Biology 9700/22 — May/June 2024

Background Concept

An eyepiece graticule is a small glass disc, etched with a scale, that fits into the eyepiece of a light microscope. It appears superimposed on the specimen when the user looks down the microscope. However, the divisions on the graticule are arbitrary — they correspond to a different actual length depending on which objective lens is being used. To convert graticule divisions into real units (µm or mm), the eyepiece graticule must be calibrated for each objective.

Understanding the Question

The stem says the student used a microscope fitted with a calibrated eyepiece graticule and used it to estimate the length of a smooth muscle cell as 250 µm. The question (b)(i) asks what type of microscope slide was used in the calibration step. The calibration step requires a scale of known length to be placed on the stage so its actual length can be matched to the graticule divisions.

Approach

Recall the standard CIE terminology: the slide placed on the stage of the microscope that carries a scale of known length (typically 1 mm divided into 100 units of 10 µm each) is called a stage micrometer.

Step-by-Step Reasoning

• A stage micrometer is a specially etched slide with a precisely known scale (commonly 1 mm in 100 divisions of 10 µm each).
• It is placed on the stage and used to calibrate the eyepiece graticule at each magnification by aligning the two scales and counting how many eyepiece divisions correspond to one stage micrometer division.
• Once calibrated, the eyepiece graticule can be used to measure the specimen (here, the smooth muscle cell), giving the 250 µm result.

The term "stage micrometer" is the precise CIE term; "graticule" alone would refer to the eyepiece disc and is not credit-worthy here. "Micrometre slide", "calibration slide" or similar vague terms are also not credited.

Key Takeaways

• Calibration of an eyepiece graticule requires a scale of known length on the stage.
• The apparatus for this is called a stage micrometer.
• Calibration must be repeated for each objective lens.

Common Mistakes

• Saying "graticule" — this is the eyepiece component, not the stage slide.
• Saying "ruler" or "scale" — not the correct scientific term.

Things to Be Careful About

• The wording must be "stage micrometer" (or "stage micrometer scale" / "stage micrometer slide"), exactly.

Background Concept

The unaided human eye has a resolving power of about 0.1–0.2 mm — that is, the smallest objects we can distinguish as separate are roughly 100–200 µm across. The question gives the student's own practical limit as 0.2 mm = 200 µm.

Units:

So:

Understanding the Question

The stem gives us a cell length of 250 µm (from the calibrated eyepiece graticule in part (b)). The question tells us the smallest object the student can resolve unaided is 0.2 mm. We must decide whether a 250 µm cell is large enough for the student to see without a microscope, and justify the answer with the numbers and units.

Approach

Convert one of the values so both are in the same units, then compare numerically.

Step-by-Step Reasoning

• Convert 0.2 mm into µm: 0.2 mm × 1000 = 200 µm.
• Compare: 250 µm (cell) vs 200 µm (smallest visible object).
• Since 250 µm > 200 µm, the cell is longer than the student's unaided resolution limit, so it can be seen without a microscope.

The mark scheme accepts several equivalent justifications: that 250 µm = 0.25 mm and 0.2 mm < 0.25 mm; or that the cell is 50 µm (0.05 mm) longer than the resolution limit; or a standard-form comparison (2 × 10⁻⁴ m vs 2.5 × 10⁻⁴ m). The numerical values and the units must both appear.

Key Takeaways

• The unaided eye can resolve objects down to roughly 0.1–0.2 mm (100–200 µm).
• Always convert to common units before comparing two lengths.
• 1 mm = 1000 µm is the key conversion here.

Common Mistakes

• Saying "yes" without converting units or without quoting numbers.
• Confusing which way the inequality goes (250 µm is bigger, not smaller, than 200 µm).
• Quoting a number without units — the mark scheme explicitly requires the units.

Things to Be Careful About

• Both a numerical value AND the units must be present to earn the mark.
• The answer must include "yes" with the qualifying comparison; a bare "yes" or "no" is not credited.

Background Concept

Capillaries are the smallest blood vessels (typically 5–10 µm in diameter — comparable to the size of a single red blood cell). Their wall consists of a single layer of flattened endothelial cells, with no muscle or elastic tissue, no collagen layer, and no outer tunica. This is much thinner than the walls of arterioles, arteries or veins, all of which have multiple layers.

Because the lumen is so narrow, only one red blood cell at a time can pass through, giving the characteristic "single file" appearance seen in histological sections and micrographs.

Understanding the Question

Part (c)(i) shows a photomicrograph (Fig. 1.2) of smooth muscle tissue with one vessel visible. The student is asked to outline the features that allow that vessel to be identified as a capillary — i.e., the distinguishing visible features in the image.

Approach

Recall the structural hallmarks of a capillary and check which are visible in the micrograph:

1. Wall is a single layer of cells (one cell thick).
2. Lumen is narrow — about the same diameter as a red blood cell.
3. Red blood cells travel in single file because only one fits at a time.
4. No muscle or elastic layer is visible around the vessel.

Step-by-Step Reasoning

• The vessel in the centre of the field shows a row of pale, rounded red blood cells in a narrow channel. The cells are arranged one after another — this is the "single file" appearance.
• Comparing the diameter of a red blood cell to the width of the channel, they are roughly the same — the lumen is about as wide as one RBC.
• The wall cannot be resolved as multiple layers — there is no visible smooth muscle or connective tissue coat around the vessel, consistent with a one-cell-thick endothelium.

The mark scheme also accepts features like: red blood cells deforming slightly to squeeze through (showing flexibility) — this is visible in some micrographs where the RBCs appear biconcave/curved.

Key Takeaways

• Capillaries are the smallest vessels; their lumen is barely wider than an RBC.
• One-cell-thick wall is a defining structural feature of capillaries.
• Single-file RBC flow is the most easily recognised identifying feature in a micrograph.

Common Mistakes

• Saying "capillaries are thin" without specifying "one cell thick".
• Confusing the identifying features with those of arteries or veins (thick muscular wall, multiple layers).
• Saying "the wall has one cell" rather than "the wall is one cell thick" — both work, but the precise form is clearer.

Things to Be Careful About

• The features must be visible in the image (or implied by what's visible) — do not credit features you would only see at much higher magnification.

Background Concept

Capillaries are the site of exchange between blood and the surrounding tissues. Their structure is uniquely suited to this role:

• The wall is a single layer of squamous (flattened) endothelial cells, often only 1 µm thick in places.
• There are small gaps (pores/fenestrations) between adjacent endothelial cells, and sometimes through the cells themselves.
• The lumen is very narrow (5–10 µm), only just wide enough for a red blood cell.

Fick's law tells us that the rate of diffusion is inversely proportional to the distance, and proportional to the surface area and the concentration gradient. So a thin wall maximises the rate of exchange, and a large total surface area (achieved by vast numbers of capillaries in a network) does the same.

Understanding the Question

Part (c) has introduced Fig. 1.2, a photomicrograph showing a capillary running through smooth muscle tissue. The smooth muscle cells are highly active (continuous contraction and relaxation) and therefore have a high demand for oxygen and nutrients, and a high production of waste products (CO₂, lactate). Part (c)(ii) asks the candidate to explain how the structure of a capillary relates to its function in supplying (and draining) such a tissue.

The command word "explain" means each point must pair a structural feature with a functional consequence — saying "thin wall" alone is not enough; it must be tied to "short diffusion distance" or similar.

Approach

Go through the key structural features of a capillary one at a time and, for each, state the functional consequence for exchange with smooth muscle cells.

Step-by-Step Reasoning

1. Function stated first: a capillary supplies oxygen, glucose, amino acids and other nutrients to smooth muscle cells and removes carbon dioxide and waste products (e.g. lactate from respiration).
2. Thin wall → short diffusion distance: the wall is one cell thick (≈ 1 µm). According to Fick's law, this shortens the diffusion path between the blood and the muscle cells, increasing the rate of exchange.
3. Endothelial pores → efficient passage of substances and cells: gaps between adjacent endothelial cells (pores/fenestrations) allow rapid movement of substances between the plasma and the tissue fluid bathing the muscle cells. They also allow phagocytes (neutrophils, monocytes, macrophages) to squeeze out of the blood to reach any site of damage or infection in the tissue.
4. Narrow lumen (5–10 µm) → cells are close to a blood supply: because the capillary is so narrow, it can branch through the tissue between the muscle cells, so no smooth muscle cell is far from a capillary. The narrow lumen also slows the flow of blood, giving more time for exchange.

The mark scheme accepts slightly different framings — for example, "endothelial pores allow formation of tissue fluid around the cells" (since tissue fluid forms by filtration of plasma through these pores). It explicitly does not credit "blood leaves to make tissue fluid" — the blood stays inside the capillary; only fluid and small solutes pass out.

Key Takeaways

• Capillary structure is the classic example of "form follows function" in biology.
• The three features to learn are: thin wall, pores, narrow lumen.
• Each must be tied to a functional consequence when you answer an "explain" question.
• The narrow lumen slowing blood flow is what gives red blood cells time to off-load oxygen.

Common Mistakes

• Stating "the wall is thin" without specifying "one cell thick" and without linking it to a short diffusion distance.
• Saying "blood leaves the capillary to make tissue fluid" — blood stays inside; fluid leaves.
• Naming only the substances without linking to a structural feature (or vice versa).
• Confusing capillaries with veins/arteries (which have thick muscular walls and would prevent exchange).

Things to Be Careful About

• Each marking point pairs STRUCTURE with FUNCTION; do not separate them.
• "Oxygen in / carbon dioxide out" is acceptable shorthand; the mark scheme rejects "oxygen out, carbon dioxide in".
• Plasma proteins do NOT pass through capillary pores (they are too large), so "plasma proteins into tissue fluid" is not credited.

Background Concept

A gene is a sequence of DNA bases (A, T, C, G) that codes for a polypeptide. The order of the bases determines the order of amino acids in the polypeptide, via the genetic code (read as triplets/codons during translation).

A mutation is a permanent alteration in the DNA sequence of a gene. It can take several forms: a substitution (one base replaced by another), a deletion (one or more bases removed), or an insertion (one or more bases added). Because the code is read in non-overlapping triplets, deletions and insertions cause a frameshift, which alters every downstream codon; a substitution usually alters only one codon.

Understanding the Question

The stem of part (d) introduces caldesmon, a protein coded for by the gene CALD1, and notes that two different forms of this protein exist. Part (d)(i) asks the student to define a gene mutation. This is a definition-style "explain what is meant by" question, which the mark scheme typically marks as two points: the cause (the DNA change) and the consequence (an altered product).

Approach

Build the definition in two parts:

1. State what changes at the molecular level.
2. State the consequence for the polypeptide.

Step-by-Step Reasoning

• A gene mutation is a change in the sequence of bases (nucleotides / base pairs) along the DNA molecule that makes up the gene. The mark scheme accepts "change in the DNA base sequence" or "change in the DNA nucleotide sequence". It explicitly does NOT credit "change in RNA base sequence" — the change must be in DNA.
• Because the base sequence of the gene determines the order of amino acids in the polypeptide, a change in the base sequence can result in an altered/different polypeptide. The mark scheme accepts "protein / amino acid sequence / primary structure" as alternatives to "polypeptide".

The mark scheme does NOT credit naming types of mutation (substitution, deletion, insertion) — these are examples, not the definition.

Key Takeaways

• A mutation is at the DNA level, not at the RNA or protein level.
• It is a permanent change in base sequence.
• The consequence (and the second mark) is that the polypeptide coded for may be different.

Common Mistakes

• Saying "a change in the bases" without specifying that they are in DNA.
• Saying "a change in amino acids" — this is the consequence, not the definition; the change must be in the DNA.
• Saying "a change in a chromosome" — too vague; the change is specifically in the base sequence.

Things to Be Careful About

• The two marking points must both appear: the change AND the consequence for the polypeptide.
• Use the precise CIE word "polypeptide" (or accepted equivalent "protein / amino acid sequence / primary structure").

Background Concept

When a protein-coding gene is transcribed, the RNA polymerase first produces a primary transcript (pre-mRNA). This transcript contains both:

• Exons — sequences that code for amino acids (the "expressed" sequences).
• Introns — non-coding sequences that lie between exons.

Before the mRNA leaves the nucleus, the introns are removed and the exons are joined together by a process called RNA splicing. A single gene can therefore give rise to different mature mRNAs in different cell types because different combinations of exons may be retained. This is called alternative splicing and is the molecular basis for producing multiple protein isoforms (such as H- and L-caldesmon) from a single gene.

Understanding the Question

The stem of part (d) tells us that caldesmon exists as two isoforms (H-caldesmon in smooth muscle, L-caldesmon in non-muscle cells) but that the gene mutation is NOT the cause. It also tells us that the difference arises from events occurring directly after transcription, involving the primary transcript. Part (d)(ii) asks the student to suggest how two different polypeptides can be made from the same primary transcript.

Approach

Recall that post-transcriptional processing in eukaryotes includes RNA splicing (intron removal and exon joining). Apply the concept of alternative splicing: the same primary transcript can be spliced differently in different cell types, giving different mature mRNAs.

Step-by-Step Reasoning

• The primary transcript contains exons (coding) and introns (non-coding).
• Splicing removes the introns and joins the exons.
• In smooth muscle cells, one combination of exons is joined (e.g. including the middle exons that code for the repeating sequence unique to H-caldesmon).
• In non-muscle cells, a different combination of exons is joined (e.g. the middle exons coding for the repeating sequence are excluded), giving L-caldesmon.
• The two mature mRNAs are different, so the two polypeptides translated from them are different.

The mark scheme accepts any combination of: "RNA splicing", "alternative splicing" (do not say "DNA splicing" — the splicing happens to the primary RNA transcript), "removal of introns", "exons joined differently / in a different combination", "(so the) mRNA formed is different". It explicitly rejects "different number of introns removed / some introns remain" — the number of introns is fixed; it is which exons are joined that varies. It also rejects "mutation occurring during splicing" — the stem has told us there is no gene mutation.

Key Takeaways

• Alternative splicing is the main mechanism by which one gene can produce several different protein isoforms.
• The cell type determines which exons are retained in the mature mRNA.
• The process occurs to the primary transcript, before translation — i.e., at the post-transcriptional stage.

Common Mistakes

• Saying "DNA splicing" — splicing happens to RNA, not DNA.
• Saying "mutations during splicing" — the question stem rules out mutations.
• Saying "different introns removed" — the introns are the same; what differs is which exons are joined.
• Confusing introns with exons (introns are removed, exons are kept and joined).

Things to Be Careful About

• The splicing is to RNA, not DNA.
• The different forms come from different combinations of exons, not different numbers of introns.

Background Concept

A protein's function depends on its three-dimensional shape (tertiary structure), which in turn is determined by the sequence of amino acids (primary structure) and the way the R-groups interact (hydrogen bonds, ionic bonds, disulfide bridges, hydrophobic interactions).

Caldesmon is described as a large protein with multiple binding sites that attach to other proteins. These binding sites are discrete regions of the polypeptide where the local 3D shape is complementary to the partner protein (analogous to an enzyme's active site being complementary to its substrate).

Understanding the Question

The stem tells us that the primary structures of H-caldesmon and L-caldesmon differ (H has a middle repeating sequence that L lacks). Yet both proteins act as regulatory proteins that bind to other proteins. The question asks how the two forms can still have similar functions despite this difference in primary structure.

Approach

Think about what determines function: it is the 3D shape of the binding sites. A change in primary structure does not abolish function unless that change affects the binding sites themselves.

Step-by-Step Reasoning

• The two forms differ only in the middle repeating sequence.
• The binding sites (where caldesmon attaches to its partner proteins) are located elsewhere in the molecule and have the same amino acid sequence in both forms.
• Because the binding-site amino acid sequence is unchanged, R-group interactions in those regions are unchanged, so the tertiary structure (3D shape) of the binding sites is preserved.
• Therefore both forms can still bind to the same partner proteins and carry out a similar regulatory role.

The mark scheme explicitly forbids referring to "active sites" (the language of enzymes) but allows "binding sites". It accepts the idea that "removed amino acids are not part of the binding site" or that "R-group interactions (still) the same/similar".

Key Takeaways

• Protein function depends on tertiary structure at binding/active sites, not on primary structure globally.
• A change in primary structure outside the binding site need not change function.
• This is why a single gene can produce several protein isoforms with similar (or partially different) functions.

Common Mistakes

• Saying "active site" — caldesmon does not catalyse a reaction; it has binding sites, not an active site.
• Saying "the primary structure is the same" — the question stem says they are different.
• Saying "they have the same function" — the question implies they may have somewhat different functions; the mark is for explaining how similar function is preserved.

Things to Be Careful About

• Use "binding sites" (plural is acceptable), not "active site".
• The explanation must locate the difference outside the binding site.

Background Concept

Eukaryotic cells contain many membrane-bound and non-membrane-bound organelles, each with a specific function. Three that are particularly easy to confuse are the nucleolus, the smooth endoplasmic reticulum (SER) and the centrioles — partly because each is small, often hidden in diagrams, or bears some resemblance to other structures.

• Nucleolus — a dense, roughly spherical body inside the nucleus. It is not a separate organelle surrounded by a membrane; it is a region of the nucleus where ribosomal RNA (rRNA) is transcribed and combined with proteins imported from the cytoplasm to assemble the large and small ribosomal subunits. These subunits are then exported through nuclear pores to the cytoplasm.
• Smooth endoplasmic reticulum (SER) — a network of flattened sacs and tubules continuous with the rough ER but lacking ribosomes on its cytoplasmic surface. It synthesises lipids (including triglycerides, phospholipids and steroids), metabolises carbohydrates and detoxifies some drugs and poisons. By contrast, rough ER is studded with ribosomes and is the site of synthesis of proteins destined for secretion or for membranes.
• Centrioles — a pair of short, hollow cylinders made of microtubules. The pair is arranged at right angles to each other (perpendicular) and sits in the centrosome near the nucleus. They organise the microtubules of the cytoskeleton and nucleate the spindle fibres that separate chromosomes during mitosis. Animal cells have centrioles; higher plant cells do not.

Understanding the Question

Part (a)(i) supplies a blank three-row table in which each row gives a function and asks for the cell structure that performs it. The instructions explicitly forbid abbreviations, so "ER" alone would not be accepted. The candidate must recall the full name of each organelle that performs the given function in a typical animal cell (here an alveolar macrophage).

Approach

Read each function and decide which organelle performs it, then write the full name of that organelle in the left-hand column.

Step-by-Step Reasoning

• "Manufactures ribosomal subunits from proteins and ribosomal RNA" → This is the definition of the nucleolus (the site of rRNA transcription and ribosomal subunit assembly, found inside the nucleus).
• "Synthesises triglycerides and other lipids" → This is the role of the smooth endoplasmic reticulum. Lipid synthesis is the textbook lipid-related function of the SER. The word "triglycerides" is the giveaway — triglycerides are lipids, and SER (not RER) makes them.
• "Organise microtubules of the cell cytoskeleton" → A pair of centrioles acts as the microtubule-organising centre (MTOC) in animal cells, nucleating the cytoskeletal microtubules and the mitotic spindle.

Key Takeaways

• Nucleolus ≠ nucleus. The nucleus stores DNA; the nucleolus makes ribosomes inside it.
• Smooth ER vs Rough ER is decided by the absence or presence of surface ribosomes, and hence by function (lipid synthesis vs protein synthesis).
• Centrioles always come as a pair of perpendicular cylinders — never a single centriole.

Common Mistakes

• Writing "nucleus" instead of "nucleolus" — confuses the container with its internal compartment.
• Writing "endoplasmic reticulum" without specifying "smooth" — loses the mark because RER also fits the description.
• Writing "centrosome" instead of "centrioles" — the centrosome is the surrounding matrix; the centrioles are the pair of cylinders inside it.
• Using abbreviations such as "ER" or omitting the word "pair" before centrioles.

Things to Be Careful About

The mark scheme rejects "endoplastic" (misspelling) and "centrosomes". Always spell out organelle names in full and use the precise term.

Background Concept

This part of the question tests whether a candidate can recognise and reproduce cellular ultrastructure on a diagram, not merely name it. Many organelles look superficially similar at the EM level, so drawings must include the distinguishing features listed in the mark scheme:

• The nucleolus is recognisable only because it is inside the nucleus and is a smaller dense body within it.
• The smooth ER is recognisable because it is a membrane-bound tubular or flattened sac without ribosomes (no dots on its cytoplasmic face). If ribosomes (dots) are drawn on the membrane, the structure looks like rough ER and the mark is lost.
• Centrioles are recognisable as two short cylinders perpendicular to each other. They must be in the cytoplasm, not inside the nucleus (which would suggest a nucleolus or chromatin), and they must not be parallel (which would simply look like a piece of ER).

Understanding the Question

Fig. 2.1 shows an alveolar macrophage with its nucleus, mitochondria, rough ER and the phagosome already drawn. The three structures from Table 2.1 (nucleolus, SER, pair of centrioles) are deliberately not shown. The candidate must add them to the figure and label them with leader lines.

Approach

Decide where on the figure each structure will be drawn so that it is in a sensible position and obviously distinct:

• Nucleolus → inside the existing nucleus.
• Smooth ER → somewhere in the cytoplasm, drawn as smooth curved or tubular sacs with no dots.
• Centrioles → in the cytoplasm near the nucleus, drawn as two short cylinders at right angles.

Draw each, then add a label connected by a clear leader line.

Step-by-Step Reasoning

• Nucleolus. Place a small circle inside the boundary of the nucleus. Label it "nucleolus" with a leader line.
• Smooth ER. Draw at least one curved/tubular membrane-bound sac in the cytoplasm. Do not put dots on it — that would make it look like rough ER and lose the mark. Label it "smooth endoplasmic reticulum" or "smooth ER".
• Centrioles. Draw two short hollow cylinders (or two pairs of short lines) at right angles to each other in the cytoplasm, not inside the nucleus. Label it "centrioles".

According to the mark scheme: "all three correct = 2 marks; one or two correct = 1 mark; all three drawn correctly but no labels = 1 mark". So a single attempt that labels nothing but draws all three correctly still earns 1 mark — but only labelling all three correctly (with at least some attempt at distinguishing detail) earns the full 2 marks.

Key Takeaways

• Drawings of organelles must include the features that distinguish them from similar-looking organelles.
• Labels need leader lines that end on the structure, not in empty space nearby.
• Recognising what isn't on a printed figure is part of the skill — read the stem to know what to add.

Common Mistakes

• Drawing the nucleolus outside the nucleus — the nucleolus is inside the nucleus; placing it in the cytoplasm is wrong.
• Putting dots on the smooth ER membrane so it resembles rough ER.
• Drawing centrioles as parallel lines or as a single structure rather than a perpendicular pair.
• Drawing centrioles inside the nucleus (rejected by the mark scheme).
• Forgetting the labels altogether.

Things to Be Careful About

• Use single lines or a simple shape for each structure; do not attempt to draw fine ultrastructural detail.
• Keep the drawings clearly distinguishable from each other and from the organelles already present in Fig. 2.1 (rough ER, mitochondria, nucleus).
• Add leader lines and labels for full marks.

Background Concept

Phagocytosis is a form of endocytosis in which specialised cells (phagocytes — macrophages and neutrophils) engulf solid particles such as bacteria, dead cells or debris. The process involves:

1. Detection / attraction — phagocytes are attracted to the site of infection by chemicals released by pathogens or by damaged host cells (chemotaxis). Receptors on the phagocyte cell surface membrane bind to antigens (often after opsonisation — coating of the pathogen by antibodies or complement proteins).
2. Engulfment — the phagocyte extends finger-like cytoplasmic projections called pseudopodia around the particle. The cell surface membrane surrounds the particle and fuses behind it.
3. Vesicle formation — the membrane pinches off internally to release a membrane-bound phagocytic vacuole (phagosome) inside the cytoplasm, enclosing the engulfed material.
4. Digestion — lysosomes fuse with the phagosome and release hydrolytic enzymes to destroy the contents.

Understanding the Question

The question asks the candidate to outline the sequence of events leading to the formation of the phagocytic vacuole after the macrophage has detected the bacteria. So the focus is on engulfment plus one earlier (recognition) or later (vesicle release) event. The two marks can be earned by: (1) describing endocytosis / engulfment (mandatory), and (2) describing a before- or after-event (chemotaxis, receptor binding, opsonisation, or membrane fusion).

Approach

State the engulfment step (pseudopodia / membrane surrounds / pinches off) for one mark, then add either a recognition event (chemotaxis / antigen–receptor binding) or the vesicle pinch-off event for the second mark.

Step-by-Step Reasoning

• Mark 1 — endocytosis. "Endocytosis occurs: the cell surface membrane (pseudopodia) surrounds / engulfs the bacterial cells." This is the core of phagocytosis.
• Mark 2 — a recognisable before- or after-event. Either before — chemotaxis / antigen binding to surface receptors / opsonisation by antibodies; or after — the surrounding membrane fuses and pinches off internally to release the phagocytic vacuole into the cytoplasm.

Either combination is acceptable; the mark scheme allows the second mark for a "before" or "after" event.

Key Takeaways

• Phagocytosis is endocytosis of a solid particle by a phagocyte.
• The cell surface membrane is the active component — it surrounds the particle and pinches off internally.
• Recognition (chemotaxis / antigen binding / opsonisation) precedes engulfment; lysosomal fusion follows it.

Common Mistakes

• Describing only the recognition event and not the engulfment — loses the core "endocytosis" mark.
• Calling pseudopodia "cilia" or "flagella" — pseudopodia are the cytoplasmic projections of phagocytes, not motile organelles.
• Saying the bacteria enter the cell — it is the macrophage that actively engulfs them.
• Describing events that happen after the phagosome has formed (e.g. lysosome fusion, digestion) — these are outside the scope of this question.

Things to Be Careful About

• "Pseudopodia surround / form a ring around / engulf" the bacteria is the credit-worthy phrasing.
• The mark scheme ignores binding "to the macrophage" alone (it requires binding to a receptor or "membrane" of the macrophage).

Background Concept

After a phagocytic vacuole (phagosome) forms, lysosomes fuse with it. Lysosomes are small, spherical organelles bounded by a single membrane that contain hydrolytic (digestive) enzymes such as lysozyme, proteases, lipases and nucleases, active at the low internal pH (~pH 4.5–5.0). When a lysosome fuses with a phagosome, the combined structure is called a phagolysosome, and the hydrolytic enzymes digest the engulfed material — in this case the M. tuberculosis bacterial cells.

Understanding the Question

The question asks for the name of the cell structures that fuse with the phagocytic vacuole and release hydrolytic enzymes. This is a one-mark recall item.

Approach

Recall that the organelle responsible for intracellular digestion of engulfed material is the lysosome.

Step-by-Step Reasoning

• Lysosomes contain hydrolytic enzymes.
• Lysosomes fuse with phagosomes to form phagolysosomes, in which the engulfed bacteria are digested.

The mark scheme accepts "lysosomes", "lysosome" or "lysosomal vesicles".

Key Takeaways

• Lysosomes are the digestive organelles of the cell.
• They fuse with phagosomes to form phagolysosomes, enabling destruction of pathogens such as M. tuberculosis.

Common Mistakes

• Writing "phagosome" or "vacuole" — these are the structures formed by engulfment, not the ones that fuse with it.
• Writing "Golgi apparatus" — the Golgi packages hydrolytic enzymes into lysosomes but does not itself fuse with the phagosome.
• Writing "ribosome" or "mitochondrion" — these have unrelated functions.

Things to Be Careful About

The mark scheme accepts "lysosomal vesicles" as well, so if the term "lysosome" is not to hand, "lysosomal vesicle" is acceptable.

Background Concept

Gene expression flows from DNA → mRNA → protein. The quantity of an enzyme a cell produces is determined by how much mRNA for that enzyme is available to be translated by ribosomes. Cells can therefore control which enzymes are present (and in what quantity) by regulating transcription of the relevant gene — only producing the mRNA (and hence the enzyme) when the substrate of that enzyme is available.

In this question, the gene NR codes for the enzyme nitrate reductase, which catalyses the first step of leaf nitrate metabolism (NO₃⁻ → NO₂⁻). The supply of nitrite is the rate-limiting step that determines how fast nitrogen can be fed into amino acid synthesis in the mesophyll.

Understanding the Question

A single mark is available. The question asks for one benefit to the leaf cell of having more NR mRNA after nitrate is added. Focus on what the EXTRA mRNA allows the cell to do, not on how the mRNA is made.

Approach

Connect 'more mRNA' → 'more translation' → 'more enzyme' → 'faster reduction of nitrate to nitrite' → 'more nitrogen for amino acid synthesis'. One focused sentence is enough; pick the most biologically meaningful link.

Step-by-Step Reasoning

• The mark scheme credits several equivalent points: increased rate of translation of the enzyme, increased enzyme synthesis/concentration, the idea that the enzyme is only produced when needed (metabolic efficiency), and downstream effects (more nitrite, more amino acid synthesis).
• A complete answer would link the mRNA to the enzyme to the metabolic outcome. The strongest single point is: 'More nitrate reductase enzyme is synthesised, so the rate of reduction of nitrate to nitrite increases.'
• The question is worth only 1 mark, so do not write an essay — one focused sentence is sufficient.

Key Takeaways

• More mRNA for a gene generally means more of the corresponding protein.
• Cells regulate gene expression to make enzymes only when their substrate is available — an example of metabolic efficiency (inducible enzyme systems).

Common Mistakes

• Describing what happens at transcription ('more transcription occurs') rather than at translation — the benefit of mRNA is realised at translation.
• Writing about general growth or protein synthesis rather than nitrate reductase specifically.
• Failing to link the benefit to the metabolic outcome (nitrogen supply for amino acids).

Things to Be Careful About

• The mRNA increase is the CAUSE; the benefit is the CONSEQUENCE. Make sure the answer states the benefit (what the cell gains), not the cause (mRNA is transcribed).

Background Concept

Colorimetry is a quantitative analytical technique. A colorimeter shines light of a specific wavelength (chosen to be maximally absorbed by the coloured product) through a sample and measures how much light is transmitted (or absorbed) by the solution. The more coloured the solution, the less light passes through and the higher the absorbance. A calibration curve, made from standard solutions of known concentration, lets you convert an absorbance reading into the actual concentration of the substance in an unknown sample.

In this experiment, nitrite reacts with the colourless reagent (Griess-type reagent) to produce a magenta (red-purple) colour. The intensity of this colour is proportional to the amount of nitrite present, which in turn reflects the activity of nitrate reductase (and hence the rate of nitrate metabolism).

Understanding the Question

Two marks are available. The command word 'suggest' means you should think about WHY a colorimeter is better than the alternative (visual judgement of colour) in this specific method. You are being asked to evaluate the technique, not just to describe it.

Approach

Think about the limitations of judging colour by eye, and how a colorimeter overcomes each one. The two most important advantages are: (1) the measurement is objective/numerical rather than subjective, and (2) the data can be used quantitatively (with a calibration curve) to find the actual concentration of nitrite.

Step-by-Step Reasoning

• Visual judgement: 'Is the solution magenta? How magenta? Very magenta? Slightly magenta?' — these judgements are subjective, vary between observers, and give only a yes/no or a rough 'more'/'less' answer.
• Colorimeter: gives a numerical absorbance (or percentage transmission) value. This is objective, repeatable between observers and instruments, and far more sensitive than the eye — small differences in colour intensity that would be invisible to a person become measurable differences in absorbance.
• Because absorbance is proportional to concentration (Beer–Lambert law), a calibration curve (absorbance vs. known nitrite concentration) lets the experimenter convert the absorbance reading into an actual concentration of nitrite, and therefore a true rate of nitrate metabolism. Without a colorimeter, only a relative comparison could be made.
• A colorimeter can also detect very low concentrations (faintly coloured samples) that the eye would miss entirely.

Key Takeaways

• Colorimetry converts a qualitative colour change into a quantitative, numerical measurement.
• The key advantages are objectivity (no observer bias), sensitivity (detects very small colour differences), and the ability to use a calibration curve to find actual concentrations.

Common Mistakes

• Vague answers like 'it is more accurate' or 'it gives better results' — the mark scheme wants the REASON it is more accurate (numerical values, not subjective judgement).
• Confusing 'transmission' and 'absorbance' — they are inversely related, and colorimeters can measure either.
• Forgetting to mention the calibration curve, which is the key to producing actual concentration values.
• Saying 'measures colour' rather than 'measures absorbance' — absorbance is the technical term.

Things to Be Careful About

• The mark scheme explicitly accepts 'improves accuracy' but better answers explain WHY (no subjective judgement, numerical values, calibration curve).
• A common CIE rule: a single statement such as 'gives a numerical value' is only worth 1 mark — you also need the consequence (it can be used to find concentration, OR it removes subjectivity).

Background Concept

Plants move assimilates (sugars, amino acids) from 'source' tissues (e.g. mesophyll, where they are made) to 'sink' tissues (e.g. roots, fruits, growing tips) through the phloem. To load sugars and amino acids into the phloem against a concentration gradient, plant cells use an elegant indirect active-transport mechanism: the proton pump.

The proton pump is a membrane protein (a $\text{H}^+$-ATPase) that uses the energy from ATP hydrolysis to pump protons ($\text{H}^+$) out of the cell against their concentration gradient. This builds up a high concentration of protons in the apoplast (cell wall space) and a negative membrane potential inside the cell. The combined electrochemical gradient (higher $\text{H}^+$ outside, lower $\text{H}^+$ inside, with a more negative potential inside) is then used to drive the uptake of other solutes by COTRANSPORT — protons diffuse back into the cell through a cotransporter, and the energy released pulls another molecule (e.g. sucrose or amino acid) in against its own gradient.

The same mechanism is used for both sucrose and amino acid loading into companion cells, which is why the question states that 'amino acids can move from the apoplast into the cytoplasm of a companion cell using the same transport mechanism that is used for sucrose transport'.

Understanding the Question

Four marks are available. The command words 'outline AND explain' mean: state each step of the mechanism in sequence (outline) AND give the reason why each step is needed (explain). A good answer will describe the cycle: pump out, gradient, diffuse back, cotransport — and explain the role of ATP and the concentration gradient.

Approach

Visualise the process as a four-step cycle:

1. Protons are pumped out (active transport, ATP used).
2. A proton gradient builds up in the apoplast.
3. Protons diffuse back into the companion cell down the gradient (facilitated diffusion through a cotransporter).
4. As protons re-enter, amino acids are pulled in with them (cotransport) — this is how amino acids move against their own gradient.

Step-by-Step Reasoning

Step 1 — Active extrusion of protons:

• The companion cell's plasma membrane contains a proton pump (a $\text{H}^+$-ATPase) that hydrolyses ATP and uses the released energy to move $\text{H}^+$ from the cytoplasm into the apoplast / cell wall.
• ATP is provided by the many mitochondria in companion cells (aerobic respiration).
  Step 2 — A proton gradient is established:
• This pumping creates a high $\text{H}^+$ concentration (and lower pH) in the apoplast, and a relatively low $\text{H}^+$ concentration inside the cell. The membrane potential is also more negative inside.
• This is an electrochemical gradient — both a concentration gradient AND a voltage gradient driving $\text{H}^+$ back into the cell.
  Step 3 — Protons return by facilitated diffusion:
• $\text{H}^+$ ions diffuse back into the cytoplasm of the companion cell, down their electrochemical gradient, through a specific transport protein (a cotransporter / symporter).
• This is facilitated diffusion (no ATP needed for the proton movement itself, but the energy stored in the gradient comes from the ATP used in step 1).
  Step 4 — Cotransport of amino acids:
• The same cotransporter has a binding site for an amino acid. As the proton moves into the cell, the protein undergoes a conformational change that pulls an amino acid molecule in with it.
• The energy released by the proton moving down its gradient is used to move the amino acid AGAINST its own concentration gradient — into the cell.
• The result is that amino acids accumulate in the companion cell at a higher concentration than in the apoplast.

Key Takeaways

• The proton pump creates a proton gradient using ATP (primary active transport).
• The energy stored in this gradient is then used to drive the uptake of other solutes (cotransport — a form of secondary active transport).
• Cotransport explains how plant cells can load large quantities of sugars and amino acids into the phloem against steep concentration gradients.

Common Mistakes

• Saying 'amino acids are actively transported' without explaining the mechanism — active transport is the OUT pump, not the cotransport step.
• Confusing the direction of proton movement — protons go OUT first, then BACK IN. Many candidates describe only one direction.
• Stating that amino acids 'use' the proton gradient without saying HOW (cotransport, against their own gradient).
• Writing 'diffusion' alone for the inward movement of protons — the mark scheme requires 'facilitated diffusion' or reference to a transport protein.

Things to Be Careful About

• A maximum of 3 marks is awarded if there is no mention of amino acids — every step must be tied to what happens to the amino acid.
• 'Protons', 'hydrogen ions' and '$\text{H}^+$' are all accepted — pick whichever you find easiest to write.
• The question asks for the sequence of events, so use clear, ordered steps and time-marker words like 'first', 'this creates', 'then', 'as a result'.

Background Concept

Plant cells are connected to one another by plasmodesmata — narrow cytoplasmic channels that pass through the cell walls, linking the cytoplasm of adjacent cells into a continuous network called the symplast. Small molecules (and even some larger ones) can move from cell to cell through plasmodesmata without having to cross a plasma membrane.

Companion cells and phloem sieve tube elements are intimately connected by many plasmodesmata, which is essential because the sieve tube element has lost most of its organelles (including the nucleus) during development and depends on the companion cell to keep it alive and to load assimilates.

Understanding the Question

One mark is available. The question asks why amino acid transporters are NOT needed at the companion cell → sieve tube interface. The context (amino acids moving from mesophyll to phloem) is provided by the parent stem.

Approach

Think about what alternative route exists between the two cells that does not require crossing a membrane. The answer is: plasmodesmata, which form a symplastic (cytoplasm-to-cytoplasm) bridge.

Step-by-Step Reasoning

• The companion cell and the phloem sieve tube element are connected by plasmodesmata, which are direct cytoplasmic channels between the two cells.
• Through these plasmodesmata, amino acids can move by simple diffusion from the companion cell (where they are at high concentration after the cotransport step in c.i) into the sieve tube element (where they are at lower concentration).
• Because no membrane is crossed, no transport protein is required — the amino acids simply diffuse down their concentration gradient through the plasmodesmata.

The mark scheme also accepts these alternative one-mark answers:

• 'Movement is down the concentration/diffusion gradient.'
• 'Because there is a higher concentration in the companion cell' (or the opposite stated for the sieve tube sap).

Key Takeaways

• Plasmodesmata are the cytoplasmic bridges that allow direct cell-to-cell movement in plants.
• They make membrane transporters unnecessary between cells that are symplastically connected.
• This is one of the key structural features that makes the phloem loading system so efficient: the active loading step (proton pump + cotransport) occurs across the companion cell membrane, and the short symplastic step from companion cell to sieve tube is passive.

Common Mistakes

• Suggesting that no transport is needed because 'the membrane is permeable' — this is wrong; the plasma membrane is selectively permeable, but the route here is not across the membrane at all (it is through plasmodesmata).
• Confusing the apoplast–symplast distinction; the question is about the symplastic route.
• Saying 'because the cells are joined' without naming plasmodesmata or the mechanism (diffusion down a gradient).

Things to Be Careful About

• Either naming plasmodesmata OR stating that movement is down the concentration gradient is sufficient for the mark — but combining both is the strongest answer.

Background Concept

The heart is supplied by two coronary arteries that arise from the aorta just above the aortic valve:

• The left coronary artery runs down the left side of the heart (appearing on the right of an anterior view).
• The right coronary artery runs down the right side of the heart (appearing on the left of an anterior view).

When a coronary artery is narrowed by atherosclerotic plaque, blood flow to the heart muscle is restricted. Surgeons use a healthy vessel (a 'graft') to bypass the blocked segment, restoring flow to the tissue beyond.

Understanding the Question

Fig. 6.1 shows the heart from the front (anterior view) with a double bypass graft in place. The candidate must mark a cross on the part of the right coronary artery that the graft has bypassed.

A critical orientational point: in an anterior view of the heart, the heart's own right side appears on the LEFT of the page as the viewer sees it (mirror-image view). So the right coronary artery is on the left side of the figure.

Approach

1. Locate the right coronary artery on Fig. 6.1 — it runs along the right atrioventricular groove, which appears on the left side of the diagram.
2. Identify the junction where the bypass graft joins onto the coronary artery (the distal anastomosis).
3. Place the X above (proximal to) this junction, on the coronary artery itself, in the region that the graft is bypassing.

Step-by-Step Reasoning

• The right coronary artery is the vessel descending along the AV groove on the anatomical right of the heart.
• In Fig. 6.1, the right coronary artery is the vessel on the left-hand side of the diagram.
• The bypass graft (the great saphenous vein, shown running up from the lower part of the figure) joins onto the right coronary artery at a junction.
• The X should be placed on the right coronary artery between its origin (at the aorta) and that junction — that is the diseased segment that has been bypassed.
• It must not be placed at the actual junction itself, because that is the surgical join, not the diseased segment.

Key Takeaways

• Anterior heart view = mirror image: anatomical right appears on viewer's left.
• A bypass graft diverts blood around the diseased segment, which lies between the aorta and the distal anastomosis.
• Anatomical identification of vessels on a heart diagram is a recurring Paper 2 skill.

Common Mistakes

• Marking the X on the right side of the figure (which is the left coronary artery, not the right).
• Placing the X at the junction between the graft and the native artery (the mark scheme explicitly rejects this).
• Placing the X on the graft itself rather than on the diseased native artery.
• Confusing the great saphenous vein (a low-pressure vessel used as a graft) with the right coronary artery.

Things to Be Careful About

The mark scheme is strict: 'X is positioned above the join on the coronary artery on the left-hand side of Fig. 6.1 / right heart' earns the mark. Wrong vessel, wrong side, or placement on the junction all fail.

Background Concept

Blood vessels are classified by the structure of their walls, which reflects the pressure they normally experience:

• Arteries: thick tunica media rich in smooth muscle and elastic fibres, plus a tough outer tunica externa (adventitia) of collagen. This lets them withstand the high, pulsatile pressure generated by ventricular systole and recoil during diastole to maintain flow.
• Veins: thin walls with much less smooth muscle and elastic tissue, because venous blood is at low pressure. Forward flow relies on skeletal muscle contraction and one-way valves rather than arterial recoil.

When a vein (the great saphenous vein) is grafted into the coronary circulation, it is suddenly exposed to arterial pressures. Its wall must adapt to avoid bursting.

Understanding the Question

This part asks the candidate to suggest AND explain why the wall of the great saphenous vein must thicken once it is functioning as a bypass graft in the arterial system. Two marks are awarded — one for the functional reason (higher pressure) and one for the structural adaptation (more muscle/elastic tissue).

Approach

Make the structure–function link explicit: the vessel must handle higher pressure → therefore needs more structural tissue (smooth muscle / elastic fibres / thicker wall layers).

Step-by-Step Reasoning

1. Functional demand: in its native position the great saphenous vein carries low-pressure blood back to the heart. Once it is connected to the coronary artery it is exposed to high, pulsatile arterial pressure. A normal thin venous wall would be likely to burst.
2. Structural adaptation: in response, the wall thickens. Specifically, the tunica media gains more smooth muscle and elastic fibres (giving it strength to withstand pressure and recoil to push blood onward), and the tunica externa gains more collagen. With these reinforcements the grafted vessel can survive long term in the arterial circulation.
3. Linking the two: a thicker wall + more elastic fibres = a vessel that can withstand higher pressure without bursting AND that can recoil to maintain the rapid, pulsatile flow required in the coronary circulation.

Key Takeaways

• Veins and arteries are structurally different because of the different pressures they carry.
• When a vessel is moved from a low-pressure to a high-pressure environment, its wall remodels (smooth muscle / elastic tissue proliferate) to handle the new load.
• This is a classic structure–function application in a clinical context.

Common Mistakes

• Saying the wall thickens 'because more blood flows through it' — pressure, not volume, is the issue.
• Saying the wall thickens 'to push blood back to the heart' — irrelevant in a grafted arterial position.
• Saying the vein would 'collapse' if it did not thicken — the mark scheme rejects this; the real risk is bursting under high pressure.
• Saying 'more blood' without specifying 'higher pressure'.
• Naming only one wall layer without linking to the functional reason.

Things to Be Careful About

The mark scheme lists two clearly distinguishable marking routes (pressure-resistance AND vessel durability) and explicitly accepts the structural detail of more smooth muscle / elastic fibres / collagen. At least one mark must combine the higher-pressure observation with the consequence (bursting or reduced durability). The other mark must name the wall layer and tissue type that increase.