Biology 9700/22 — February/March 2024

Background Concept

A phospholipid is built from one glycerol molecule ester-linked to two fatty acid tails (which form the hydrophobic, non-polar region) and to one phosphate group (which forms the hydrophilic, polar head). The glycerol is the three-carbon backbone that physically joins the two fatty acids to the phosphate head. In the bilayer, the heads face the aqueous environments on either side of the membrane while the fatty acid tails are sequestered in the hydrophobic core.

Understanding the Question

The question points to label A in Fig. 1.1, which sits between the phosphate head and the two fatty acid tails of a phospholipid. The candidate is asked to name this connecting region.

Approach

A quick structural identification: read the position of A in the diagram, recall the three components of a phospholipid, and pick the component that lies in that position.

Step-by-Step Reasoning

• A is drawn as the small square that connects the circular phosphate head above to the two zigzag fatty acid tails below.
• The only part of a phospholipid that bonds both to the phosphate head AND to two fatty acid tails is the three-carbon alcohol glycerol.
• Two of glycerol's –OH groups form ester bonds with the carboxyl groups of the two fatty acids; the third –OH bonds with the phosphate group.

Key Takeaways

• A phospholipid = glycerol + 2 fatty acids + 1 phosphate group.
• The glycerol backbone is the linker between the polar head and the non-polar tails.

Common Mistakes

• Writing "phosphate" or "fatty acid" — these are the parts that bond TO glycerol, not the part labelled A.
• Spelling "glycerol" as "glycerine" (a different compound) or "glyceral".

Things to Be Careful About

The mark scheme awards the precise term "glycerol"; the spelling matters.

Background Concept

Cholesterol is an amphipathic molecule: it has a small polar region (one –OH group) and a large non-polar region (four fused hydrocarbon rings plus a hydrocarbon tail). The fluid-mosaic bilayer is itself arranged so that polar groups face water and non-polar groups face each other in the centre. Amphipathic molecules always orient themselves so that "like meets like" — polar with polar, non-polar with non-polar.

Understanding the Question

Fig. 1.2 shows cholesterol's two parts explicitly. Fig. 1.1 shows how cholesterol sits in the bilayer. The candidate must explain WHY the molecule takes up the position shown, using the structure in Fig. 1.2.

Approach

Match the two parts of cholesterol to the two parts of the bilayer using polarity:

• polar –OH ↔ polar phosphate heads / aqueous environment
• non-polar rings/tail ↔ non-polar fatty acid tails / hydrophobic core

Step-by-Step Reasoning

• Cholesterol's –OH group is polar, so it is attracted to the polar phosphate heads and/or water. In Fig. 1.1 the –OH is drawn pointing towards the phosphate-head surface — this is the energetically favourable arrangement.
• The hydrocarbon ring system and the branched chain are non-polar (hydrophobic), so they associate with the hydrophobic fatty-acid tails in the middle of the bilayer.
• Because one end is polar and the rest is non-polar, cholesterol is held in a specific vertical orientation in the bilayer, exactly as shown in Fig. 1.1.

Key Takeaways

• Cholesterol is amphipathic; its position in the bilayer is dictated by the polarity of its two regions.
• Polar groups orient towards water; non-polar groups orient towards the hydrophobic core.

Common Mistakes

• Saying cholesterol is "trapped" or "held in place" without referring to the polar/non-polar match.
• Reversing the orientation in the answer (e.g. claiming –OH lies among the fatty acid tails).

Things to Be Careful About

The mark scheme accepts either an explicit "polar / non-polar" pairing or a "hydrophilic / hydrophobic" pairing, but BOTH ends of the molecule must be addressed for a complete explanation.

Background Concept

Cholesterol is wedged between phospholipid molecules and acts as a "membrane moderator". Its rigid steroid ring system restricts the movement of the fatty acid tails when the membrane would otherwise be too fluid (high temperature) and, by disrupting the regular packing of the tails, prevents the membrane from becoming too rigid at low temperatures. It also mechanically reinforces the bilayer.

Understanding the Question

A single-marker "state" question, so only one clear, accurate role is needed.

Approach

Pick the most commonly credited role and write it in the mark-scheme wording.

Step-by-Step Reasoning

• Most frequently rewarded answer: regulates membrane fluidity.
• Alternative correct answers: mechanical stability; reduces permeability to ions / polar molecules; restricts lateral movement of phospholipids; prevents membrane rupture.
• Any one of these, phrased concisely, scores the mark.

Key Takeaways

• Cholesterol is both a fluidity buffer and a stabiliser of the bilayer.
• A "state" command word requires only one point but it must be the precise term from the mark scheme.

Common Mistakes

• Giving a vague answer such as "it helps the membrane" — this is too imprecise to score.
• Writing "forms the bilayer" — cholesterol is a minor component; it does NOT form the bilayer.

Things to Be Careful About

"Fluidity" is the precise term; avoid the looser "keeps the membrane flexible" unless paired with a clear reference to fluidity regulation.

Background Concept

The interior of a phospholipid bilayer is hydrophobic because it is lined by the long hydrocarbon fatty acid tails. Simple diffusion across a membrane requires a substance to dissolve in this hydrophobic core. Polar molecules (water-soluble) and ions (charged) cannot dissolve in a non-polar environment and so cannot pass through by simple diffusion; they must be helped across by transport proteins.

Understanding the Question

The command word is "explain", so the candidate must give a reason, not just an observation.

Approach

State the property of the sodium ion (it is charged) and the property of the membrane interior (it is hydrophobic/non-polar), then link them with a verb of repulsion / inability to dissolve.

Step-by-Step Reasoning

• Na⁺ carries a positive charge.
• The bilayer core consists of hydrophobic (non-polar) fatty acid tails.
• Like charges and hydrophobic regions are mutually incompatible — the charged ion is repelled by / cannot dissolve in the non-polar core.
• Therefore Na⁺ cannot simply diffuse across.

Key Takeaways

• Ions and polar molecules are excluded from the hydrophobic core.
• "Repelled by the hydrophobic tails" is the mark-scheme wording — both ideas (charged AND hydrophobic tails) must be present.

Common Mistakes

• Saying only "Na⁺ is charged" with no reference to the membrane — incomplete.
• Saying "the membrane is solid" or "Na⁺ is too big" — both biologically wrong and not credited.

Things to Be Careful About

Acceptable alternatives to "hydrophobic" include "non-polar"; the mark scheme lists "hydrophobic tails / non-polar core / non-polar tails" as interchangeable.

Background Concept

Both processes move substances across the cell surface membrane via transport proteins, but they differ in three important ways: the direction of movement relative to the concentration gradient, the source of energy used, and (in the case of channel proteins) the type of protein involved.

• Facilitated diffusion is passive: substances move down a concentration gradient through channel or carrier proteins; no ATP is needed.
• Active transport moves substances against the concentration gradient via carrier proteins and requires energy from ATP (often supplied by co-transport of ions or by direct ATP hydrolysis).

Understanding the Question

The question is a structured comparison requiring: one similarity (1 mark) and two differences (1 mark each). The candidate must phrase contrasts as point-by-point pairs, not as two separate descriptions.

Approach

• For the similarity, identify what is true of BOTH processes.
• For each difference, state the situation in facilitated diffusion AND the contrasting situation in active transport in the same sentence (or as a paired row of a table).

Step-by-Step Reasoning

Similarity — the mark-scheme-credited common feature is that both processes use transport proteins (channel and/or carrier) embedded in the membrane. Alternatives accepted by the mark scheme: both are specific to the molecule/ion; both can involve a conformational change in a carrier protein; both can transport substances into and out of the cell.

Difference 1 — direction of movement. The contrast is down the gradient (facilitated diffusion) versus against the gradient (active transport). The mark scheme phrases this as "down the concentration gradient" vs "against a concentration gradient".

Difference 2 — energy. The contrast is no ATP / passive (facilitated diffusion) versus ATP required / metabolic energy (active transport). The mark-scheme wording is "does not require ATP" vs "requires ATP".

(A third difference, also accepted: facilitated diffusion can use both channel and carrier proteins, while active transport only uses carrier proteins.)

Key Takeaways

• A good comparison states the shared feature first, then pairs the contrasts clearly.
• Phrase contrasts so facilitated diffusion is named first, then active transport — or present them as a two-column comparison.

Common Mistakes

• Saying facilitated diffusion "uses ATP" — wrong; it is passive.
• Confusing the direction of the gradient: "active transport goes down the gradient" is a common error.
• Writing similarities that are actually differences (e.g. "facilitated diffusion doesn't need ATP, active transport does" — this is a difference, not a similarity).
• Omitting the protein aspect of the similarity.

Things to Be Careful About

• One mark is for the similarity, one each for the two differences — three separate ideas, three separate sentences is safest.
• Avoid the term "facilitated diffusion is faster" — it is not a mark-scheme point and may not be credited.

Background Concept

The smooth endoplasmic reticulum (SER) is a network of membrane-bound flattened sacs (cisternae) and tubules continuous with the nuclear envelope. Unlike the rough ER, it lacks ribosomes. The SER is the main site of lipid and steroid synthesis (including phospholipids and cholesterol), and it also stores calcium ions and helps to detoxify drugs.

Because the SER is enclosed by a membrane, it separates the reactions inside from those in the surrounding cytoplasm. This compartmentalisation is a key feature of eukaryotic cells: different reactions can occur in different locations with their own optimal conditions and without interference from competing reactions.

Understanding the Question

The stem describes a pathway (phospholipid → arachidonic acid → prostaglandin) catalysed partly by COX, and states that the pathway occurs in the SER. The candidate is asked to suggest an advantage of this location, compared with the cytoplasm.

Approach

Two main families of argument are credited:

1. Substrate/product supply — the SER is the site of phospholipid synthesis, so the starting material is right there, and being lipid-based the products can be transported through / stored in the SER membrane.
2. Compartmentalisation — keeping the pathway in a separate compartment gives optimum conditions (pH, substrate concentration) and prevents the intermediates from interfering with other cytoplasmic reactions.

Step-by-Step Reasoning

• The first substrate of the pathway is a phospholipid, which is itself made in the SER; locating the pathway here means the substrate is readily available.
• Prostaglandins are lipids and so are soluble in membranes; they can be transported within or released from the SER easily.
• The enzymes (e.g. COX) of the pathway can be at high local concentration inside the SER, speeding up the reaction.
• Separating the pathway from the cytoplasm prevents unwanted side reactions and concentrates the substrates.

Any ONE of these points is sufficient for the single mark.

Key Takeaways

• Eukaryotic compartmentalisation allows specific reactions to occur in optimal conditions.
• The SER is a lipid-handling organelle — pathways involving lipids are commonly located there.

Common Mistakes

• Saying only "the SER has ribosomes" — the SER does NOT have ribosomes; that is the RER.
• Writing "for protection" without saying what is protected from what — too vague.
• Saying the SER "produces ATP" — that is mitochondria.

Things to Be Careful About

This is a "suggest" question — there is no single correct answer, but the answer must specifically link the structure of the SER (membrane-bound) to a feature of the prostaglandin pathway (lipid substrates/products).

Background Concept

The tertiary structure of an enzyme is held together by interactions between the R-groups of amino acids far apart in the chain: hydrogen bonds, ionic bonds, disulfide bridges and hydrophobic interactions. The precise 3-D shape of the active site depends on this tertiary structure. If the chemistry of an R-group is changed, the interactions it makes with neighbouring R-groups change, the tertiary structure distorts, and the active site loses its complementary shape. The substrate can then no longer bind (lock-and-key) or the active site can no longer mould around the substrate (induced fit). Without a productive enzyme–substrate complex, the enzyme cannot lower the activation energy and the reaction slows or stops.

Aspirin is a classic non-competitive / irreversible inhibitor of COX — it acetylates a serine residue inside the active site, blocking the channel leading to it. The question's wording is generic, however: "modifying the R-group of one of the amino acids".

Understanding the Question

The stem states that aspirin reduces the catalytic activity of COX by modifying an R-group. The candidate must explain the cascade of consequences from this single molecular change.

Approach

Step through the logic in the order in which the mark scheme rewards it:

1. State which R-group interactions are affected.
2. State the consequence for tertiary structure / active site shape.
3. State the consequence for substrate binding (ES complex).

Step-by-Step Reasoning

• Amino-acid R-groups interact through hydrogen bonds, ionic bonds and hydrophobic interactions to maintain the 3-D shape of the protein and therefore the precise shape of the active site.
• Modifying an R-group changes its chemistry (e.g. adds a charged or bulky group). This disrupts the original interactions — for example, an ionic bond may break, or a new steric clash may appear.
• The tertiary structure of the enzyme changes, and with it the shape / conformation of the active site.
• The active site is no longer complementary in shape to arachidonic acid (the substrate for COX), so by the lock-and-key / induced-fit model the enzyme–substrate complex cannot form, or forms only at a much reduced rate.
• With no (or fewer) ES complexes, COX cannot lower the activation energy of the conversion of arachidonic acid to prostaglandin, so catalytic activity is reduced.

Key Takeaways

• Primary structure (R-group sequence) determines tertiary structure, which determines active-site shape, which determines substrate specificity.
• Changing a single R-group can therefore abolish enzyme activity.
• This is a non-competitive / irreversible form of inhibition: the substrate can still approach, but the binding site no longer fits.

Common Mistakes

• Saying "the enzyme is denatured" — denaturation usually refers to disruption by heat / pH across the whole protein; here a single R-group is changed, so "shape of the active site changes" is more accurate.
• Saying "the substrate is broken down" — aspirin is not a competitive substrate; it modifies the enzyme, not the substrate.
• Failing to link R-group change → tertiary structure → active-site shape → substrate binding; any break in this chain loses marks.

Things to Be Careful About

• The mark scheme requires any three of the listed ideas for the 3 marks; cover at least three.
• "Complementary" is the precise term to use when describing the active site and substrate relationship.
• "Enzyme–substrate complex" should be written in full (or ES complex) — do not just say "they don't bind".

Background Concept

Cell signalling is the process by which cells communicate. A typical sequence is:

1. A signalling cell secretes a signalling molecule (a ligand, e.g. a hormone, neurotransmitter or local mediator such as a prostaglandin).
2. The ligand travels (by diffusion through tissue fluid, or in the blood) to a target cell.
3. The ligand binds to a specific receptor on the target cell's surface membrane (or, for steroid-like lipids, inside the cell).
4. Receptor binding triggers a response inside the cell — often via a second messenger (e.g. cyclic AMP) and an enzyme cascade of phosphorylation events, which amplify the signal and produce the cell's response.

Prostaglandins are local mediators (paracrine signals) — they act on neighbouring cells rather than being carried long distances in the blood.

Understanding the Question

The stem reminds the candidate that prostaglandins are cell-signalling molecules that lead to inflammation. The candidate must outline (give the main steps of) the signalling process.

The mark scheme rewards any two of: release/transport, receptor binding, and a valid example of the triggered response.

Approach

Write a tight, ordered outline covering: source and travel of the signal, binding to a receptor, and the resulting cellular response. Two well-chosen points are enough for 2 marks; covering all three makes the answer bullet-proof.

Step-by-Step Reasoning

• Step 1 — release: Prostaglandins are secreted (released) by the cells that produce them. Because they are lipids they can simply diffuse out through the cell surface membrane and travel through the tissue fluid to neighbouring target cells (paracrine signalling).
• Step 2 — receptor binding: The prostaglandin binds to a specific receptor protein on the surface membrane of the target cell involved in inflammation. The receptor is specific — only cells carrying the appropriate receptor respond.
• Step 3 — response: Receptor binding activates a signal-transduction pathway inside the cell — typically a second messenger (e.g. cAMP, Ca²⁺), an enzyme cascade and phosphorylation events. This amplifies the signal and produces the inflammation response: vasodilation, increased capillary permeability, redness, swelling, pain, recruitment of immune cells.

The mark scheme accepts any valid example at step 3 (second messenger, enzyme cascade, phosphorylation cascade or signal transduction).

Key Takeaways

• Cell signalling = ligand release → travel → receptor binding → cellular response.
• Prostaglandins are local (paracrine) lipid mediators, not classical hormones.
• A "response" in cell signalling is usually a change in cell behaviour triggered by an intracellular cascade.

Common Mistakes

• Writing "the prostaglandin enters the cell and changes the DNA" — prostaglandins bind to surface receptors, not intracellular ones (unlike steroid hormones).
• Saying "the prostaglandin is an antigen" — antigens are recognised by lymphocytes in immunity; this is a different process.
• Omitting the receptor step — the mark scheme specifically rewards "binds to receptors".

Things to Be Careful About

• The mark scheme rejects "antigens" — do not confuse cell signalling with immune recognition.
• A concrete example of the triggered response (second messenger, enzyme cascade, phosphorylation, signal transduction) is rewarded; a vague "the cell responds" is not.

Background Concept

The oxygen dissociation curve plots the percentage saturation of haemoglobin with oxygen against the partial pressure of oxygen ($\text{pO}_2$). The curve is sigmoidal because haemoglobin is a tetramer with four binding sites that show cooperativity: once one oxygen molecule binds, the others bind more easily.

Key regions of the curve:

• The flat upper plateau (above about $8\ \text{kPa}$) corresponds to alveolar conditions in healthy lungs at sea level, where haemoglobin is ~$95-98\%$ saturated.
• The steep middle portion (around $1-6\ \text{kPa}$) corresponds to tissue conditions, where small changes in $\text{pO}_2$ lead to large changes in saturation — ideal for oxygen unloading.

The position of the curve reflects the affinity of haemoglobin for oxygen:

• Curve to the LEFT = HIGHER affinity (binds oxygen more readily at lower $\text{pO}_2$).
• Curve to the RIGHT = LOWER affinity (releases oxygen more readily at tissues).

At high altitude the atmospheric pressure falls, so the $\text{pO}_2$ in the alveoli is lower than at sea level. This means that at a sea-level $\text{pO}_2$ of about $13-14\ \text{kPa}$ a mammal's haemoglobin is near saturation, but at $3500\ \text{m}$ the alveolar $\text{pO}_2$ is only $6.4\ \text{kPa}$, which on the human curve corresponds to a saturation of roughly $80\%$ rather than $97\%$ at sea level. Animals native to high altitudes have evolved haemoglobin with a left-shifted curve to compensate.

Understanding the Question

The question provides a graph (Fig. 2.3) with two oxygen dissociation curves: adult llama and adult human. The llama curve is to the LEFT of the human curve, meaning llama haemoglobin has a higher affinity for oxygen. The candidate is told the partial pressure in the lungs at $3500\ \text{m}$ is $6.4\ \text{kPa}$.

The command word is 'explain' with the requirement to 'refer to Fig. 2.3'. This means the answer must quote specific features of the graph — curve position, saturation values, or partial pressures — and use them to support the conclusion.

Approach

1. Read the two curves at the relevant $\text{pO}_2$ ($6.4\ \text{kPa}$).
2. Compare the saturation values.
3. State the direction of the difference (which curve is to the left, which haemoglobin has higher affinity).
4. Link this to oxygen loading in the lungs and delivery to tissues.

The mark scheme is 'any three from' five or six possible points — the candidate does not need all of them, but does need at least three well-supported points.

Step-by-Step Reasoning

1. Curve position: The llama curve lies to the LEFT of the human curve across the entire $\text{pO}_2$ range. This means that at any given $\text{pO}_2$, llama haemoglobin is more highly saturated with oxygen than human haemoglobin.

2. Reading at 6.4 kPa: At a $\text{pO}_2$ of $6.4\ \text{kPa}$ (the alveolar partial pressure at $3500\ \text{m}$), the llama haemoglobin is approximately $86\%$ saturated, whereas the human haemoglobin is approximately $79-80\%$ saturated. That is a difference of about $6-7\%$ more oxygen carried by llama haemoglobin per unit volume of blood.

3. Affinity: Because the curve is to the left, llama haemoglobin has a higher affinity for oxygen — it picks up oxygen more readily at low $\text{pO}_2$. This is exactly what is needed at high altitude where the available oxygen is reduced.

4. Loading in lungs: The higher affinity means that even when the alveolar $\text{pO}_2$ is reduced to $6.4\ \text{kPa}$, llama haemoglobin still loads a high percentage of its capacity with oxygen. A human at the same altitude would be at a much lower saturation.

5. Delivery to tissues: Despite the higher affinity, the curve is still sigmoidal, so the llama's tissues (with a tissue $\text{pO}_2$ of about $5\ \text{kPa}$ or less) still receive a substantial oxygen delivery. The llama gains at the loading end without losing too much at the unloading end.

Key Takeaways

• Left-shifted curve = higher affinity = better loading at low $\text{pO}_2$.
• High-altitude animals have evolved haemoglobin with a left-shifted dissociation curve.
• Read specific values from the graph to support the explanation — generic statements are not enough.
• The shape of the curve still permits unloading at tissues because the curve remains sigmoidal.
• The 'explanation' is about oxygen LOADING in the lungs; the mark scheme specifically ignores comments about higher saturation in tissues (because that is not where the advantage lies).

Common Mistakes

• Saying the llama curve is 'higher' rather than 'to the left' — the curve is shifted, not simply elevated.
• Confusing affinity with carrying capacity — they are different concepts.
• Citing values from the wrong curve (e.g. quoting human saturation when describing the llama).
• Stating that the llama haemoglobin carries 'more oxygen' without specifying per unit volume or per molecule of haemoglobin.
• Referring to tissue unloading advantages rather than lung loading — the mark scheme explicitly ignores references to higher saturation at tissue partial pressures.
• Failing to mention affinity at all — the mark scheme treats this as a separate point worth crediting.

Things to Be Careful About

• Always read the x-axis in the units shown ($\text{kPa}$) and the y-axis in percentage saturation.
• The mark scheme accepts either quoting specific values OR stating the relative positions of the curves — both approaches are valid.
• 'Explain' requires more than describing the graph: it requires linking the observation to the adaptive advantage.
• The numbers ($86\%$, $79-80\%$, $6-7\%$) are read from the graph; the mark scheme accepts values within the reading tolerance of the candidate.

Background Concept

The Bohr effect (or Bohr shift) describes the rightward shift of the oxygen dissociation curve that occurs when the partial pressure of carbon dioxide rises (and/or pH falls). The molecular basis is that $\text{CO}_2$ reacts with water inside red blood cells to form carbonic acid (catalysed by carbonic anhydrase), which dissociates to give $\text{H}^+$ and $\text{HCO}_3^-$. The $\text{H}^+$ ions bind to haemoglobin and stabilise the deoxygenated (T) state, decreasing its affinity for oxygen. A small fraction of $\text{CO}_2$ also binds directly to the N-terminal amino groups of globin chains, forming carbamino compounds, with a similar effect.

A rightward shift means that at any given $\text{pO}_2$, haemoglobin is LESS saturated with oxygen — so it unloads oxygen more readily to the tissues. This is biologically useful because metabolically active tissues produce $\text{CO}_2$, so the Bohr shift delivers more oxygen precisely where it is needed most.

The shift is small in absolute terms but physiologically very significant.

Understanding the Question

The question asks the candidate to draw a new curve on Fig. 2.3 showing the effect of increased $\text{CO}_2$ on the percentage saturation of adult HUMAN haemoglobin. The original human curve (dashed line) is already on the graph, and the candidate must add a second sigmoidal curve shifted to the right of it.

The command word is 'sketch' — a quick, clear drawing that captures the essential feature (rightward shift) is all that is required. The mark scheme allocates 1 mark, which is given for the curve being to the right of the human curve.

Approach

1. Identify the position of the existing human curve (dashed line).
2. Sketch a new sigmoidal curve parallel to it, but shifted to the right.
3. The new curve should still plateau at high $\text{pO}_2$ and pass through the lower-left region near the origin.
4. Do not shift the curve up or down — it is a horizontal shift only.

Step-by-Step Reasoning

1. The Bohr effect shifts the dissociation curve to the right when $\text{pCO}_2$ rises (or pH falls). This is the established convention in physiology.
2. A rightward shift means the curve is moved along the x-axis (the $\text{pO}_2$ axis) so that it lies at HIGHER $\text{pO}_2$ values than the original at any given saturation.
3. The curve retains its general sigmoidal shape — start near the origin, rise steeply in the middle, and plateau at the top right.
4. For the human curve specifically, the new curve should pass to the right of the dashed line at every point.

Key Takeaways

• Bohr shift = rightward shift of the dissociation curve.
• Increased $\text{CO}_2$ / decreased pH → decreased affinity → rightward shift.
• A 'sketch' should capture the essential feature (the shift), not be a precise curve.
• The shift is horizontal (along the $\text{pO}_2$ axis), not vertical.

Common Mistakes

• Drawing the curve to the LEFT (wrong direction).
• Drawing the curve ABOVE the human curve (a vertical shift, not a horizontal one) — the Bohr shift is a horizontal shift along the $\text{pO}_2$ axis.
• Changing the shape of the curve (e.g. making it shallower or steeper) — only the position should change.
• Forgetting that the question specifies ADULT HUMAN haemoglobin, not llama — the new curve is added to the right of the human curve, not the llama curve.

Things to Be Careful About

• The mark is awarded solely for correct position. A curve that is sigmoidal, starts near the origin and plateaus at the top right, drawn clearly to the right of the human dashed line, scores the mark.
• Do not let the new curve cross the human curve — the entire new curve should be to the right of the human one at every shared $\text{pO}_2$ value.
• A neat, smooth freehand curve is fine; ruler-drawn straight lines are not appropriate.

Background Concept

The Bohr effect (or Bohr shift) describes how the oxygen-carrying behaviour of haemoglobin responds to its chemical environment. Three factors shift the oxygen dissociation curve to the right (lower affinity, easier unloading):

1. Increased $\text{pCO}_2$ (partial pressure of carbon dioxide).
2. Decreased pH (increased $\text{H}^+$ concentration).
3. Increased temperature.

All three are characteristic of METABOLICALLY ACTIVE tissues:

• Aerobic respiration produces $\text{CO}_2$ as a waste product.
• $\text{CO}_2$ reacts with water inside red blood cells (catalysed by carbonic anhydrase) to form carbonic acid, which dissociates to $\text{H}^+$ and $\text{HCO}_3^-$.
• The $\text{H}^+$ ions bind to haemoglobin and stabilise its deoxygenated (T) state.
• A small fraction of $\text{CO}_2$ also binds directly to haemoglobin to form carbaminohaemoglobin.
• Respiration releases energy as heat, so active tissues are slightly warmer.

The net result is that where the oxygen demand is highest, the curve is shifted right, so haemoglobin releases more oxygen — exactly the right amount, exactly where it is needed.

Metabolically active organs in this context include the liver (which carries out many energy-demanding biosyntheses and detoxification reactions), exercising skeletal muscle, the brain, the heart, and the kidneys.

The relevant chemical equation inside the red blood cell is:

The $\text{H}^+$ ions are the immediate cause of the decreased oxygen affinity.

Understanding the Question

The question asks the candidate to explain the importance of the Bohr shift IN METABOLICALLY ACTIVE ORGANS, using the liver as the example. The command word is 'explain' — the candidate must state the mechanism (what happens) and link it to the consequence (why it matters).

The mark scheme gives 'any three from' four possible points. The candidate needs to mention the cause (more $\text{CO}_2$), the mechanism (decreased affinity), the consequence (more $\text{O}_2$ released), and the biological purpose (more aerobic respiration / ATP). Three of these four are sufficient for full marks.

Approach

Walk through the causal chain in the correct physiological order:

1. Active organ → high respiration rate → high $\text{CO}_2$ output.
2. High $\text{CO}_2$ → low pH (or direct effect on haemoglobin).
3. Low pH / high $\text{CO}_2$ → decreased affinity of haemoglobin for $\text{O}_2$ (Bohr shift).
4. Decreased affinity → more $\text{O}_2$ released from haemoglobin to the tissue.
5. More $\text{O}_2$ → more aerobic respiration → more ATP → meets the high metabolic demand of the organ.

The mark scheme rewards each logical step; the candidate should aim for three or four of them.

Step-by-Step Reasoning

1. Cause: A metabolically active organ such as the liver carries out many energy-demanding processes (biosynthesis, detoxification, etc.) at a high rate. This requires a high rate of aerobic respiration in its cells, which produces a correspondingly high output of $\text{CO}_2$. The local $\text{pCO}_2$ in the tissue is therefore higher than in less active tissues.

2. Mechanism: The increased $\text{pCO}_2$ lowers the local pH (because of the carbonic anhydrase-catalysed reaction). The lower pH (or the $\text{CO}_2$ itself) decreases the affinity of haemoglobin for oxygen — the Bohr shift. The mark scheme specifically allows reference to 'haemoglobinic acid' (the protonated form of haemoglobin) as the mechanistic link.

3. Consequence: As a result of the lower affinity, more oxygen is released from haemoglobin to the tissue. The dissociation curve effectively shifts to the right at the tissue level.

4. Purpose: The extra oxygen supports the high rate of aerobic respiration, which produces more ATP to meet the metabolic demand of the active organ. Without the Bohr shift, the oxygen supply would be insufficient for the high demand.

Key Takeaways

• The Bohr shift is a NEGATIVE FEEDBACK mechanism: high demand → high $\text{CO}_2$ → more $\text{O}_2$ released → demand met.
• The liver, heart, brain, kidneys and exercising muscle are all metabolically active and benefit from the Bohr shift.
• The mechanism involves $\text{CO}_2$, water, carbonic anhydrase, $\text{H}^+$ and haemoglobin — knowing the chemistry is not required at AS, but the link between $\text{CO}_2$ and $\text{O}_2$ release is.
• The Bohr shift is a LOCAL effect: the blood in capillaries supplying an active tissue experiences the shift; blood elsewhere does not.
• The mark scheme requires the comparative 'more' to be implicit or explicit in the answer.

Common Mistakes

• Stating that the Bohr shift 'increases the affinity of haemoglobin for oxygen' — the shift DECREASES affinity.
• Reversing cause and effect: saying 'more $\text{O}_2$ causes more $\text{CO}_2$' rather than the correct 'more $\text{CO}_2$ causes more $\text{O}_2$ to be released'.
• Confusing the Bohr shift with the effect of high altitude (which is a different curve shift mechanism involving different haemoglobin).
• Failing to mention the CONSEQUENCE for respiration/ATP — the mark scheme explicitly looks for 'more aerobic respiration' or 'more ATP production'.
• Treating the Bohr shift as something that 'helps the blood carry more oxygen' — it actually helps RELEASE more oxygen to the tissues.
• Stating only that the organ 'produces $\text{CO}_2$' without indicating the comparison with less active tissues.

Things to Be Careful About

• The mark scheme requires 'more' (a comparative) — simply saying the organ 'produces $\text{CO}_2$' without saying 'more' is not enough; the comparison with less active tissues must be implicit or explicit.
• The cause is increased $\text{CO}_2$ / decreased pH; the consequence is more $\text{O}_2$ released. Do not reverse the order.
• 'Explain' questions at AS need the WHY behind each point, not just a list of facts.
• The Bohr shift should be described as decreasing affinity, not as some other change.

Background Concept

The transverse section of a young root reveals the tissues arranged from the outside in: epidermis → cortex → endodermis → pericycle → vascular tissue (xylem and phloem) → central pith (parenchyma in monocots, xylem in dicots).

• Endodermis is a single ring of tightly packed cells around the vascular cylinder. The radial and transverse walls are impregnated with suberin, forming the Casparian strip, and the cells appear as a continuous dark ring in section.
• Xylem conducts water and mineral ions from root to shoot. Xylem vessel elements are dead, hollow tubes with lignified walls; in cross-section they appear as large, open, thick-walled cells.
• Phloem translocates organic solutes (mainly sucrose). Sieve tube elements are living, lack a nucleus, and are connected end-to-end; in cross-section they appear as small cells with thin walls, often accompanied by companion cells.

In a monocot root the xylem forms a ring of separate vessels around a central parenchyma pith, with phloem strands lying between the xylem vessels.

Understanding the Question

Fig. 3.1 is a transverse section through the root of an iris (Iris germanica), a herbaceous monocotyledon, at $\times 120$ magnification. The stem tells the candidate that the central tissue is parenchyma and that R, S and T are cells in three different tissues. They must name each tissue from its visible appearance in the labelled micrograph.

Approach

Read off where each labelled line points in Fig. 3.1 and match it to the description of that tissue:

• R points to the conspicuous dark, ring-like layer of cells just outside the vascular tissue → endodermis.
• S points to a large, open, thick-walled vessel → xylem.
• T points to a smaller cell lying between xylem vessels → phloem.

Step-by-Step Reasoning

• R: the line ends at the inner boundary of the cortex. The cells here form an unbroken ring and have thickened radial walls — characteristic of the endodermis. The endodermis separates the cortex from the stele.
• S: the line ends at a large, hollow cell with a thick wall. This is a xylem vessel element. Xylem conducts water and dissolved mineral ions upwards.
• T: the line ends at a smaller, thinner-walled cell between xylem vessels. This is a sieve tube element (phloem), which translocates organic solutes such as sucrose.

Key Takeaways

In any root TS, look for:

• the dark Casparian-strip ring = endodermis;
• large hollow vessels = xylem;
• small thin-walled cells between xylem vessels = phloem.

In monocots the xylem forms a ring of separate bundles around a parenchyma pith (unlike dicots, where the xylem forms a central star with phloem between the arms).

Common Mistakes

• Confusing the pericycle with the endodermis: the pericycle lies inside the endodermis and is usually one or two cells thick.
• Calling R "cortex" or "epidermis": the cortex lies outside the endodermis and consists of larger, loosely packed parenchyma cells.
• Confusing xylem and phloem in cross-section: xylem vessels are much larger and have thicker walls than the small sieve tube elements of the phloem.

Things to Be Careful About

The order of labels on this micrograph is unusual — R is the outer ring (endodermis), not a xylem vessel. Always read the position of the label line on the image, not the letter order on the page.

Background Concept

Water entering a root from the soil can move through two parallel pathways:

• the apoplast — through the continuous network of porous cell walls and intercellular spaces, without crossing any membrane;
• the symplast — through the cytoplasm of cells connected by plasmodesmata, with everything enclosed by plasma membranes.

The endodermis is the boundary layer whose radial and transverse walls are impregnated with suberin, forming the waterproof Casparian strip. This blocks the apoplast pathway at the inner edge of the cortex.

Understanding the Question

Cell R is in the endodermis. Part (a)(ii) asks the candidate to outline the role of that tissue — i.e. why its structure matters for water and solute movement into the vascular cylinder.

Approach

The command word "outline" only requires a short summary of the key effects. Two marking points are available: (1) the apoplast-blocking effect and (2) the resulting symplast-only route, which allows control of what enters the xylem. The mark scheme accepts as an alternative valid point a reference to passage cells or to the Casparian strip / suberin itself.

Step-by-Step Reasoning

• Water moving in the apoplast would otherwise pass freely through cell walls and reach the xylem without crossing any membrane. The Casparian strip is waterproof, so the apoplast route is blocked at the endodermis.
• To continue inwards, water must cross the plasma membrane of an endodermal cell and enter the symplast. Crossing a selectively permeable membrane allows the cell to control which ions and molecules are admitted to the xylem.
• Optional supporting detail — passage cells (endodermal cells without a Casparian strip) allow specific solutes to pass, or the strip itself is made of suberin and is the structural feature responsible.

Key Takeaways

• Casparian strip → blocks apoplast → forces symplast → membrane crossing gives control of xylem contents.
• This is the main reason roots can selectively absorb mineral ions.

Common Mistakes

• Saying only that the endodermis "controls what enters" without explaining the apoplast/symplast mechanism — this is too vague to earn both marks.
• Describing the endodermis as "impermeable" — it is not impermeable overall, only at the Casparian strip, and passage cells exist.
• Confusing "apoplast" with "symplast" — apoplast = walls + spaces, symplast = cytoplasm + plasmodesmata.

Things to Be Careful About

The mark scheme explicitly accepts AVP including references to passage cells and to the Casparian strip / suberin — both are useful, accurate supplementary points to include if room allows.

Background Concept

Phloem translocates organic solutes — the products of photosynthesis (assimilates) and other metabolites — from sources (e.g. mature leaves, storage organs being mobilised) to sinks (e.g. roots, fruits, growing tips, storage organs). The main transported sugar in most plants is sucrose, because it is non-reducing, highly soluble and metabolically inert in transit, so it does not react with other solutes in the phloem sap.

Understanding the Question

The question asks for a single example of an organic compound moved in the phloem of an iris root. The compound must be organic (carbon-containing) and translocated (transported in the phloem), not something carried in the xylem.

Approach

Choose the most common, textbook example: sucrose. The mark scheme lists sucrose, amino acids, peptides/polypeptides/proteins, RNA, and plant hormones as acceptable answers, but explicitly rejects minerals — these are xylem-transported and would not be credited here.

Step-by-Step Reasoning

• Sucrose is the principal assimilate translocated in most plant phloem.
• It is synthesised in mesophyll cells (sources) and loaded into sieve tubes at the source end.
• It is then carried by mass flow to sinks such as roots, where it may be respired or stored as starch.

Key Takeaways

• The default answer to "what is translocated in phloem?" is sucrose.
• Other acceptable answers exist (amino acids, hormones, RNA, some proteins), but minerals are rejected — they travel in xylem.

Common Mistakes

• Writing "glucose" — glucose is rarely the major transported form in phloem because it is a reactive reducing sugar; sucrose is preferred.
• Writing a mineral ion (e.g. "nitrate", "magnesium") — explicitly rejected by the mark scheme; minerals travel in xylem.

Things to Be Careful About

Any named organic compound that is genuinely translocated in phloem earns the mark — sucrose is the safest, most widely credited answer.

Background Concept

Transcription is the first stage of protein synthesis, in which a length of DNA is copied into a single-stranded molecule of messenger RNA (the primary transcript). Only one of the two DNA strands — the template (antisense) strand — is read by RNA polymerase. The other strand is the coding (sense) strand and has the same base sequence as the mRNA (except that T in DNA corresponds to U in RNA).

The base-pairing rules for transcription are:

• A (DNA) pairs with U (mRNA)
• T (DNA) pairs with A (mRNA)
• C (DNA) pairs with G (mRNA)
• G (DNA) pairs with C (mRNA)

Crucially, RNA uses the base uracil (U) in place of thymine (T), so any A in the template DNA appears as U in the mRNA.

Understanding the Question

The question gives the 12-nucleotide sequence of a DNA template strand running from position 1 to 12. We are asked to write the corresponding 12-nucleotide primary transcript (mRNA). No further processing (5′ capping, poly-A tail, splicing) is being asked for — just the raw RNA copy.

Approach

Work position by position, writing the RNA complement of each DNA nucleotide. The two strands are antiparallel, but because the table gives the template nucleotides in the order in which they are read by RNA polymerase, the mRNA can simply be written position-by-position beneath.

Step-by-Step Reasoning

Applying the base-pairing rules to each position:

• Pos 1: C → G
• Pos 2: A → U
• Pos 3: C → G
• Pos 4: T → A
• Pos 5: A → U
• Pos 6: C → G
• Pos 7: T → A
• Pos 8: C → G
• Pos 9: C → G
• Pos 10: A → U
• Pos 11: A → U
• Pos 12: C → G

So the completed primary transcript is G–U–G–A–U–G–A–G–G–U–U–G.

Key Takeaways

• The primary transcript is a complementary RNA copy of the template DNA strand.
• Uracil (U) replaces thymine (T) in RNA, so A in the DNA template is transcribed as U in the mRNA.
• RNA polymerase reads the template 3′→5′ and synthesises the new RNA strand 5′→3′.

Common Mistakes

• Writing T instead of U in the mRNA — the molecule is RNA, so every complementary base to a DNA A must be U, not T.
• Pairing the wrong way round (e.g. writing A opposite T) — only complementary base pairing is allowed; there is no T–T or A–A pair.
• Skipping a position or miscounting columns in the table.

Things to Be Careful About

• This is transcription (DNA → RNA), not replication (DNA → DNA) and not translation (mRNA → protein). The base-pairing partners in transcription use U where DNA would have used T.
• The question is part of a sequence, so getting position 3 correct matters for parts (iii) and (iv) which refer back to it.

Background Concept

The genetic code is read in non-overlapping groups of three nucleotides called codons. A codon in mRNA corresponds to a triplet on the template DNA strand. The codon table is degenerate: most amino acids are specified by more than one codon, but each codon specifies only one amino acid. Table 4.2 in this question lists, for each amino acid, every possible DNA template triplet that codes for it.

The relationship between a DNA template triplet and its mRNA codon is complementary and antiparallel. In this question, Table 4.2 already gives the template strand triplets, so each codon in mRNA can be deduced from the complement of the DNA triplet shown — but the exam shortcut is to read Table 4.2 directly: any DNA template triplet listed there corresponds to that amino acid.

Understanding the Question

The 12-nucleotide template sequence must be split into four consecutive, non-overlapping triplets starting at position 1. Each triplet in turn codes for one amino acid, labelled aa1, aa2, aa3 and aa4. The amino acid labels in the original Table 4.1 sit at positions 2, 5, 8 and 11 — the middle nucleotide of each triplet — which is a useful cross-check that the reading frame begins at position 1.

Approach

• Slice the template into triplets: positions 1–3, 4–6, 7–9, 10–12.
• Look each triplet up in Table 4.2 and read off the corresponding amino acid.

Step-by-Step Reasoning

Template DNA: C A C | T A C | T C C | A A C

• Triplet 1 (positions 1–3): CAC — listed under val. → aa1 = val
• Triplet 2 (positions 4–6): TAC — listed under met. → aa2 = met
• Triplet 3 (positions 7–9): TCC — listed under arg. → aa3 = arg
• Triplet 4 (positions 10–12): AAC — listed under leu. → aa4 = leu

Cross-check via the mRNA codons (GUG, AUG, AGG, UUG): all four are standard codons and match the amino acids above.

Key Takeaways

• The genetic code is read in non-overlapping triplets from a fixed start position.
• A codon table can be used to translate a DNA template (or its mRNA) directly into an amino acid sequence.
• The code is degenerate: more than one triplet can specify the same amino acid, but the converse is not true.

Common Mistakes

• Reading the triplets with an off-by-one error (e.g. starting at position 2 instead of 1) — the amino acid labels at positions 2, 5, 8 and 11 should be the middle base of each codon.
• Confusing the DNA template with the mRNA codon when using the table; the table here lists template DNA triplets, so just match directly.
• Mixing up the rows of Table 4.2 — for example, mistaking TCC for leu (it is arg).

Things to Be Careful About

• A single mistake in the triplet grouping cascades into wrong answers for all four amino acids, costing the only mark for this sub-part (it is an "all four correct" mark).
• The amino acid label sits at the middle nucleotide of each codon, which can be used as a sanity check.

Background Concept

A substitution mutation replaces one nucleotide with another at a single position. The consequence depends on whether the new triplet still codes for the same amino acid. The genetic code is degenerate (or redundant): most amino acids are specified by more than one codon, often differing only in the third (wobble) position. Substitutions at this third position are frequently silent — they change the codon but not the amino acid, leaving the protein unchanged.

Understanding the Question

The nucleotide at position 3 of the DNA template is the C that completes the first codon (CAC, coding for val). The question asks what happens if this single nucleotide is substituted — i.e. replaced by a different base — and why.

Approach

• Identify which codon the affected nucleotide belongs to: position 3 is the third base of the first triplet (CAC).
• List every possible substitution at that position.
• Look each new triplet up in Table 4.2 and check whether the amino acid changes.

Step-by-Step Reasoning

The first triplet is C A C. The first two bases (CA) are unaffected, so any substitution at position 3 only changes the third base of this codon. The four possibilities are:

• C → A: gives CAA → still val (in Table 4.2)
• C → G: gives CAG → still val
• C → T: gives CAT → still val
• C (unchanged): CAC → val

Every possible substitution produces a triplet that still codes for valine. So the first amino acid is unchanged, and because this is the only codon that position 3 lies within, the rest of the protein is also unaffected.

The reason this works is the degeneracy (redundancy) of the genetic code — specifically, the fact that valine has four codons differing only in the third base, and that this third-base "wobble" position often does not change the amino acid specified.

Key Takeaways

• Substitution mutations are not always harmful; many are silent because the code is degenerate.
• Changes at the third base of a codon are particularly likely to be silent — a feature of the genetic code sometimes called the "wobble" position.
• Whether a substitution matters depends on which amino acid, if any, the new codon specifies, and how similar the new amino acid is to the original.

Common Mistakes

• Saying the protein will change without checking Table 4.2 — the data table is the key reference, and it shows the protein does not change.
• Calling the code non-overlapping or universal instead of degenerate / redundant — these are different properties of the code.
• Confusing this with the deletion mutation in part (iv), which does change the reading frame.

Things to Be Careful About

• Position 3 is inside the first codon; the substitution is contained within that one codon and does not shift the reading frame, so only that one amino acid could possibly be affected.
• A mark is available for stating that the code is degenerate — this is the underlying biological reason a substitution can be silent.

Background Concept

A deletion mutation removes one or more nucleotides from the DNA. If the number removed is not a multiple of three, the deletion shifts the reading frame of every codon downstream of the mutation — a so-called frameshift mutation. From the point of the deletion onwards, the ribosome reads entirely new triplets, almost always producing a completely different amino acid sequence. Frameshifts very often introduce a premature stop codon, truncating the polypeptide and almost always abolishing its function.

Understanding the Question

The question asks what happens if the single nucleotide at position 3 of the template is removed. We need to (1) re-read the new sequence in triplets from position 1, (2) check what the new codons code for, and (3) explain the consequences for the protein.

Approach

• Delete the C at position 3.
• Re-write the 11 remaining nucleotides.
• Re-read in triplets from position 1 — note the new first codon and all subsequent codons.
• Compare with Table 4.2 and explain why the protein changes.

Step-by-Step Reasoning

Original template: C A C T A C T C C A A C
Delete position 3: C A T A C T C C A A C

Re-read in triplets from position 1:

• New codon 1: CAT → val (still valine, because CAT is one of the valine codons in Table 4.2)
• New codon 2: ACT → not in Table 4.2; codes for a different amino acid (threonine)
• New codon 3: CCA → not in Table 4.2; codes for a different amino acid (proline)
• New codon 4: AC_ → incomplete; the sequence ends here

So the first amino acid is unchanged, but from codon 2 onwards every codon is read differently. In a real, longer sequence this would:

• alter the primary structure (the amino acid sequence after the mutation),
• very likely introduce a premature stop codon, producing a shorter polypeptide,
• and consequently alter the tertiary structure and the shape of the active site (if the protein is an enzyme), usually abolishing function.

The key point is that a single-nucleotide deletion is a frameshift mutation — it is not contained within a single codon the way a substitution is, and so it has effects that propagate all the way to the end of the protein.

Key Takeaways

• A deletion of a number of nucleotides that is not a multiple of three causes a frameshift.
• Frameshifts change every codon downstream of the mutation, almost always destroying protein function.
• The first amino acid may or may not be unchanged — here it happens to be unchanged because CAT still codes for val, but this is a coincidence and is not generally the case.

Common Mistakes

• Saying the protein is unchanged — this is a deletion, not a substitution, and the reading frame has shifted.
• Saying only one amino acid is affected — frameshifts affect all amino acids from the mutation point onwards.
• Forgetting to mention the possibility of a premature stop codon and a shorter polypeptide.
• Confusing deletion (this part) with substitution (part iii).

Things to Be Careful About

• The fact that the first amino acid is still val is a "lucky" consequence of the new codon CAT also being a val codon; it is not a general property of frameshifts.
• The mark scheme accepts any three of: first AA unchanged, reading frame change, all downstream AAs altered, possible premature stop / shorter polypeptide, altered tertiary structure / active site. Aim to give at least three of these for full marks.

Background Concept

Two types of electron microscope are used in biology:

• Transmission electron microscope (TEM) — a beam of electrons is passed through an ultra-thin specimen. The image is a 2D cross-section showing internal structures (e.g. ribosomes, internal membranes, chromosome).
• Scanning electron microscope (SEM) — electrons scan the surface of a specimen (which is usually coated in a thin layer of heavy metal, e.g. gold). Secondary electrons emitted from the surface are collected to build a 3D image of surface topography.

Both have much higher resolution than the light microscope because the wavelength of electrons is far shorter than that of visible light.

Understanding the Question

The question asks which type of electron microscope was used to produce Fig. 5.1. The image shows several curved, rod-shaped cells with long thin flagella clearly projecting from one end, all rendered with a strong 3D appearance — you can see the rounded contours and shadows of the cell surface.

Approach

Look at the image and decide whether it shows internal 2D structure (TEM) or external 3D surface detail (SEM). The 3D shaded appearance of whole cells on a textured surface is the giveaway for SEM.

Step-by-Step Reasoning

• The image is not a flat 2D slice; whole cells are shown in three dimensions.
• Surface features (texture of the cell wall, the curve of the flagella, the rounded ends of the rods) are clearly visible.
• This is exactly what an SEM produces — secondary electrons from the specimen's surface are used to construct a 3D topographical image.
• A TEM image of the same cells would appear as flat, electron-dense outlines with no surface shadowing.

Key Takeaways

• SEM → 3D surface images; useful for whole cells, surface structures (flagella, pili, cilia), external morphology.
• TEM → 2D internal images; useful for organelles, viruses, internal cell structure.

Common Mistakes

• Writing just "electron microscope" without specifying scanning — this does not earn the mark.
• Confusing SEM with TEM because the word "microscope" is in both — remember SEM is for surfaces, TEM is for thin sections.

Things to Be Careful About

• The clue is the 3D surface appearance. The mark scheme requires the word "scanning" (or the abbreviation SEM) — "transmission" would be wrong.

Background Concept

Cholera is an acute diarrhoeal disease of the small intestine caused by the bacterium Vibrio cholerae. The genus name Vibrio refers to the curved, comma-like shape of the cells (visible in Fig. 5.1). Pathogenic strains produce the cholera toxin (CT), which causes the secretion of large volumes of water and electrolytes into the gut lumen, leading to severe watery diarrhoea ('rice-water stools') and rapid dehydration.

Understanding the Question

The question gives an image of curved, flagellated bacteria and asks for the species name of the prokaryote that causes cholera. "Species" means the full binomial (genus species).

Approach

Recall the binomial name of the cholera pathogen. Write it in the accepted format: genus capitalised, species lower-case, both in italics.

Step-by-Step Reasoning

• Cholera is caused by the bacterium Vibrio cholerae.
• Vibrio = genus (capital V, italicised).
• cholerae = specific epithet / species (lower-case c, italicised).
• Writing only "Vibrio" would refer to the whole genus (which includes non-pathogenic marine species) and would not score the mark.

Key Takeaways

• The four major pathogens named on this syllabus are: Vibrio cholerae (cholera), Plasmodium falciparum / P. vivax / P. ovale / P. malariae (malaria), Mycobacterium tuberculosis / M. bovis (TB), and HIV (a virus, not given a binomial name).
• Always use binomial nomenclature — both genus and specific epithet — and italicise.

Common Mistakes

• Writing only "Vibrio" — genus alone does not specify the species.
• Capitalising the species: Vibrio Cholerae — the specific epithet is lower-case.
• Forgetting italics — binomials should be italicised (or underlined if handwritten).
• Spelling the species as colera or kolera — the correct spelling is cholerae.

Things to Be Careful About

• The cell in the image is described as a prokaryote — a useful sanity check that the answer should be a bacterium (binomial name) rather than a virus (which has no binomial, e.g. HIV is a name not a species).