Biology 9700/24 — May/June 2025

Background Concept

The arrangement of vascular tissues in a transverse section (TS) is a fundamental diagnostic feature distinguishing roots from stems:

• Root TS: xylem and phloem are found in the centre of the root, not at the edges. The xylem typically forms a star-shaped (or cross-shaped) mass in the very centre, with separate strands of phloem lying between the arms (points) of the xylem star. This arrangement is described as a radial vascular bundle, and the whole vascular cylinder is called the stele.
• Stem TS: vascular bundles are arranged in a ring near the edge of the section, with xylem on the inside of each bundle and phloem on the outside (a collateral arrangement). The mark scheme rejects any answer showing vascular bundles around the edge for a root.

Understanding the Question

The question asks for a diagram showing the distribution of vascular tissue in a TS of a root in the zone of differentiation. You need to draw the typical root vascular pattern: a central xylem star with phloem between the arms.

Approach

Think of the classic textbook image of a dicotyledonous root TS: a roughly circular outline, with a star-shaped (or X-shaped) xylem in the middle and small phloem strands wedged between the points of the star. This is not a stem — do not put the bundles near the edge.

Step-by-Step Reasoning

1. Draw a roughly circular outline representing the TS of the root (the epidermis/cortex can be omitted or simply suggested).
2. In the centre, draw the xylem as a star or cross (typically with 4–6 arms radiating outwards).
3. Between each pair of arms of the xylem star, draw a phloem strand.
4. Make sure the whole vascular tissue is shown centrally within the root, not at the periphery.

The mark scheme awards one mark for the correct shape and relative position of xylem and phloem, and one mark for the whole vascular cylinder being shown centrally.

Key Takeaways

• Root vascular tissue is central, in a radial arrangement.
• Xylem forms a star; phloem lies between the arms of the star.
• This is the opposite of a stem, where vascular bundles are at the periphery.

Common Mistakes

• Drawing vascular bundles around the edge of the root (this is the stem arrangement) — rejected.
• Drawing phloem on the outside and xylem on the inside of a single ring (still a stem-like arrangement) — rejected.
• Drawing the xylem as a complete circle with phloem just outside it — this is not the root arrangement and is not credited.

Things to Be Careful About

This is a drawing question: the candidate must show the arrangement, not just describe it in words. A plan diagram is appropriate (no individual cells needed) — clear lines, correct proportions, and correct labels are what score marks.

Background Concept

The pressure-flow hypothesis (originally proposed by Ernst Münch, 1930) explains how organic assimilates such as sucrose move through the phloem from a source (where they are made or stored — here, the starch-storing roots after conversion to sucrose) to a sink (where they are used or stored — here, the growing areas of the stem).

The mechanism depends on four interlinked ideas:

1. Sucrose loading at the source (often via a proton-sucrose co-transporter, but the AS syllabus does not require the molecular detail). Sucrose is actively loaded into the sieve tubes, raising the solute concentration inside them.
2. Lower water potential at the source. More solute means a more negative $\Psi$, so water enters the sieve tubes by osmosis from the surrounding xylem water.
3. High hydrostatic (turgor) pressure at the source. Incoming water raises the volume and hence the pressure inside the sieve tube at the source end.
4. Sucrose unloading at the sink. At a growing region, sucrose is removed from the sieve tubes (used in respiration or converted to other compounds). Water also leaves osmotically. Pressure at the sink end is therefore low.
5. Bulk flow / mass flow down the pressure gradient. Because pressure is high at the source and low at the sink, the phloem sap moves en masse from source to sink through the sieve tubes.

Understanding the Question

The question asks you to explain how phloem sap is moved within the sieve tubes from the root storage areas (source) to the growing areas of the stem (sink) by mass flow. The command word "explain" means each point must include the reason (the link between cause and effect), not just a list of events. The stem already tells you that sucrose loading happens and that movement is by mass flow — your job is to explain the mechanism that drives the mass flow.

Approach

Build the chain of cause-and-effect in the order in which the events happen: sucrose loading → water potential change → osmotic water entry → volume/pressure increase at source → unloading at sink → pressure decrease at sink → mass flow down the pressure gradient.

Step-by-Step Reasoning

Point 1 — Loading lowers the water potential.
When sucrose (and other assimilates) is loaded into the sieve tube at the source, the solute concentration in the sieve tube rises, so the water potential inside the sieve tube falls (becomes more negative).

Point 2 — Water enters the sieve tube by osmosis.
Because the sieve tube now has a lower (more negative) water potential than the surrounding cells/xylem water, water moves down the water potential gradient into the sieve tube by osmosis across the partially permeable sieve tube membrane.

Point 3 — Volume increases inside the sieve tube.
The entry of water increases the volume of the fluid inside the sieve tube at the source end.

Point 4 — Hydrostatic pressure rises at the source.
The increased volume cannot be accommodated indefinitely (the sieve tube walls are rigid and the plant cells around resist expansion), so a high hydrostatic (turgor) pressure builds up inside the sieve tube at the source.

Point 5 — Unloading at the sink lowers the pressure there.
At the sink (the growing areas of the stem), sucrose is unloaded and used (in respiration or for making new cell materials). As solute is removed, water also leaves. The hydrostatic pressure at the sink end of the sieve tube is therefore lower than at the source end.

Point 6 — Mass flow down the pressure gradient.
Because pressure is high at the source and low at the sink, the phloem sap (water + dissolved sucrose) moves in bulk from source to sink — this is mass flow, driven by the pressure gradient.

Key Takeaways

• The driving force for phloem transport is a pressure gradient (not a water potential gradient in the sieve tube itself, although a water potential gradient is the mechanism by which that pressure gradient is created).
• Osmosis sets up the pressure; the pressure gradient then drives mass flow.
• Source = where assimilates are loaded (roots in this case); sink = where they are unloaded (growing stem in this case).
• The sieve tubes must be alive but have reduced cytoplasm to minimise resistance to flow; the mature sieve tube element has lost its nucleus and most organelles, and the end walls are perforated to form sieve plates.

Common Mistakes

• Saying "water moves down a concentration gradient" — rejected; the correct term is water potential gradient.
• Saying "water potential gradient decreases" — ignored; the water potential itself decreases, not the gradient.
• Describing mass flow as "diffusion" — diffusion is too slow to account for the rapid transport observed in phloem; mass flow is bulk movement driven by a pressure difference.
• Implying that sucrose itself moves by osmosis — only water moves by osmosis; sucrose is loaded/unloaded by other mechanisms.
• Saying pressure "creates" or "forms" (instead of "builds up" or "increases") — the mark scheme ignores "creates" / "forms" and accepts "builds up" or "high(er)".

Things to Be Careful About

• Hydrostatic pressure must be mentioned by name at least once (mark scheme: "must have hydrostatic once").
• "Mass flow" or "bulk flow" must be linked to the pressure gradient (from high to low pressure).
• The unloading step must explicitly lower the pressure at the sink — not just "sucrose is used".
• The word "down a water potential gradient" is preferred over "from high to low water potential gradient" (the latter is ignored).
• Ecf applies: if a candidate gives a wrong direction but consistent reasoning, marks can still be awarded for the consequence steps.

Background Concept

Enzymes are biological catalysts and are classified by where they act:

• Intracellular enzymes are made and used inside the same cell — e.g. the enzymes of glycolysis, the Krebs cycle or DNA replication.
• Extracellular enzymes are synthesised inside a cell but secreted across the plasma membrane to act on substrate outside the cell — e.g. digestive enzymes such as pancreatic lipase, amylase and trypsin, which are released into the small intestine.

Understanding the Question

The stem tells us that pancreatic lipase is made in the pancreas but has its effect in the small intestine — i.e. it leaves the cell that made it. The question asks for the single term that describes an enzyme secreted out of the cell to do its work.

Approach

This is a one-word recall item. The accepted CIE term is extracellular. The mark scheme explicitly rejects exocellular, so the standard word must be used.

Step-by-Step Reasoning

1. The enzyme works in the small intestine, outside the pancreatic cells that secreted it.
2. Enzymes that are secreted from a cell and act externally are called extracellular enzymes.
3. The single required answer is extracellular.

Key Takeaways

• Extracellular = synthesised inside a cell, secreted, and active outside the cell.
• Intracellular = made and used inside the same cell.

Common Mistakes

• Writing exocellular or exoenzyme — neither is the accepted CIE term.
• Confusing with endoplasmic (a subcellular organelle, not an enzyme class).

Things to Be Careful About

• One mark only — give the single word; no explanation is needed or credited.

Background Concept

The Michaelis–Menten model describes how the rate of an enzyme-catalysed reaction depends on substrate concentration. As [S] rises, the rate rises and approaches a maximum value, $V_{\text{max}}$, at which all active sites are saturated. The substrate concentration at which the rate is exactly half of $V_{\text{max}}$ is defined as the Michaelis–Menten constant, $K_m$.

$K_m$ is often used as an inverse measure of the apparent affinity of the enzyme for its substrate: a low $K_m$ means the enzyme reaches half its maximum rate at a low substrate concentration (high affinity), while a high $K_m$ means a higher substrate concentration is required (lower affinity).

Understanding the Question

The question gives a specific $K_m$ value (with units) for a bacterial lipase acting on olive oil at $37^\circ\text{C}$. The candidate must explain what this number actually means in a biological context, not just recite the textbook definition.

The mark scheme requires two ideas:

1. Identification of the variable: substrate (olive oil) concentration.
2. The condition under which this concentration applies: the rate of reaction is half the maximum rate, i.e. $\frac{1}{2}V_{\text{max}}$.
3. The context (temperature of $37^\circ\text{C}$) is part of the experimental setup.

Approach

1. Identify the variable from the units ($\text{mmol dm}^{-3}$ are concentration units): the substrate (olive oil) concentration.
2. Identify the operational meaning: at this concentration, the enzyme works at half its maximum rate, $\frac{1}{2}V_{\text{max}}$.
3. Reference the experimental temperature: $37^\circ\text{C}$ (as stated in the stem).

Step-by-Step Reasoning

• The $K_m$ value is given in $\text{mmol dm}^{-3}$, which are units of concentration. This tells us $K_m$ is a concentration of the substrate.
• The value is the concentration at which the rate of reaction = $\frac{1}{2}V_{\text{max}}$ — i.e. half the maximum rate achievable when substrate is saturating.
• The investigation was carried out at $37^\circ\text{C}$, so this $K_m$ value applies at that temperature only.
• The mark scheme rejects a statement that $K_m$ is half the maximum rate — $K_m$ is a concentration, not a rate.

Key Takeaways

• $K_m$ has units of concentration (e.g. $\text{mmol dm}^{-3}$), not units of rate.
• $K_m$ is the substrate concentration at which rate = $\frac{1}{2}V_{\text{max}}$.
• $K_m$ is a property of a particular enzyme-substrate pair at a particular temperature and pH.

Common Mistakes

• Stating "$K_m$ is half the maximum rate" — this is a rate, not a concentration; the mark scheme rejects it.
• Omitting the substrate — saying only "the concentration at which rate is half Vmax" without specifying that it is the substrate (olive oil) concentration.
• Forgetting the experimental temperature context.
• Confusing $K_m$ with $V_{\text{max}}$ (which is a maximum rate, not a concentration).

Things to Be Careful About

• Two separate ideas are required: (1) the quantity being measured (substrate concentration), and (2) the rate condition ($\frac{1}{2}V_{\text{max}}$). A single sentence that omits either half scores at most one mark.

Background Concept

Magnification describes how much larger (or smaller) an image is than the actual specimen. The relationship is:

Rearranged to find the actual size of a specimen from an electron micrograph:

The image size must always be measured in the same units as the desired answer (or converted to those units). For a chromosome typically measured in micrometres ($\text{µm}$), the image length taken from the micrograph (in mm) must be converted to µm by multiplying by 1000 before dividing by the magnification. Alternatively, the actual size can be expressed in mm by dividing the image size (in mm) by the magnification, and then converted to µm.

Understanding the Question

The question asks the candidate to find the real (actual) length of the chromosome highlighted by line X–Y in Fig. 3.1, a scanning electron micrograph at a stated magnification of $\times 8625$. The image provides the magnified length, the magnification is printed on the micrograph, and the candidate must use these to back-calculate the true length of one chromatid in µm to the nearest $0.1\ \text{µm}$.

Approach

1. Measure the length of line X–Y on the printed micrograph in millimetres (this depends on the candidate's measurement of the printed page).
2. Convert mm into µm so units match the required answer.
3. Divide the converted image size by the magnification stated on the micrograph ($8625$).
4. Round to the nearest $0.1\ \text{µm}$.

Step-by-Step Reasoning

• A representative measurement of X–Y on the printed micrograph is $45\ \text{mm}$. Converting to µm: $45\ \text{mm} \times 1000 = 45\ 000\ \text{µm}$.
• The magnification is $\times 8625$ (printed on the figure).
• Substituting into the rearranged formula:

• Rounded to the nearest $0.1\ \text{µm}$, this is $5.2\ \text{µm}$.
• The mark scheme accepts answers from $5.0$ to $5.4\ \text{µm}$ to allow for small differences in measurement of the printed line X–Y.

Key Takeaways

• The magnification formula and its rearrangement are essential tools in any microscopy-based question.
• Unit conversion (mm ↔ µm) is a common source of error and must always be shown.
• The number of significant figures / decimal places in the final answer must match what the question requests.

Common Mistakes

• Forgetting to convert mm to µm before dividing by magnification — this would give a tiny answer in the wrong units.
• Multiplying instead of dividing (using the image size $\times$ magnification), which is a common error when the candidate is unsure which way round the formula works.
• Not writing the formula explicitly, even though the question demands it.
• Quoting an unrounded answer (e.g. $5.22\ \text{µm}$) when one decimal place is requested, or rounding to the wrong number of decimal places.

Things to Be Careful About

• The line X–Y is along a single chromatid, not the full chromosome. The candidate must measure the line as drawn, not the whole chromosome.
• Always write the formula before the substitution — the mark scheme awards a mark for the formula alone.
• Be consistent with units: if the question asks for µm, the numerator must also be in µm.

Background Concept

Electron microscopes use a beam of electrons (rather than light) to image specimens, allowing much higher resolution than light microscopes. There are two main types:

• Transmission electron microscope (TEM): electrons pass through an ultra-thin section of the specimen. The image is 2D and shows internal structures as varying shades according to how many electrons pass through.
• Scanning electron microscope (SEM): electrons scan across the surface of the specimen, and secondary electrons emitted from the surface are detected. The image shows 3D surface topography (contours, bumps, texture) but no internal detail.

Understanding the Question

The candidate is shown a scanning electron micrograph and asked to identify one observable feature of Fig. 3.1 that tells us the microscope used was a SEM and not a TEM. The image is provided, so the answer must be something visible in the figure itself, not a general fact about SEMs.

Approach

Look at Fig. 3.1 and decide which characteristic of SEM (and not TEM) is apparent. The figure shows two chromosomes that look solid, with shadowed surfaces, depth, and visible texture on the outer surface of each chromatid.

Step-by-Step Reasoning

• Each chromosome in Fig. 3.1 has visible 3D shape: light and shadow on the chromatid surfaces give a strong impression of depth.
• The surface texture (small bumps, fuzzy projections from the chromosome surface) is visible — this is surface topography, exactly what SEM reveals.
• No internal structure is visible (no dark/light contrast showing the inside of the chromatid), which is consistent with SEM (a TEM would show internal detail in a thin section).
• Any one of these observations is sufficient to score the mark.

Key Takeaways

• SEM = 3D, surface, opaque, depth of field, no internal detail.
• TEM = 2D, internal, thin sections, dark/light contrast from electron transmission.
• When asked to identify the type of EM from an image, pick an observable visual feature — not an invisible property.

Common Mistakes

• Stating a fact about SEMs that is not actually visible in the figure (e.g. "it has a higher resolution than a light microscope") — this would not be evident from the image and would not score.
• Confusing the two types: describing a SEM feature as if it were a TEM feature (e.g. "you can see internal structure") — this is the opposite of the truth and scores zero.
• Vague wording such as "it looks more detailed" or "it looks better" — the mark scheme wants specific features (3D, surface contours, depth of field, no internal detail).

Things to Be Careful About

• The image is small, so the candidate must look carefully for the 3D / surface appearance. The fuzzy outline and the shading on the chromatids are the strongest visual clues.
• The mark scheme accepts several equivalent phrasings (e.g. "3D image", "good depth of field", "surface contours visible", "no internal detail"); any one of these scores the mark.

Background Concept

During the cell cycle, DNA is replicated during S phase (the synthesis phase) of interphase, before mitosis begins. The result of replication is that each chromosome now consists of two identical sister chromatids held together at the centromere. Each chromatid contains one DNA double helix (with associated histone proteins), and the two chromatids are genetically identical because they are copies of the same original DNA molecule.

This two-chromatid structure persists through prophase, metaphase and into anaphase of mitosis, when the sister chromatids finally separate and are pulled to opposite poles of the cell.

Understanding the Question

The candidate is shown a micrograph of two chromosomes, each clearly made of two chromatids joined at a centromere. The question asks for the observable feature in the figure that shows DNA replication has already taken place. The key is to link what is visible (two chromatids) to the event that must have produced it (S phase of interphase).

Approach

Identify the most direct piece of evidence visible in the figure: the presence of two sister chromatids per chromosome. Connect this to the event that produces sister chromatids — DNA replication during S phase.

Step-by-Step Reasoning

• A unreplicated chromosome (e.g. in G1 phase) consists of a single DNA molecule and appears as a single chromatid under the microscope.
• In Fig. 3.1, each chromosome clearly shows two sister chromatids joined at a centromere.
• Sister chromatids are formed only when the original DNA molecule is copied, which happens in S phase.
• Therefore, the two-chromatid appearance is direct evidence that DNA replication has already occurred.
• Alternative acceptable answers (from the mark scheme) are: a centromere is visible in each chromosome; or the cell is at late prophase / prometaphase / metaphase — all stages that occur after S phase has been completed.

Key Takeaways

• The two-chromatid structure of a chromosome is the visible signature of completed DNA replication.
• S phase (synthesis) occurs in interphase, before any of the visible stages of mitosis.
• Linking visible structure to an invisible molecular event is a key skill in cell biology questions.

Common Mistakes

• Saying "two DNA molecules are present" without mentioning that they are the result of replication. The mark scheme explicitly ignores "two identical DNA molecules" as a standalone answer — the candidate should refer to the visible chromatids.
• Confusing the cell cycle stage: the candidate may say "the cell is in anaphase" or "the chromosomes are separating" — but in Fig. 3.1 the chromatids are still joined at the centromere, so the cell is in prophase / metaphase, not anaphase.
• Vague answers such as "it has chromosomes" or "the nucleus has divided" — these do not specifically indicate that DNA replication has occurred.

Things to Be Careful About

• Use the term sister chromatids (the two identical copies joined at the centromere), not "two chromosomes" — chromosomes and chromatids are distinct.
• DNA replication is the copying of the DNA; the appearance of two chromatids is the consequence of that copying.
• The centromere is a feature of an already-replicated chromosome; its presence is also accepted by the mark scheme as evidence.

Background Concept

Plasmodium is the genus of single-celled parasites responsible for malaria in humans. Although Plasmodium is eukaryotic (it has a true nucleus and membrane-bound organelles such as mitochondria and an endoplasmic reticulum), it does NOT belong to the kingdoms Animalia, Plantae, Fungi, Prokaryotae or Virus. The kingdom to which it is assigned at AS Biology is the Protoctista (sometimes written Protoctista or Protista) — the group containing mostly single-celled eukaryotes that are not fungi, animals or plants (e.g. Amoeba, Paramecium, Plasmodium and Trypanosoma). Other important diseases caused by protoctists include sleeping sickness (Trypanosoma) and amoebic dysentery (Entamoeba).

Understanding the Question

The command word is "state" — give a single, factual, mark-scheme term. The question asks for the type (i.e. the kind of organism) that causes malaria, not a description of what it does.

Approach

Recognise that Plasmodium is not a bacterium (so not Prokaryotae), not a fungus, not an animal in the multicellular sense, and not a virus. Recall its kingdom and give the single word/term expected.

Step-by-Step Reasoning

• "Type of organism" at AS Biology level means the kingdom (or equivalent grouping) of the pathogen.
• Plasmodium is a single-celled eukaryote, so it is neither prokaryote nor virus.
• The kingdom is Protoctista (also acceptable: Protoctista, with the capital P, or Protoctistae). The mark scheme explicitly accepts "Protoctist" and "Protoctista".
• The mark scheme also IGNORES species names (so writing Plasmodium is not enough) and IGNORES "eukaryote" alone (too vague — a protoctist IS a eukaryote but the question asks for the type of organism, i.e. the kingdom).

Key Takeaways

• Malaria is caused by protoctists of the genus Plasmodium (mainly P. falciparum, P. vivax, P. ovale, P. malariae and P. knowlesi).
• At CIE AS, write the kingdom, not the species or the cell type.

Common Mistakes

• Writing Plasmodium (the genus) — credit is for the type/kingdom, not the species.
• Writing "parasite" — true but too vague; it does not name the kingdom.
• Writing "protozoan" — this is a colloquial sub-group within protoctists; "protoctist" is the accepted term in this syllabus.
• Writing "animal" — Plasmodium belongs to Kingdom Protoctista, not Animalia.

Things to Be Careful About

Spell "protoctist" with a 'c' (proto-ctist), not "protist". The mark scheme accepts both spellings but "protoctist" is the form used in the CIE syllabus.

Background Concept

A healthy human red blood cell (erythrocyte) is a biconcave disc — a flattened disc with a central depression (dimple) on each face, and no nucleus. The biconcave shape gives a high surface area to volume ratio for gas exchange and allows the cell to deform and squeeze through capillaries. When Plasmodium merozoites invade and develop inside a red blood cell, the cell is changed both in shape and contents. The parasite digests haemoglobin and uses red blood cell proteins for its own growth, and the cell often swells as the parasite develops.

Understanding the Question

You are given a scanning electron micrograph (SEM) of a single red blood cell at ×23 000 magnification. The cell shows one large rounded bulge in its centre and a generally lumpy/irregular outline, plus some small debris particles from specimen preparation. The question asks for a single difference between the appearance of this cell and that of a healthy red blood cell.

Approach

1. Recall the appearance of a normal red blood cell (biconcave disc, central dimple, smooth round outline).
2. Compare that to the image: there is a clear rounded bulge in the centre and the cell is misshapen/lumpy overall.
3. State the single most obvious visible difference, in the precise wording the mark scheme credits.

Step-by-Step Reasoning

• Healthy red blood cell: smooth circular outline, biconcave (a clear depression/dimple on each side).
• Image of infected cell: the centre is bulged out (not dimpled); a large spherical bulge is obvious in the middle of the cell, and the cell is also irregular/misshapen.
• The mark scheme credits: "bulge in centre / swollen in centre / lost (full) biconcave shape" — and accepts "misshapen / shape not regular" as alternatives.
• A single, clear visible difference is all that is required for the 1 mark.

Key Takeaways

• Merozoite-infected red blood cells change shape: the central dimple is lost and replaced by a bulge or an irregular outline.
• SEM images show 3-D surface topography, so a bulge in the centre is the most obvious feature.

Common Mistakes

• Stating the difference is the presence of merozoites on the surface — they are not visibly sitting on the surface in this image; the visible feature is the bulge.
• Stating vague things like "the cell looks different" or "it is infected" — the mark requires a specific visible feature.
• Confusing the cellular bulge with the central dimple of a normal cell.

Things to Be Careful About

The question asks for one difference, so a single clear point is enough. The mark scheme rejects anything that is not directly visible in the image.

Background Concept

Haemoglobin is the oxygen-carrying protein that fills a healthy red blood cell. Plasmodium merozoites, once inside the red blood cell, cannot make all the amino acids they need — they obtain many of them by digesting the host's haemoglobin. The parasite digests haemoglobin inside a food vacuole, releasing free amino acids (and toxic haem, which the parasite sequesters as haemozoin). The released amino acids are then used by the parasite to make its own proteins and to grow. As a result, the concentration of haemoglobin in the infected red blood cell falls over the parasite's intra-erythrocytic cycle.

Understanding the Question

The stem says haemoglobin concentration decreases inside infected red blood cells. The command word is "suggest" — this means propose a reasonable biological explanation. Only 1 mark is available, so one clear point is needed.

Approach

Think about what the parasite needs (amino acids for protein synthesis) and what resource is plentiful inside the red blood cell (haemoglobin, which is a protein). Connect the two: the parasite breaks down haemoglobin to obtain the amino acids it needs.

Step-by-Step Reasoning

• Haemoglobin is a protein — when broken down, it yields amino acids.
• The parasite must grow and synthesise new proteins (e.g. MSP1, membrane proteins) inside the host cell.
• Therefore, the parasite digests haemoglobin → amino acids → used to build Plasmodium proteins / for parasite growth.
• The net result is a falling concentration of haemoglobin inside the infected cell.
• A mark-scheme-acceptable point is therefore: "(merozoites) break down (haemoglobin) to produce amino acids" AND/OR "(merozoites) use the amino acids for growth / to synthesise other proteins".

Key Takeaways

• Plasmodium is a haemoglobin-consumer inside the red blood cell.
• A fall in host protein is a direct consequence of the parasite's feeding strategy.

Common Mistakes

• Writing only "the parasite uses haemoglobin for energy" — rejected by the mark scheme. The protein part of haemoglobin is the source of amino acids; Plasmodium does not significantly oxidise haemoglobin for ATP — it digests the globin for amino acids.
• Writing "the parasite destroys the red blood cell" — too vague, and haemoglobin falls well before the cell bursts.
• Writing "haemoglobin leaks out" — the cell membrane is intact during most of the intra-erythrocytic stage, so leakage is not the main cause.

Things to Be Careful About

Only 1 mark is available, so give a single tight point. "Suggest" questions often reward any biologically reasonable idea but in this case the mark scheme lists specific accepted answers — phrase yours as either "break down haemoglobin to amino acids" or "use amino acids to make proteins / for growth".

Background Concept

Protein synthesis occurs in two stages:

1. Transcription in the nucleus: a section of the DNA double helix unwinds; RNA polymerase synthesises a complementary pre-mRNA strand by forming phosphodiester bonds between free RNA nucleotides; after splicing, the mature mRNA leaves the nucleus through a nuclear pore and enters the cytoplasm.
2. Translation in the cytoplasm: the mRNA binds to a ribosome; aminoacyl-tRNAs (tRNAs that have been "charged" with their amino acid by aminoacyl-tRNA synthetases, a process called amino acid activation) enter the ribosome; the codon on the mRNA pairs with the anticodon on the tRNA (codon–anticodon binding); the ribosome catalyses peptide bond formation between adjacent amino acids; this repeats along the mRNA until a stop codon is reached, when the completed polypeptide is released.

MSP1 is a single polypeptide, so once it is released, the chain folds into the secondary and tertiary structure described in part (e).

Understanding the Question

Table 4.1 lists eight steps (A–H) of protein synthesis out of order. Two of the steps (H and C) are already placed in the sequence table (Table 4.2). You need to place the remaining six steps (A, B, D, E, F, G) in their correct positions. There are 4 marks — one mark per correct line that is filled in, and the mark scheme notes a paired credit: ";(for steps 2 and 3)" and ";(for steps 5 and 6)", meaning if you place BOTH 2 and 3 correctly you get one of the four marks.

Approach

Mentally split the steps into two phases — transcription (in the nucleus) and translation (in the cytoplasm). Then arrange each group chronologically:

• Transcription: DNA unwinds (D) → phosphodiester bonds form as RNA polymerase makes the mRNA (F) → mRNA leaves the nucleus (B).
• Translation, after the mRNA binds the ribosome (H, given): activated tRNA attaches (A) → codon–anticodon binding (E) → peptide bond forms (G) → stop codon releases the chain (C, given).

Step-by-Step Reasoning

• The mRNA has to be made before it can leave the nucleus, and it leaves the nucleus before it binds a ribosome in the cytoplasm.
• Transcription steps in the nucleus, in order: D (unwind DNA) → F (RNA polymerase forms phosphodiester bonds to make the mRNA) → B (mRNA leaves the nucleus through a nuclear pore).
• Step 4 is given as H (mRNA binds the ribosome), so step 3 must end transcription.
• After H, the first cytoplasmic event is an activated tRNA arriving at the ribosome (A). Note that tRNAs are charged with their amino acid by aminoacyl-tRNA synthetase in the cytoplasm BEFORE they bind the ribosome — "amino acid activation" is the energy-requiring step that attaches the correct amino acid to its tRNA.
• Once the tRNA is in place, its anticodon pairs with the codon on the mRNA (E — codon–anticodon binding).
• The ribosome then catalyses the peptide bond between the new amino acid and the growing chain (G).
• The cycle of A → E → G repeats as the ribosome moves along the mRNA until a stop codon is reached (C, given).
• Mark-scheme check: steps 2 & 3 form one linked credit, and steps 5 & 6 form another linked credit — so getting each pair correct together is the route to all 4 marks.

Key Takeaways

• Transcription order: DNA unwinds → RNA polymerase joins RNA nucleotides by phosphodiester bonds → mRNA leaves through nuclear pores.
• Translation order: mRNA binds ribosome → aminoacyl-tRNA enters → codon–anticodon pairing → peptide bond formation → stop codon releases the polypeptide.
• The amino acid is activated (charged onto its tRNA) BEFORE the tRNA enters the ribosome, so step A precedes E.

Common Mistakes

• Placing A before B (i.e. before the mRNA is in the cytoplasm) — translation cannot begin until mRNA has left the nucleus.
• Putting E before A — codon–anticodon pairing only happens once a charged tRNA is in the ribosome.
• Putting B after H — the mRNA must reach the cytoplasm before binding a ribosome, so B is the last transcription event and comes before H.
• Treating F (phosphodiester bonds) as a translation step — it is the mRNA being synthesised in the nucleus, not peptide bonds.
• Forgetting that mRNA has to be exported from the nucleus to the cytoplasm in a eukaryote like Plasmodium.

Things to Be Careful About

The mark scheme groups steps 2–3 and 5–6 as paired marks: 1 mark for both correct, 1 mark for both correct. Therefore, if you swap the two steps in either pair (e.g. write B then F, or E then A), you lose that pair's mark. Within the pairs the order is fixed because RNA polymerase must act on the unwound DNA BEFORE the finished mRNA can leave the nucleus, and a charged tRNA must be in the ribosome BEFORE its anticodon can pair with the codon.

Background Concept

A newly synthesised polypeptide is a linear chain of amino acids joined by peptide bonds. To become a functional protein it must fold into a specific 3-D shape.

• Primary structure = the linear sequence of amino acids held by peptide bonds.
• Secondary structure = regular, repeating patterns of folding of the polypeptide backbone, stabilised by hydrogen bonds between the H of an N–H group and the O of a C=O group further along the chain. The two main motifs are the α-helix (a right-handed coil, with H-bonds every 3.6 amino acids) and the β-pleated sheet (polypeptide strands lying side-by-side, held by H-bonds between them).
• Tertiary structure = the further folding of an already-folded region into a specific 3-D shape, held by interactions between the R-groups (side chains) of amino acids. The four main types of R-group interaction are:
  • Hydrogen bonds between polar R-groups (e.g. –OH groups).
  • Ionic / electrostatic bonds between charged R-groups (e.g. –NH₃⁺ and –COO⁻).
  • Disulfide bridges / bonds between two –SH groups of cysteine residues (covalent S–S bonds).
  • Hydrophobic interactions between non-polar R-groups, which cluster in the interior of soluble proteins.

(Quaternary structure, when several polypeptide chains combine, is not relevant for MSP1 because it is a single polypeptide — the question already notes this.)

Understanding the Question

The stem tells you MSP1 has three levels of structure (primary, secondary, tertiary). The command word is "outline" — give the main, organised points without lengthy detail. 4 marks means four clear, distinct points. The mark scheme specifically REJECTS R-group bonds when linked to secondary structure, and REJECTS peptide bonds when describing the bonds that hold tertiary structure together.

Approach

Plan your answer in two clean blocks:

1. A block on secondary structure — folding, the named motifs, and that they are stabilised by H-bonds in the backbone.
2. A block on tertiary structure — further folding, bonds between R-groups, and at least two of the four named R-group bond types.

Step-by-Step Reasoning

• Point 1 (1 mark): the polypeptide folds / coils — and is therefore now in secondary / tertiary structure. "Folding" is the headline idea.
• Point 2 (1 mark): the secondary structure includes named motifs — α-helices OR β-pleated sheets. Either is enough; the mark scheme accepts both and any equivalent spelling (β-pleated sheets / β-pleats; R "beta-plates").
• Point 3 (1 mark): in tertiary structure, bonds / interactions form between the R-groups (side chains) of different amino acids. The mark scheme rejects answers that put R-group bonds into secondary structure.
• Point 4 (1 mark): at least two named bonds from the list — hydrogen, disulfide, ionic/electrostatic, hydrophobic. Don't include peptide bonds or "covalent bonds" as a generic answer.
• A bonus point (AVP) is available for additional correct detail, e.g. describing exactly where H-bonds in the α-helix form (between the H of N–H and the O of C=O) or the pattern of H-bonds (every 3.6 residues).

Key Takeaways

• Secondary structure is about the backbone; tertiary structure is about the R-groups.
• The four R-group interactions are: hydrogen, ionic, disulfide, hydrophobic.
• "Folding" is the one-word headline for what happens after translation.

Common Mistakes

• Calling the secondary-structure H-bonds "R-group bonds" — they are backbone H-bonds.
• Including "peptide bonds" in the list of tertiary-structure interactions — peptide bonds hold the primary structure, not the tertiary.
• Writing "van der Waals forces" or "hydrophilic interactions" — both are ignored in the mark scheme for this question.
• Spelling "β-pleated sheet" as "β-plates" or "β-pleats" alone is acceptable (β-pleated sheets is safer).
• Confusing secondary and tertiary structure — the question explicitly asks for both, so two distinct blocks of answer are needed.

Things to Be Careful About

• Use the technical terms (R-group, hydrogen bond, disulfide bridge, ionic bond, hydrophobic interaction). "Bonding" alone is too vague.
• The mark scheme REJECTS bonds credited to the wrong level — get the level right.
• If you only give one named R-group bond, you only score the bond point for the first one named (you need two for the mark).

Background Concept

Active immunity is immunity produced by the body's own immune system, in response to antigens. The body makes its own antibodies and memory cells.

• Natural active immunity follows infection (e.g. catching measles gives life-long immunity).
• Artificial active immunity follows vaccination — the body is deliberately exposed to a harmless form of the antigen.

Passive immunity is immunity "borrowed" from another source, e.g. maternal antibodies crossing the placenta, or injection of antivenom. The body does NOT make its own memory cells, so protection is short-lived.

A vaccine is a preparation of antigen (a whole weakened/killed organism, or part of an organism — e.g. a surface protein) that stimulates a primary immune response:

1. Antigen-presenting cells (e.g. macrophages) take up and present the antigen.
2. Specific B-lymphocytes (with complementary surface receptors) bind the antigen, are activated with T-helper-cell help, and undergo clonal selection.
3. Selected lymphocytes divide many times (clonal expansion) by mitosis, producing a large clone of effector cells (plasma cells that secrete antibody) and long-lived memory cells.
4. On a later encounter with the real pathogen (which has the same antigen), the memory cells are quickly reactivated — this is the secondary immune response. It is faster, larger and produces much more antibody than the primary response, so the pathogen is destroyed before it causes disease.

Understanding the Question

The stem tells you that the vaccine contains MSP1 — a non-self protein — and that the resulting immunity is "artificial active". The command word is "explain" — give reasons, not just definitions. 4 marks are available for explaining the mechanism by which a vaccine produces artificial active immunity.

Approach

Structure your answer around the sequence of immune events caused by the vaccine, and finish by linking the speed/size of the secondary response to the absence of symptoms:

1. Antigen recognition + clonal selection (the trigger).
2. Clonal expansion (the production of effectors and memory cells).
3. Memory cells are long-lived (so protection persists).
4. On real infection, the secondary response is faster/larger — pathogen is cleared before disease (this is what immunity means in practice).

Step-by-Step Reasoning

• Mark point 1 (memory cells formed) — the headline outcome of vaccination: long-lived B- and T-memory cells are produced and persist in the body.
• Mark point 2 (clonal selection details) — the antigen is processed/antigen presented; specific B- and T-lymphocytes recognise and bind the antigen; this triggers the lymphocytes to be activated. ("MSP1 stimulates B-cell to divide" alone counts as point 2 not point 3 — credit it as the activation step.)
• Mark point 3 (clonal expansion) — once activated, the specific lymphocytes divide many times by mitosis, making a large clone of plasma cells (which secrete antibody) and memory cells.
• Mark point 4 (memory cells are long-lived) — the B- and T-memory cells remain in circulation for years ("remain in the body" alone is rejected; "long-lived" or "in circulation" is needed).
• Mark point 5 (secondary response on real exposure) — when the person is later bitten by an infected mosquito, the real Plasmodium antigen (MSP1 on the merozoite) is recognised, triggering a secondary immune response.
• Mark point 6 — because the clone of specific lymphocytes is already present and very large, the response is faster / greater; more plasma cells, more antibody, more T-helper cells. The mark scheme also accepts "increased chance of fast recognition of antigen".
• Mark point 7 — therefore higher levels / faster production of antibody, which neutralises the merozoite before it can establish a malaria infection.
• Mark point 8 (AVP) — e.g. antibodies binding MSP1 prevent the merozoite entering the red blood cell (the precise protective mechanism in this case).
• Finally, tie the chain back to the type of immunity: because the antigen is deliberately injected as a vaccine (not caught naturally) and the body makes its own antibodies and memory cells, the immunity is artificial and active.

Key Takeaways

• A vaccine works by generating memory cells in a primary response, so that the real pathogen triggers a larger, faster secondary response.
• Artificial = the antigen is administered deliberately (vaccination).
• Active = the body makes its own antibodies and memory cells.
• The MSP1 protein in the subunit vaccine is a surface antigen of the merozoite, so antibodies raised against it should block merozoite entry into red blood cells.

Common Mistakes

• Confusing artificial active with artificial passive — passive would mean injecting ready-made antibodies, which the vaccine does NOT do.
• Saying "vaccines give you antibodies" — the vaccine makes the body produce its own antibodies, that is the active component.
• Failing to mention memory cells — these are the essential output of vaccination and the basis of long-term protection.
• Saying the immune system "kills the pathogen immediately on vaccination" — the vaccine contains only a non-self protein (subunit), not the whole pathogen, so it does not cause disease; protection only becomes visible on later exposure.
• Stating that antibodies are produced by T-lymphocytes — antibodies are produced by plasma cells, which are differentiated B-lymphocytes.

Things to Be Careful About

• The mark scheme rejects "kept in the body" alone — memory cells must be described as long-lived or remaining in circulation.
• "Live longer" is rejected unless it is made clear that they are present when the real antigen arrives (i.e. they persist long enough).
• Make sure the link from memory cells to faster / more antibody on re-exposure is explicit — the question asks for an explanation of immunity, not just a list of events.

Background Concept

Malaria is transmitted by the bite of an infected female Anopheles mosquito. The mosquito is therefore the vector. The life cycle of the mosquito has four stages: egg → larva → pupa → adult, with the first three in water and only the adult flying. The female needs a blood meal to mature her eggs. Plasmodium itself completes its sexual reproduction inside the mosquito, so the mosquito is essential to the parasite's life cycle.

Vector control aims to break the chain of transmission. It can be attacked at different points:

• Before the mosquito bites a person — by repelling or killing adults, or by physical barriers.
• Before the mosquito breeds — by removing or treating the water where eggs are laid and larvae develop.
• By making matings unproductive — e.g. releasing sterile males.
• By killing the parasite inside the mosquito — e.g. endectocides (drugs that kill Plasmodium in the mosquito when the mosquito feeds on a treated person).

Understanding the Question

The command word is "discuss" — at AS Biology this means present the main approaches with an example each. 3 marks are available, so the mark scheme expects three distinct points, each usually paired with a named example.

Approach

Group the methods by what they target:

1. The adult mosquito before it bites (barriers/repellents).
2. The adult mosquito with chemicals (insecticides).
3. The aquatic stages (eggs, larvae, pupae) by removing or treating the water.
4. Optional AVP — biological/genetic methods (sterile males, drug-treated water).

Step-by-Step Reasoning

• Mark point 1 (prevent blood meals) — only the female mosquito takes blood (to mature eggs), and Plasmodium is transmitted by her bite. Stop her feeding, and the life cycle is broken. Examples that earn the linked mark: mosquito nets (used at night / while sleeping); long-sleeved clothing (covering bare skin as a barrier); insect repellents (qualified — on body or clothes); indoor residual spraying of walls.
• Mark point 2 (use of insecticides / pesticides) — kill adult mosquitoes. Examples: insecticide-treated bed nets (ITNs / LLINs), indoor residual spraying (IRS), aerial sprays over wetlands. Mosquito coils and insect repellents count as repellents, not as insecticides, so they are credited under mark point 1 only once.
• Mark point 3 (treat water to disrupt egg-laying, eggs, larvae or pupae) — the aquatic stages are particularly vulnerable because they are concentrated and unable to fly away. Examples: remove or cover standing water (water tanks, ponds — NOT rivers); spray oil on ponds / small lakes / swamps to suffocate larvae; use biological control such as mosquitofish (Gambusia) that eat mosquito larvae; release Bti (a bacterial larvicide).
• Mark point 4 (AVP) — more advanced methods: release sterile (irradiated) males so matings produce no offspring; treat the mosquito's water with drugs that kill Plasmodium inside the mosquito.

Key Takeaways

• Vector control hits the mosquito at different points: adult (before she bites), aquatic (eggs, larvae, pupae), and reproduction (sterile males).
• The most effective measures target the adult female because she is the only one that bites and transmits Plasmodium.
• Insecticide-treated bed nets (ITNs / LLINs) are one of the most cost-effective public-health interventions against malaria.
• Standing water is the breeding ground — environmental management that removes it is a cheap, long-term control method.

Common Mistakes

• Mentioning only one method (e.g. "use a mosquito net") — "discuss" expects breadth, so cover both adult and aquatic stages.
• Calling the male mosquito the one that transmits the disease — only the female takes a blood meal and therefore only the female transmits Plasmodium.
• Confusing vectors (the mosquito) with pathogens (the Plasmodium protoctist). Vector control targets the mosquito; treatments that target the parasite (e.g. artemisinin) are not vector control.
• Suggesting the use of antibiotics — antibiotics kill bacteria, not mosquitoes.
• Releasing sterile females (ignored by the mark scheme); it must be sterile males.
• Suggesting "covering rivers" (ignored) — mosquitoes do not lay in flowing water; only still/stagnant water.

Things to Be Careful About

• For each control method, give a named example or qualification — the mark scheme marks the example alongside the principle. "Use a net" alone is borderline; "mosquito net at night" is safer.
• Mosquito coils and insect repellents count as repellents/barriers (point 1), not as insecticides (point 2) — credit once, not twice.
• "Aerial sprays" must be qualified (where, e.g. over wetlands) to earn the mark.

Background Concept

The atrioventricular node (AVN) sits between the atria and ventricles. Its role is twofold: (1) it is the only electrical bridge between atria and ventricles (the fibrous skeleton of the heart insulates the rest), and (2) it introduces a short delay (~0.1 s) between atrial contraction and ventricular contraction. This delay allows the atria to empty their blood fully into the ventricles before the ventricles contract. If this delay is removed, the atria and ventricles contract almost simultaneously, which is inefficient for pumping.

Understanding the Question

The stem describes Wolff–Parkinson–White syndrome: an extra bundle of fibres (bundle of Kent) allows impulses to pass directly from atria to ventricles, bypassing the AVN. Using your knowledge of the AVN's role, you must (1) state the change in heart rate, and (2) explain why that change happens. Two marks are available: one for identifying the change (heart rate increases), and one for explaining the mechanism (AVN delay is bypassed → premature ventricular contraction).

Approach

First identify what the AVN does. Then consider what removing it from the conduction route (via the bundle of Kent) would do to the timing of ventricular contraction relative to atrial contraction. Finally, translate that earlier ventricular contraction into an increase in heart rate.

Step-by-Step Reasoning

1. The AVN's normal role is to delay the impulse travelling from atria to ventricles so that the ventricles contract after the atria have finished contracting.
2. In Wolff–Parkinson–White syndrome, impulses from the atria can travel down the bundle of Kent directly to the ventricles, without going through the AVN.
3. Because the AVN's delay is bypassed, impulses reach the ventricular muscle earlier than they would normally.
4. The ventricles therefore contract too soon — while the atria are still contracting — shortening the cardiac cycle.
5. A shorter cardiac cycle means more contractions per minute: heart rate is increased (the ventricles contract at a faster rate).

Key Takeaways

• The AVN delays the impulse to allow sequential atrial-then-ventricular contraction.
• Bypassing the AVN causes premature ventricular contraction and an increased heart rate.
• This illustrates why the AVN delay is essential for efficient, coordinated cardiac action.

Common Mistakes

• Saying the heart rate decreases — wrong; bypassing a delay speeds up the cycle.
• Just saying "the ventricles contract too soon" without linking it to a faster heart rate.
• Describing the bundle of Kent as "another AVN" — it is an abnormal accessory pathway, not a duplicate AVN, and crucially it does not provide the AVN's delay.

Things to Be Careful About

The question says "suggest and explain the change that occurs in the heart rate" — so you must explicitly state the direction of change (increase) AND explain it. Use precise language: "AVN delays the impulse", "bypasses the AVN", "premature ventricular contraction". Avoid vague phrases like "the rhythm is disrupted".