
• Note the orientation: the question is a TS through a root, not a stem — in a root the xylem is typically central, in a stem it is peripheral (in bundles).

Background Concept

Procambium (also called procambial tissue or vascular cambium precursor) is a primary meristem — a tissue made up of undifferentiated, actively dividing cells. Meristems are the source of all new cells in a plant: their cells retain the capacity to divide by mitosis indefinitely, and the daughter cells produced can differentiate into the mature cell types of a particular tissue.

In a young root, the procambium lies between the developing xylem and phloem. Its cells divide and the inner derivatives differentiate into more xylem elements while the outer derivatives differentiate into more phloem sieve elements and companion cells. This allows the vascular cylinder to grow thicker and to repair damaged vessels.

Understanding the Question

The question says the procambium "consists of stem cells" and asks the candidate to suggest a role of this tissue in the root. The command word "suggest" allows a reasonable deduction, but the mark-scheme answers are tightly constrained around two ideas: (1) the cells divide by mitosis, and (2) they produce cells that differentiate into xylem/phloem for growth/repair.

Approach

Recognise procambium as a meristem and apply the standard role of a meristem: mitosis → new cells → differentiation into the tissue types of the surrounding organ. Tie this to Ranunculus root context — growth of the vascular cylinder and repair of xylem/phloem.

Step-by-Step Reasoning

• The cells of the procambium are unspecialised and divide continuously by mitosis, generating a pool of new cells.
• These new cells differentiate into cells of the vascular tissues — most importantly xylem vessel elements and phloem sieve tube elements.
• This supplies the root with new vascular tissue for growth of the root and for repair of damaged xylem/phloem.

Any one of these points is sufficient to earn the single mark, but combining the division idea with a function (e.g. "divide to form new xylem for growth") is the cleanest answer.

Key Takeaways

• Procambium is a primary meristem; its defining feature is continuous mitotic division.
• Daughter cells of a meristem differentiate into the functional cell types of the surrounding tissue.
• Meristems sustain growth and enable repair of plant tissues.

Common Mistakes

• Stating only "growth" without linking it to new vascular tissue: too vague for the mark.
• Saying the procambium "transports water" — it does not; it produces the cells that later become water-transporting xylem.
• Confusing procambium with the vascular cambium of a woody root: in a young herbaceous root the procambium is the relevant meristem producing primary vascular tissue.

Things to Be Careful About

• "Suggest" questions still demand precise biological vocabulary — name the tissue formed (xylem / phloem) and the process (mitosis + differentiation); do not give a hand-wavy answer.

Background Concept

All amino acids share a common backbone: a central α-carbon (alpha carbon) bonded to four different groups — an amino group ($-NH_2$), a carboxyl group ($-COOH$), a hydrogen atom ($-H$) and a variable side chain called the R group. The R group is the only part that differs between amino acids, and it is what gives each amino acid its unique chemical properties (size, charge, polarity, ability to form bonds).

Lysine is one of the 20 standard amino acids. Its R group has the structure $-CH_2-CH_2-CH_2-CH_2-NH_2$, which is a chain of four $CH_2$ groups ending in a primary amine ($-NH_2$). Lysine is therefore classified as a basic, positively-charged amino acid at physiological pH.

Understanding the Question

Fig. 2.1 shows a partially-drawn molecule. The vertical chain below the topmost carbon is the R group of lysine (the side chain). The topmost carbon — the one with a single H drawn above it — is the α-carbon position. The candidate must count how many carbon atoms are in the R group itself, NOT including the α-carbon.

The command word is "state", so a single number is the entire required answer.

Approach

Identify the boundary of the R group and count only the carbon atoms inside it. The α-carbon is the carbon that would carry the amino group, the carboxyl group and an H; the R group is everything attached to the α-carbon except those three groups.

Step-by-Step Reasoning

Looking at Fig. 2.1 from top to bottom below the α-carbon position:

1. The first carbon below the α-carbon is shown as $CH_2$.
2. The second carbon is shown as $CH_2$.
3. The third carbon is shown as $CH_2$.
4. The fourth carbon is shown as $CH_2$.
5. This fourth $CH_2$ is bonded to the nitrogen of the terminal $-NH_2$.

There are four $CH_2$ units in the side chain. None of them is the α-carbon (the α-carbon is the one to which the $-NH_2$ and $-COOH$ of the main amino acid backbone would be attached). Therefore, the R group of lysine contains four carbon atoms.

Key Takeaways

• The R group of an amino acid is everything attached to the α-carbon except $-NH_2$, $-COOH$ and $-H$.
• Lysine's R group is $-(CH_2)_4-NH_2$ — four carbons long.
• Always be careful not to count the α-carbon as part of the R group.

Common Mistakes

• Counting the α-carbon (the topmost C in the diagram) as part of the R group, which would give 5 instead of 4.
• Counting the nitrogen as a carbon.

Things to Be Careful About

The mark scheme accepts either "4" or "four". Do not include any units; this is a count, not a measurement.

Background Concept

Every amino acid has the same general structure: a central α-carbon bonded to four different groups — $-H$, $-NH_2$ (amino), $-COOH$ (carboxyl) and an R group. In a properly drawn amino acid, all four substituents appear on the α-carbon.

In Fig. 2.1, only the R group ($-CH_2-CH_2-CH_2-CH_2-NH_2$) and a single H are drawn. The α-carbon is shown as the topmost C, but it is missing the two functional groups that define an amino acid.

Understanding the Question

The candidate must complete Fig. 2.1 by adding the two missing functional groups to the topmost carbon. The mark scheme awards one mark for the $-COOH$ group and one mark for the $-NH_2$ group; both must be bonded to the same topmost carbon.

Approach

Identify the α-carbon position (the topmost C in Fig. 2.1) and attach the two missing functional groups — amino and carboxyl — to it. The single H drawn above this carbon is retained.

Step-by-Step Reasoning

1. The α-carbon is the topmost C in the diagram. It already has one bond shown going up to H and one going down to the first $CH_2$ of the R group.
2. Two further bonds need to be drawn on this α-carbon: one to $-COOH$ and one to $-NH_2$.
3. The completed α-carbon therefore has four substituents: $-H$, $-NH_2$, $-COOH$ and the R group.

The marking scheme accepts the $-COOH$ and $-NH_2$ with or without explicit bond lines within the functional groups (e.g. the C=O of the carboxyl may be drawn or implied). Atoms other than C and N bonded to the top carbon are not allowed.

Key Takeaways

• The general amino acid formula has four groups on the α-carbon: $-H$, $-NH_2$, $-COOH$ and R.
• Lysine is a basic amino acid because its R group contains an extra $-NH_2$, but the backbone amino and carboxyl groups are still required.

Common Mistakes

• Attaching the new groups to the wrong carbon (e.g. to the first $CH_2$ of the R group instead of the α-carbon).
• Drawing only one of the two required groups.
• Adding an extra group such as $-OH$ or another atom not in the amino acid general formula.

Things to Be Careful About

The mark scheme explicitly rejects any diagram that shows atoms other than C and N bonded to the α-carbon, and the $-COOH$ and $-NH_2$ must both be present. Writing $NH_2$ rather than $H_2N$ is fine, and the explicit double bond of the carboxyl group is not required.

Background Concept

The trachea is lined by a pseudostratified ciliated columnar epithelium with goblet cells scattered among it. Goblet cells are unicellular glands whose function is to secrete mucus — a sticky glycoprotein-rich fluid. Mucus sits on top of the epithelium and traps inhaled particles, dust and microorganisms (a non-specific, innate defence). The cilia of the neighbouring ciliated epithelial cells then beat in a coordinated rhythm to move the mucus and trapped debris upwards towards the pharynx, where it is swallowed (the mucociliary escalator).

Understanding the Question

Part (c)(i) is restricted to the role of goblet cells (not the ciliated cells) in protecting tracheal tissue from infection by S. aureus specifically. The question requires two creditable points.

Approach

Recall the structure–function relationship: goblet cells → secrete mucus. Then link the property of mucus (sticky / viscous) to its effect on bacteria (trapping them) and therefore to the protection of the underlying cells.

Step-by-Step Reasoning

• Mark 1 — secretion: Goblet cells secrete (produce / release) mucus (or mucin, the principal glycoprotein of mucus). (Mark scheme: "secrete / produce / AW, mucus / mucin".)
• Mark 2 — function: This mucus traps S. aureus cells and forms a barrier between the bacteria and the cells lining the trachea, so the bacteria cannot reach and infect those cells. (Mark scheme: "mucus traps S. aureus / acts as a barrier".)
• Optional elaboration: The mucus is then moved by cilia away from the lungs, removing the trapped bacteria.

Key Takeaways

• Goblet cells are a key component of the non-specific / innate defence of the gas exchange system.
• "Mucus traps pathogens" is the core functional point and is the one the mark scheme expects.
• Goblet cells and ciliated cells have separate but cooperating roles; the question deliberately splits them between (c)(i) and (c)(ii).

Common Mistakes

• Saying the goblet cells "trap" S. aureus without first stating that they secrete the mucus that does the trapping.
• Writing "goblet cells prevent infection by S. aureus" — too vague; the mark scheme credits the mechanism.
• Confusing goblet cells with ciliated cells (which move the mucus rather than produce it).

Things to Be Careful About

The mark scheme allows "bacteria / pathogens / microorganisms" in place of S. aureus for the second mark, so phrasing such as "mucus traps pathogens" is acceptable. Do not credit answers about viruses, because the mark scheme explicitly says R virus here.

Background Concept

The lumen of the trachea is lined by a pseudostratified ciliated columnar epithelium (sometimes referred to as "respiratory epithelium"). Among these cells are scattered goblet cells and basal cells. The dominant cell type other than goblet cells is the ciliated epithelial cell. Each ciliated cell bears numerous cilia on its apical surface; the cilia beat in synchrony to move mucus (and trapped debris) upwards — the mucociliary escalator.

Understanding the Question

The question is a single-mark recall: name a cell type, other than goblet cells, that lines the lumen of the trachea.

Approach

Pick the most prominent and obvious lining cell of the tracheal lumen — the ciliated epithelial cell. The stem of the question (3(c)) even mentions that S. aureus can pass in between cells lining the lumen, hinting at an epithelium made up of multiple cell types.

Step-by-Step Reasoning

• The two principal cell types in the tracheal lining are goblet cells and ciliated epithelial cells. Since goblet cells are excluded, the answer is ciliated epithelial cell.

Key Takeaways

• The trachea is lined by ciliated epithelium plus mucus-secreting goblet cells.
• Knowing the cell types of a typical gas-exchange epithelium is a standard CIE requirement.

Common Mistakes

• Writing just "ciliated cell" — the mark scheme explicitly ignores this wording.
• Writing "epithelial cell" — too vague, doesn't capture the ciliated feature that distinguishes it.
• Naming an immune cell such as a phagocyte — these patrol the tissue but are not part of the lining of the lumen.

Things to Be Careful About

Use the full name: ciliated epithelial cell (the scheme accepts "ciliated epithelium cell"). "Ciliated cell" alone is ignored.

Background Concept

Bacterial cell walls contain peptidoglycan — long glycan chains crosslinked by short peptide bridges. The crosslinks are made by enzymes such as transpeptidases (also called penicillin-binding proteins, PBPs). Without crosslinks the wall is weak, the bacterium cannot withstand osmotic pressure, and it lyses.

Penicillin is a structural analogue of the D-Ala-D-Ala terminus of the peptidoglycan precursor. It binds irreversibly to the active site of the transpeptidase, blocking it. The enzyme can no longer form crosslinks, so the wall is not properly assembled. Penicillin is therefore described as an enzyme inhibitor.

Vancomycin works differently (Fig. 3.2). It binds directly to the D-Ala-D-Ala terminus of the peptidoglycan subunit (the free precursor), physically blocking the subunit so that the transpeptidase cannot use it. The crosslink is therefore not formed, but the blockage is at the substrate, not at the enzyme.

Understanding the Question

Part (d)(i) is a structured compare-and-contrast ("similarities and differences"). The candidate must consult Fig. 3.2 for vancomycin's mechanism and recall penicillin's mechanism from prior knowledge. The mark scheme allocates three marks to be drawn from any of the credit-worthy points; the answer must be clearly organised into similarities and differences.

Approach

1. Similarities — ask: what is the net effect of both antibiotics on the cell? Both ultimately prevent the formation of crosslinks between peptidoglycan chains, and therefore stop / prevent the synthesis (and repair) of the cell wall.
2. Differences — ask: how does each antibiotic achieve this? Penicillin binds to the enzyme (transpeptidase) that makes the crosslinks; vancomycin binds to the peptidoglycan subunit so the enzyme has nothing to work on.
3. Keep the language tight and the contrast explicit. Use terms such as "enzyme" and "substrate / peptidoglycan component" to make the difference unambiguous.

Step-by-Step Reasoning

• Similarity 1 (mark-scheme point): Both prevent the formation of crosslinks (cross-bridges) between peptidoglycan chains. This is the mechanistic similarity.
• Similarity 2 (mark-scheme point): Both therefore stop / prevent the synthesis and repair of the cell wall. (Without new crosslinks the wall is not made, and existing damage cannot be patched — the bacterium is doomed by osmotic lysis.)
• Difference 1 (mark-scheme point): Penicillin binds to and inhibits the enzyme that catalyses crosslink formation (transpeptidase / PBP). Mark scheme language: "penicillin binds to / is an inhibitor of enzymes / transpeptidases that catalyse formation of crosslinks".
• Difference 2 (mark-scheme point): Vancomycin binds to / acts on the peptidoglycan (subunit / crosslink component) itself. Mark scheme language: "vancomycin binds to / acts on peptidoglycan / crosslink components".
• Optional elaboration (also worth a mark, AVP): Vancomycin blocks access of the peptidoglycan subunit to the growing chain, or it may prevent the enzyme that joins subunits together from binding.

Key Takeaways

• "Prevents crosslinks" is the common end-point of both antibiotics; the means differ — enzyme inhibition (penicillin) vs. substrate binding (vancomycin).
• A clear compare-and-contrast structure makes it obvious to the examiner that both halves of the question have been addressed.
• Reading Fig. 3.2 carefully tells you exactly what vancomycin is binding to (the free peptidoglycan subunit), which is the key piece of information for the difference.

Common Mistakes

• Saying only that both "break down" or "destroy" the cell wall — the action is on crosslink formation, not on direct wall destruction.
• Failing to distinguish the levels of action (enzyme vs. substrate) when describing the difference.
• Skipping the similarities and listing only differences (or vice-versa) — the question requires both.
• Confusing penicillin with vancomycin: e.g. saying penicillin "binds to peptidoglycan components" (it binds to the enzyme).

Things to Be Careful About

The mark scheme permits ecf (error carried forward): if a candidate mis-states penicillin's mechanism, an answer that correctly describes vancomycin from Fig. 3.2 may still earn one mark. So even with a partial misconception, some credit is recoverable. The phrase "prevent formation of crosslinks" must come from a correct understanding of both antibiotics for that similarity mark to be awarded.

Background Concept

Antibiotic resistance arises when random mutations or horizontal gene transfer give individual bacteria the ability to survive an antibiotic. In the presence of the antibiotic these resistant cells are strongly selected for and come to dominate the population. Resistant strains of S. aureus (notably MRSA — meticillin-resistant Staphylococcus aureus, and VRSA — vancomycin-resistant) are major clinical problems.

Reducing the impact of resistance relies on two complementary strategies:

• Slowing the emergence and spread of resistance (stewardship, hygiene, vaccination).
• Replenishing the armoury (new antibiotics, alternative antibacterials, combination therapy).

The mark scheme provides a long menu of acceptable points, so any three of them earn the three marks.

Understanding the Question

Part (d)(ii) asks for steps that can be taken to reduce the impact of antibiotic resistance. This is broader than just "stop using antibiotics" — it covers the full set of clinical, public-health and research actions.

Approach

Group the ideas into three or four natural categories and pick the three most concrete, distinct points:

• Prescribing behaviour (when and what to prescribe).
• Patient behaviour (completing the course).
• Drug development (new antibiotics / alternatives).
• Agricultural use (reducing non-medical antibiotic exposure).
• Transmission control (vaccines, hygiene, isolation).

Step-by-Step Reasoning

• Point 1 — Only when necessary: Prescribing / taking antibiotics only when absolutely required; not for viral infections and not prophylactically. This reduces selection pressure.
• Point 2 — Correct antibiotic: Make sure the correct, effective antibiotic is used for the diagnosed bacterial infection (use sensitivity testing where possible). This avoids low-dose / wrong-drug exposure that breeds resistance.
• Point 3 — Complete the course: Patients must complete the full prescribed course so that partially-resistant survivors are not selected for. DOTS (Directly Observed Treatment, Short-course) is a structured example for TB.
• Point 4 (alternative): Develop new antibiotics or use other antibacterials, so there is always an effective drug available against resistant strains.
• Point 5 (alternative): Reduce or control the use of antibiotics in agriculture and in animals reared for food, since resistant organisms can spread from farms to humans.
• Point 6 (alternative): Break the transmission cycle of resistant strains through vaccination, good hospital hygiene (especially hand-washing between patients), and isolation / quarantine.
• Point 7 (AVP): Train and update healthcare professionals and the public; report patterns of antibiotic resistance; monitor whether an antibiotic remains effective; limit over-the-counter antibiotic sales; support initiatives such as the WHO Global Plan to End TB.

Key Takeaways

• Antibiotic resistance is driven by selection pressure; reducing unnecessary antibiotic use is the single most important intervention.
• Combinations of stewardship (Point 1), correct prescribing (Point 2), patient adherence (Point 3), new drugs (Point 4), agricultural controls (Point 5) and transmission control (Point 6) are all needed; no single measure is sufficient.
• A public-health-style answer (vaccines, hygiene, quarantine) is as relevant as a clinical-style answer (prescribing, courses).

Common Mistakes

• Vague answers like "be careful with antibiotics" or "use less medicine" — too non-specific to earn marks.
• Only listing clinical measures and ignoring agriculture, transmission control, or drug development.
• Confusing antibiotic resistance with vaccine resistance or antiviral resistance.

Things to Be Careful About

The mark scheme is generous (three points from a long list) and explicitly allows AVP — so any plausible, distinct, well-stated measure earns a mark. However, the wording must be specific to antibiotic resistance rather than generic infection control (e.g. "wash hands" alone is too broad; "hospital hand-hygiene to prevent spread of MRSA" is on-target).

Background Concept

Cholesterol is a small, planar lipid molecule that sits within the phospholipid bilayer of eukaryotic plasma and organelle membranes. Its polar hydroxyl (-OH) group hydrogen-bonds with the phospholipid head groups, while its rigid four-ring hydrocarbon body nestles among the fatty acid tails of the inner bilayer. Through these interactions, cholesterol modulates membrane fluidity:

• At higher temperatures it restrains the movement of the fatty acid tails, reducing membrane fluidity.
• At lower temperatures it prevents the tails from packing too closely, maintaining some fluidity.

In mammalian cells at body temperature (~37 °C), the net effect of cholesterol is to reduce membrane fluidity and to make the membrane less permeable to small water-soluble molecules and ions.

Understanding the Question

The stem tells you that lysosome membranes contain a lower concentration of cholesterol than other membranes in the same cell. The question asks you to state how this lower cholesterol level makes the lysosome membrane's properties different. The command word "state" means a single, clearly-worded fact.

Approach

Apply the principle above in reverse: if cholesterol reduces fluidity, then less cholesterol means the membrane loses some of that restraint and becomes more fluid.

Step-by-Step Reasoning

• Cholesterol normally reduces the lateral movement of phospholipids in the bilayer.
• With less cholesterol, this restraining effect is reduced.
• The membrane therefore becomes more fluid and phospholipids show greater (lateral) movement.
• The mark scheme also accepts "increased passage of polar molecules/ions" or "easier fusion with other membranes" as alternative consequences, but the core, easiest point is the increase in fluidity.

Key Takeaways

• Cholesterol is a fluidity regulator: less cholesterol → more fluid membrane.
• A physical property (fluidity) follows directly from membrane composition.
• This principle sets up the structure–function link tested in (a)(ii).

Common Mistakes

• Saying the membrane becomes "weaker", "thinner" or "less stable" — these are not properties directly affected by cholesterol.
• Saying the membrane becomes "less fluid" — the opposite of the correct answer.
• Vague answers like "membrane changes" without specifying what changes.

Things to Be Careful About

• The word "fluidity" is the precise term expected in CIE mark schemes; synonyms such as "runny" or "loose" do not score.
• The answer is one point only; do not pad with extra (unrewarded) biology.

Background Concept

Lysosomes are roughly spherical, membrane-bound organelles that contain hydrolytic (digestive) enzymes — proteases, lipases, nucleases, and glycosidases such as α-galactosidase. They work at an internal pH of about 4.5–5.0, maintained by V-type H⁺-ATPases (proton pumps) in the lysosomal membrane, while the cytoplasm sits at about pH 7.2.

A lysosome does not "eat" substrates directly. Instead, it fuses with another membrane-bounded compartment that contains material to be digested:

• Phagosomes / endosomes carrying material taken in by endocytosis or phagocytosis.
• Autophagosomes containing worn-out organelles from the cell itself.

Fusion mixes the lysosomal hydrolases with the substrate, breaking it down inside the resulting vesicle.

Understanding the Question

You have just established that lysosome membranes are more fluid than other cellular membranes. The question now asks you to suggest why this greater fluidity would help the lysosome perform its function. The command word "suggest" allows you to propose a reasoned idea rather than state a textbook fact.

Approach

Connect the physical property (fluidity) to the lysosome's mode of action (membrane fusion). More fluid bilayers bend, deform and merge with other bilayers more easily.

Step-by-Step Reasoning

• A lysosome's job is to fuse with vesicles (phagosomes, endosomes, autophagosomes) that contain material to be digested.
• Two membranes can fuse only if their phospholipid bilayers can deform, come into very close contact, and merge.
• A more fluid membrane is more deformable; its phospholipids can rearrange more readily.
• Therefore, the lysosome can more easily fuse with these vesicles, allowing the hydrolytic enzymes to be delivered to the substrate.
• Note the mark scheme's ecf concession: if you wrongly said fluidity DECREASED in (i), then the explanation here would be that the membrane becomes more impermeable, preventing the hydrolytic enzymes from leaking out of the lysosome.

Key Takeaways

• Lysosomes function by fusion, not by direct digestion.
• Membrane fluidity is the physical property that determines ease of fusion.
• A structural feature (low cholesterol) is adapted to a function (fusion).

Common Mistakes

• Saying the lysosome "digests material" without specifying the fusion step.
• Conflating fluidity with permeability — they are related but distinct; the mark scheme wants the fusion link.
• Repeating the answer from (i) without going further into the functional link.

Things to Be Careful About

• "Suggest" lets you give a reasonable proposal; it does not require quoting a textbook sentence.
• The mark scheme wants at least one of: vacuoles, vesicles, phagosomes, endosomes, or "membranes of other organelles".

Background Concept

The Michaelis–Menten equation describes the relationship between the rate ($v$) of an enzyme-catalysed reaction and the substrate concentration $[S]$:

Two key constants are:

• $V_{\text{max}}$ — the maximum rate, reached at saturating substrate concentration when essentially every active site is occupied by substrate.
• $K_m$ (the Michaelis constant) — the substrate concentration at which $v = \dfrac{1}{2} V_{\text{max}}$. It is an inverse measure of enzyme–substrate affinity: a low $K_m$ means high affinity.

On a Michaelis–Menten curve, $K_m$ is read off by:

1. Reading the plateau (which is $V_{\text{max}}$).
2. Halving it.
3. Drawing a horizontal line from the half-rate to the curve, then a vertical line down to the substrate axis.

Understanding the Question

You are given Fig. 4.1, a Michaelis–Menten-style curve for α-galactosidase at pH 4.5. You must determine $K_m$ for that pH and quote the unit. The unit of $K_m$ is always the unit of substrate concentration (the same as the x-axis).

Approach

Identify the pH 4.5 curve (open circles, the lower curve), read its plateau as $V_{\text{max}}$, halve that value, and read the substrate concentration at which the curve crosses that half-rate.

Step-by-Step Reasoning

1. The pH 4.5 curve plateaus at $V_{\text{max}} \approx 2.8\ \text{pmol min}^{-1}$.
2. Half of $V_{\text{max}}$: $\dfrac{2.8}{2} = 1.4\ \text{pmol min}^{-1}$.
3. Locate $1.4\ \text{pmol min}^{-1}$ on the y-axis and draw a horizontal line to the right until it meets the pH 4.5 curve.
4. From that point, drop a vertical line to the x-axis; the reading is $120\ \mu\text{mol dm}^{-3}$.
5. Therefore $K_m = 120\ \mu\text{mol dm}^{-3}$ at pH 4.5.

Key Takeaways

• $K_m$ is read off a Michaelis–Menten curve at half $V_{\text{max}}$.
• The unit of $K_m$ is the same as the unit of substrate concentration.
• A small $K_m$ indicates a high-affinity enzyme.

Common Mistakes

• Reading $K_m$ at the substrate concentration equal to half the x-axis range (i.e. 400) — wrong; you must halve the y-axis value first.
• Forgetting to state the unit.
• Reading the wrong curve (the pH 5.9 curve).
• Saying the unit is "mol" instead of "$\mu\text{mol dm}^{-3}$".

Things to Be Careful About

• The x-axis in the figure is in $\mu\text{mol dm}^{-3}$, not $\text{mmol dm}^{-3}$.
• The question wants the value AND the unit; both are credited separately.
• The mark scheme accepts the 110–130 range; 120 is the intended reading.

Background Concept

Enzyme activity is highly sensitive to pH. Each enzyme has an optimum pH at which:

• The ionisation state of the active-site R-groups is correct for substrate binding and catalysis.
• The ionisation state of the substrate is also correct.
• The tertiary structure (held by ionic, hydrogen and disulfide bonds) is undisturbed.

Away from the optimum, the altered H⁺ concentration disrupts these bonds, partially distorts the active site, reduces the rate of E–S complex formation, and therefore reduces $V_{\text{max}}$. The enzyme is not necessarily fully denatured; it is simply less catalytically efficient.

For the α-galactosidase curves shown in Fig. 4.1, pH 5.9 gives the higher $V_{\text{max}}$ and so is closer to the optimum, while pH 4.5 is further from the optimum and gives the lower $V_{\text{max}}$.

Understanding the Question

The question provides two Michaelis–Menten curves (pH 4.5 and pH 5.9) and asks you to:

1. Describe the differences between the two sets of results (data extraction).
2. Suggest explanations for those differences (mechanistic reasoning).

The command word "describe AND suggest explanations" tells you that both halves of the answer carry marks.

Approach

First, systematically compare the two curves on the graph:

• The overall rate at any given substrate concentration.
• The plateau height ($V_{\text{max}}$).
• The substrate concentration at which the plateau is reached.
• The apparent steepness of the initial rise.

Then turn each observation into a mechanism: what does the lower pH do to the enzyme's structure and active site?

Step-by-Step Reasoning

Differences (data extraction):

1. The rate is lower at pH 4.5 than at pH 5.9 at every substrate concentration shown.
2. $V_{\text{max}}$ at pH 4.5 is $2.8\ \text{pmol min}^{-1}$ while at pH 5.9 it is $5.9\ \text{pmol min}^{-1}$ — about half.
3. $V_{\text{max}}$ is reached at a lower substrate concentration at pH 4.5 ($\approx 330{-}400\ \mu\text{mol dm}^{-3}$) than at pH 5.9 ($\approx 700\ \mu\text{mol dm}^{-3}$).

Explanations (mechanism):

1. pH 5.9 is closer to the optimum pH of α-galactosidase; pH 4.5 is further from the optimum, which is why the latter gives the lower $V_{\text{max}}$.
2. The higher H⁺ concentration at pH 4.5 alters the ionisation of the R-groups of amino acids in the active site, disturbing ionic and hydrogen bonds and slightly changing the enzyme's tertiary structure and active-site shape.
3. The active site becomes less complementary to the substrate, so fewer successful enzyme–substrate complexes form per unit time.
4. Consequently, the rate is lower at every substrate concentration and $V_{\text{max}}$ is reduced.

Key Takeaways

• pH influences enzyme activity by altering the ionisation of R-groups and the active-site shape.
• The curve with the higher $V_{\text{max}}$ is closer to the optimum pH.
• A systematic comparison should cover rate, $V_{\text{max}}$, substrate concentration at $V_{\text{max}}$ and the shape of the initial rise.
• An enzyme is not "all or nothing" — moving away from the optimum lowers $V_{\text{max}}$ gradually rather than switching the enzyme off.

Common Mistakes

• Describing only the differences and forgetting the explanations.
• Saying pH 4.5 is the optimum — wrong, because it has the lower $V_{\text{max}}$.
• Saying the enzyme is denatured at pH 4.5 — the enzyme is still working (Vmax is reached); it is just less active.
• Vague explanations like "pH affects the enzyme" without specifying how.
• Mixing up pH 4.5 and pH 5.9 in either the description or the explanation.

Things to Be Careful About

• The mark scheme rewards up to 3 explanation marks; you don't need every detail point to score well.
• The data extraction marks require specific numerical comparisons; vague language like "lower" alone is not enough — quote values.
• An "or reverse argument" (ora) is acceptable on the differences, but you must state it the right way round for the explanation.

Background Concept

Transcription is the synthesis of an RNA molecule using one strand of DNA (the template / transcribed strand) as a guide. RNA polymerase reads the template 3′→5′ and synthesises a complementary RNA strand 5′→3′. Complementary base pairing in transcription follows the rules: A (DNA) pairs with U (RNA), T (DNA) pairs with A (RNA), C (DNA) pairs with G (RNA), and G (DNA) pairs with C (RNA). The only difference from DNA replication is that uracil (U) replaces thymine (T) in the new RNA strand.

Understanding the Question

The question states explicitly that TERRA is transcribed from the DNA sequence shown in Fig. 5.1. The strand given is therefore the template strand, and the student must write out the six complementary RNA bases underneath. One mark is awarded for the correct six-letter RNA sequence.

Approach

Apply the transcription base-pairing rules position by position across the six DNA bases shown: A A T C C C. Each DNA base has only one correct RNA partner.

Step-by-Step Reasoning

• Position 1: DNA base A → RNA base U.
• Position 2: DNA base A → RNA base U.
• Position 3: DNA base T → RNA base A (T pairs with A in RNA, not with U).
• Position 4: DNA base C → RNA base G.
• Position 5: DNA base C → RNA base G.
• Position 6: DNA base C → RNA base G.

Resulting RNA: 5′-UUAGGG-3′. This is the six-base sequence written into Fig. 5.2.

Key Takeaways

• Transcription uses complementary base pairing, but with U replacing T in the RNA product.
• The DNA strand shown is the template; do not write the same sequence back (that would be the non-template strand, not the RNA product).

Common Mistakes

• Writing the complementary DNA strand instead of RNA: AAU is wrong; the RNA must contain U, not T.
• Writing AATTCC or copying the DNA sequence with T instead of producing the RNA complement.
• Pairing DNA-T with DNA-T (using replication rules instead of transcription rules).

Things to Be Careful About

• RNA contains uracil, never thymine. Every A on the DNA template must be read as U on the RNA.
• Check the orientation: the question asks for the RNA 'transcribed from' the given sequence, so the given sequence is the template.

Background Concept

The genetic code is read in triplets (codons), each codon specifying one amino acid. On an mRNA molecule the codons are read 5′→3′, in non-overlapping groups of three. A dipeptide consists of two amino acids joined by a single peptide bond and is therefore the product of two codons (six bases) of mRNA.

In a double-stranded DNA gene, the strand that is not used as the template during transcription is called the non-transcribed (or sense / coding) strand. Its sequence matches the mRNA exactly except that T is present where the mRNA has U. The circular genetic code chart in Fig. 5.3 lists DNA triplets on this non-transcribed strand, so the codon for the first amino acid can be read directly by converting the mRNA codon back to its DNA equivalent (U → T) and using the chart.

Understanding the Question

The TERRA RNA is UUAGGG (from part (b)(i)). The question asks for the two amino acids of the dipeptide translated from TERRA. The figure provides a circular code with DNA triplets on the non-transcribed strand, read from the centre (first base) outwards. Two marks are available, one per amino acid.

Approach

1. Group the six RNA bases into two non-overlapping triplets starting at the 5′ end: UUA then GGG.
2. Convert each mRNA triplet to its equivalent DNA triplet on the non-transcribed strand (replace U with T): TTA and GGG.
3. Use the chart, starting from the centre, to read off the amino acid for each triplet.

Step-by-Step Reasoning

• Step 1 — split the mRNA into codons:
  TERRA = UUA GGG → two codons: UUA and GGG.
• Step 2 — convert to the DNA form used by the chart:
  UUA → TTA, and GGG → GGG (no change, since there is no U in this codon).
• Step 3 — read the chart:
  • TTA: start with T in the centre, then T as the second base (on the T side), then A as the third base → Leucine.
  • GGG: start with G in the centre, then G as the second base, then G as the third base → Glycine.
• The dipeptide is therefore leucine–glycine (or glycine–leucine depending on which is read as amino acid 1, but the mark scheme credits 'leucine; glycine' as the two answers).

The mark scheme offers error carried forward (ecf) for the alternative reading of asparagine and proline, which corresponds to reading the codons in a different frame or a different interpretation, but the standard correct reading is leucine and glycine.

Key Takeaways

• mRNA is read in non-overlapping triplets of three bases from the 5′ end.
• A dipeptide is the product of two codons (six bases).
• The non-transcribed (sense) DNA strand has the same sequence as the mRNA except T replaces U, which lets a DNA-based code chart be used directly.
• A circular genetic code chart is read from the centre (1st base) outwards (2nd, then 3rd base).

Common Mistakes

• Reading the chart with the wrong base first (using the 3′ end instead of the 5′ end) — codons are read 5′→3′.
• Forgetting to convert U to T before using a DNA-based code chart, leading to wrong amino acids.
• Reading overlapping triplets (UAG, GGG…) instead of non-overlapping triplets (UUA, GGG).
• Confusing the non-transcribed strand with the transcribed (template) strand, which would be the complement of the mRNA and would give a different amino acid sequence.

Things to Be Careful About

• The figure shows DNA triplets on the non-transcribed strand, so U in the mRNA must be converted to T before using the chart.
• The reading direction is 5′→3′ on the mRNA, which corresponds to a fixed direction around the circular chart (centre → middle ring → outer ring).
• The ecf in the mark scheme accepts 'asparagine and proline' if a different reading frame is used; only credit the standard 'leucine; glycine' if the correct frame is used.

Background Concept

Stem cells are unspecialised cells that retain the ability to divide and to differentiate into a range of specialised cell types. Differentiation depends on selective gene expression: different cells transcribe different subsets of their genes, producing different mRNAs and therefore different proteins (especially transcription factors, enzymes and structural proteins) that give each cell type its identity and function.

If many more genes are transcribed at once, the cell gains access to a wider repertoire of mRNAs, and therefore a wider repertoire of proteins, than would otherwise be possible. In a stem cell, this expanded protein repertoire provides the raw material from which the cell can choose a particular differentiation pathway.

Understanding the Question

The question states that TERRA interacts with genes in stem cells and that a large increase in TERRA concentration leads to a large increase in the number of genes transcribed. The student must suggest ONE plausible biological consequence for the stem cell. One mark is available, so a single well-articulated point is sufficient.

Approach

Trace the flow of information from DNA → mRNA → protein → cellular function. An increase in the number of transcribed genes means an increase in the variety of mRNAs and therefore proteins the cell can make. In a stem cell, the most biologically significant consequence is the ability to differentiate into a wider range of specialised cell types.

Step-by-Step Reasoning

• More genes transcribed → more different mRNA molecules available in the cell.
• More different mRNAs → more different proteins can be translated (the mark scheme accepts 'enzymes' as an alternative specific form of 'protein').
• In a stem cell, this expanded protein set provides the molecular machinery needed for the cell to differentiate (become specialised / take on a particular function) or to grow and produce organelles in preparation for mitosis.

The mark scheme credits any one of:

• increase in number of different proteins (or enzymes) produced;
• reference to the cell's ability to differentiate / become specialised / take on a particular function;
• increase in growth of the cell in preparation for mitosis;
• production of cell organelles in preparation for mitosis.

Key Takeaways

• Gene expression is the link between the genome and cellular phenotype.
• A wider range of transcribed genes gives a wider range of proteins, which in stem cells supports differentiation.
• This is an 'outline' / 'suggest' question, so a single well-articulated downstream consequence is sufficient for the mark.

Common Mistakes

• Saying 'more protein is produced' without specifying 'different' kinds — quantity alone is not the right idea here, the point is variety.
• Saying 'more mRNA is produced' instead of more different types of mRNA / protein.
• Suggesting that the cell becomes cancerous — the question is about stem cells, not uncontrolled division.
• Confusing transcription with translation: the question explicitly says 'genes that are transcribed', so the first consequence is at the mRNA level.

Things to Be Careful About

• The command word is 'suggest', so a plausible biological consequence is required — there is no single 'correct' answer, but the mark scheme lists the creditable options.
• 'Differentiation' or 'becoming specialised' is a high-yield answer that links gene expression to stem-cell function.