Biology 9700/41 — May/June 2024

Background Concept

Cells in a multicellular organism must communicate to coordinate their activities. Several modes of cell signalling exist, distinguished mainly by the type of chemical used and the distance it travels:

• Endocrine signalling — a gland releases a hormone into the bloodstream, which carries it to distant target cells bearing the appropriate receptor.
• Paracrine signalling — a cell secretes a chemical messenger that diffuses only a short distance through the extracellular fluid to act on cells in the immediate neighbourhood. Local mediators such as histamine, prostaglandins and many growth factors work this way.
• Autocrine signalling — a cell signals itself.
• Juxtacrine signalling — direct cell-to-cell contact.

In every mode, the secreted chemical is a ligand that binds a specific receptor on the target cell, triggering a response inside that cell.

Understanding the Question

The stem of part (a) defines paracrine signalling and contrasts it with endocrine signalling. The command word is "identify", so you simply state the points — you do not need to explain them in detail. The question is worth 2 marks: 1 for the similarity, 1 for the difference.

Approach

Place the two definitions side by side and look for:

• a feature that is shared → similarity
• a feature that clearly distinguishes them → difference

The mode of transport of the chemical is the most obvious distinguishing feature; the use of a chemical signal is the most obvious shared feature.

Step-by-Step Reasoning

1. Similarity — both endocrine and paracrine signalling rely on a chemical messenger (a ligand) that is released by one cell and binds to a receptor on a target cell. The shared feature is therefore the use of a chemical signal.
2. Difference — the two systems differ in how far the chemical travels. Endocrine hormones enter the bloodstream and are carried around the whole body, while paracrine signals simply diffuse through the tissue fluid for only a short distance before reaching their target cells. The distinguishing feature is the route / distance of travel.

Key Takeaways

• The defining features of any signalling mode are the type of chemical and the distance it travels.
• Endocrine = blood-borne, long-distance; paracrine = local diffusion, short-distance.
• Both rely on a ligand binding a specific receptor on the target cell.

Common Mistakes

• Giving only the similarity (or only the difference) — the question asks for BOTH and the mark scheme awards one mark for each.
• Stating the difference as "endocrine is slower / faster" — the mark scheme rewards the mode of transport (blood vs diffusion) as the key distinguishing point.
• Confusing endocrine signalling with the nervous system. The nervous system uses neurones and is much faster; endocrine signalling uses blood-borne hormones.

Things to Be Careful About

• Use precise biology terms: "chemical", "signalling molecule" or "ligand" are all accepted by the mark scheme; "messenger" on its own is too vague.
• The difference must mention BOTH halves — endocrine travels in the blood AND paracrine diffuses only a short distance — for full credit.

Background Concept

Acetylcholine (ACh) is a neurotransmitter released by:

• cholinergic neurones in the central and peripheral nervous systems
• motor neurones at the neuromuscular junction (NMJ)

At the cholinergic synapse, an action potential arriving at the presynaptic terminal opens voltage-gated Ca²⁺ channels. Ca²⁺ influx triggers vesicles containing ACh to fuse with the presynaptic membrane, releasing ACh into the synaptic cleft (a gap of about 20–40 nm). ACh then diffuses across the cleft and binds to nicotinic (or muscarinic) receptors on the postsynaptic membrane, depolarising the postsynaptic cell. Acetylcholinesterase in the cleft rapidly breaks ACh down to terminate the signal.

Understanding the Question

The stem of part (a) defines a paracrine signal as a chemical that "diffuses a short distance to act upon cells that are very near to the secreting cell". You must explain why ACh fits that description. The command word is "explain", so you need both the mechanism and the reason it matches the definition (2 marks).

Approach

Recall the layout of a cholinergic synapse / NMJ and identify the features that match the paracrine definition:

• a chemical (ACh) is secreted;
• it travels by diffusion;
• the distance is very short (the synaptic cleft);
• the target is an adjacent cell.

You must state at least two of these features explicitly to earn both marks.

Step-by-Step Reasoning

1. ACh is a chemical secreted by one cell — the presynaptic neurone (or motor neurone at the NMJ) packages ACh into vesicles and releases it into the synaptic cleft when triggered by Ca²⁺.
2. It diffuses only a short distance — across the synaptic cleft, which is a very narrow gap (≈ 20–40 nm). The chemical does not enter the bloodstream and is not carried any distance; it simply moves by diffusion to the other side of the cleft.
3. It acts on an adjacent cell — the postsynaptic neurone (or skeletal muscle cell at the NMJ), which is the cell immediately next to the secreting cell.
4. It matches the paracrine definition because every element of the definition (chemical messenger, short diffusion distance, adjacent target cell) is satisfied.

Key Takeaways

• A neurotransmitter such as ACh is the textbook example of a paracrine signalling molecule.
• The defining feature is the very short diffusion distance — the synaptic cleft — not the nature of the chemical.
• The same logic applies to any neurotransmitter at any chemical synapse.

Common Mistakes

• Stating only that ACh is a "chemical messenger" — you must specifically say it diffuses a SHORT distance, and across what structure.
• Naming the wrong location (e.g. "across the blood" or "across the cell" is wrong). The mark scheme accepts "synapse", "synaptic cleft" or "neuromuscular junction".
• Confusing neurotransmitter release with hormone release — neurotransmitters do NOT travel in the blood.

Things to Be Careful About

• "Synaptic cleft", "synapse" and "neuromuscular junction" are all accepted by the mark scheme. Pick whichever fits the context you want to give.
• ACh binds to a receptor on the postsynaptic membrane; it does not enter the postsynaptic cell.
• The action is local and brief because acetylcholinesterase in the cleft rapidly hydrolyses ACh.

Background Concept

Stomatal conductance is a measure of how easily water vapour (and other gases) can pass through the stomata. It depends on the size of the stomatal pores — the wider the pore, the more water vapour diffuses out (transpiration) and the more $\text{CO}_2$ diffuses in. Guard cell turgor controls pore size: turgid guard cells open the pore; flaccid guard cells close it.

Abscisic acid (ABA) is a plant hormone produced in roots and leaves in response to water stress. It is transported in the xylem from roots to leaves, where it triggers stomatal closure. This reduces transpiration and conserves water when the plant is dehydrated.

Understanding the Question

"With reference to Fig. 3.1, describe the relationship between the concentration of ABA in the xylem sap and stomatal conductance." The command word describe requires the candidate to state the trend in words and to support it with evidence from the graph. The mark scheme requires: (1) the trend, (2) two x-y pairs of figures with units, and (3) a comment on the shape (steep initial decrease, then a plateau).

Approach

First, name the trend (negative correlation / inverse relationship). Then read off two specific x-y pairs with units to support it. Finally, describe the shape of the curve: steep decrease followed by a plateau.

Step-by-Step Reasoning

1. Identify the trend: as x (ABA concentration) increases, y (stomatal conductance) decreases. This is a negative correlation.
2. Read specific values from the graph to support the trend:
   • At ABA ≈ 50 $\mu\text{mol m}^{-3}$, conductance ≈ 2.2 $\text{mol m}^{-2}\text{ s}^{-1}$ (the highest data point)
   • At ABA ≈ 800 $\mu\text{mol m}^{-3}$, conductance ≈ 0.2 $\text{mol m}^{-2}\text{ s}^{-1}$ (the lowest cluster of points)
3. Describe the shape of the curve: the decrease is not linear. From 0 to about 150 $\mu\text{mol m}^{-3}$ ABA, conductance drops steeply from ~2.2 down to ~0.7–0.8 $\text{mol m}^{-2}\text{ s}^{-1}$. Above about 400 $\mu\text{mol m}^{-3}$ ABA, additional ABA has little further effect; the curve plateaus at a low conductance of ~0.1–0.2 $\text{mol m}^{-2}\text{ s}^{-1}$.
4. The shape indicates a saturating response — once enough ABA is present to close most stomata, extra ABA cannot close them any further.

Key Takeaways

• Reading values off a graph requires identifying the closest data point and including the units.
• A describe question on a graph always needs the trend (with an appropriate correlation term) AND supporting figures.
• The shape of the curve (here, exponential-decay-like) carries biological meaning: it indicates a saturable receptor response.

Common Mistakes

• Giving a trend without figures (no credit for the supporting point).
• Giving figures without units — the mark scheme requires units at least once.
• Saying the relationship is "linear" — it is clearly non-linear.
• Confusing x and y axes (ABA is x, conductance is y).
• Saying stomatal conductance increases with ABA — this is the opposite of what the graph shows.

Things to Be Careful About

• The mark scheme wants the trend (mp 1) and at least one set of x-y figures with units (mp 2); the shape (mp 3, 4) provides the remaining marks.
• "Negative correlation", "inverse correlation", or "as ABA increases conductance decreases" are all acceptable phrasings.
• Avoid saying "inversely proportional" unless you mean it strictly — the curve is not a strict inverse proportion.

Background Concept

Abscisic acid (ABA) is a plant hormone that signals water stress. It is synthesised in roots and leaves and travels in the xylem to the guard cells. ABA acts through a receptor (the PYR/PYL/RCAR receptor in the cytosol, with downstream signalling through PP2C and SnRK2 kinases) to trigger the loss of solutes and turgor from guard cells, closing the stomata. This is the reverse of the opening mechanism: instead of $\text{K}^+$ going in, $\text{K}^+$ goes out; instead of water entering, water leaves; instead of turgor increasing, turgor decreases.

The key second messenger is $\text{Ca}^{2+}$. ABA binding leads to a rise in cytoplasmic $\text{Ca}^{2+}$ concentration, which activates anion channels that release $\text{Cl}^-$ and malate$^{2-}$, depolarising the membrane. This in turn opens outward-rectifying $\text{K}^+$ channels, so $\text{K}^+$ flows out. The loss of solutes raises the water potential of the guard cells, water leaves by osmosis, and the cells become flaccid.

Understanding the Question

"Suggest explanations for the results shown in Fig. 3.1." The command word suggest means the candidate should apply biological knowledge to explain the data. The graph shows stomatal conductance falling as xylem ABA rises. The mechanism of ABA-induced stomatal closure must be described, and a candidate should also be able to comment on the shape of the curve (the plateau suggests receptor saturation).

Approach

Link the experimental observation (high xylem ABA → low conductance) to the molecular mechanism (ABA receptor binding → $\text{Ca}^{2+}$ rise → $\text{K}^+$ efflux → water loss → flaccid cells → closed stomata). The shape of the curve (plateau at high ABA) suggests receptor saturation — this is the AVP.

Step-by-Step Reasoning

1. ABA is delivered in the xylem sap from the roots to the leaves.
2. ABA binds to specific receptors on (or in) the guard cells.
3. This activates a signalling cascade that causes $\text{Ca}^{2+}$ ions to enter the cytoplasm from outside the cell (or from internal stores such as the vacuole or endoplasmic reticulum). $\text{Ca}^{2+}$ acts as a second messenger.
4. The rise in cytoplasmic $\text{Ca}^{2+}$ opens anion channels, releasing $\text{Cl}^-$ (and malate$^{2-}$). The membrane depolarises.
5. The depolarisation opens outward-rectifying $\text{K}^+$ channels; $\text{K}^+$ ions leave the guard cells down their electrochemical gradient.
6. Loss of solutes raises (makes less negative) the water potential of the guard cells, so water leaves by osmosis.
7. The guard cells become flaccid, lose turgor, and the stoma closes. This is observed as a fall in stomatal conductance.
8. AVP: At very high ABA concentrations the receptors are saturated, so additional ABA cannot cause further closure — this is why the curve plateaus above ~400 $\mu\text{mol m}^{-3}$.

Key Takeaways

• ABA-induced closure is the reverse of the opening mechanism: $\text{K}^+$ leaves rather than enters, water leaves rather than enters, the cells become flaccid rather than turgid.
• $\text{Ca}^{2+}$ is the key second messenger — it is the link between the ABA signal at the membrane and the ion channels that execute the response.
• A saturating curve (plateau) is the signature of a receptor-limited response — a useful general concept.

Common Mistakes

• Saying the guard cells "become turgid" — they become flaccid (the opposite of opening).
• Saying $\text{K}^+$ enters the guard cells — it leaves.
• Confusing ABA signalling with auxin or gibberellin signalling — they use different mechanisms.
• Missing the link to the receptor — just saying "ABA causes closure" without the receptor step loses marks.
• Missing the second-messenger role of $\text{Ca}^{2+}$.

Things to Be Careful About

• The order matters: receptor binding → $\text{Ca}^{2+}$ rise → $\text{K}^+$ efflux → water loss → flaccid cells → closed stomata.
• The mark scheme allows up to 5 points plus one AVP; giving 4 of the listed points is enough for full marks.
• "Flaccid" or "less turgid" is acceptable for the final cell state.
• The AVP for the plateau is a strong candidate answer — it explains a feature of the data that the basic mechanism does not.

Background Concept

The polymerase chain reaction (PCR) is an in vitro technique that amplifies a specific DNA sequence exponentially. It needs: a DNA template, two short single-stranded DNA primers complementary to the flanks of the target sequence, free deoxynucleotide triphosphates (dNTPs), a heat-tolerant DNA polymerase (Taq, originally isolated from Thermus aquaticus) and a buffer with $\text{Mg}^{2+}$ ions. The reaction is carried out in a thermocycler that rapidly cycles between temperatures. Because each new strand acts as a template in the next cycle, the number of target copies roughly doubles per cycle.

Understanding the Question

The question follows the eDNA workflow used to detect surviving A. pallipes along a river. (b)(i) asks you to outline AND explain the steps of PCR — i.e. for each step you must name both what happens and why that temperature is needed. Four marks are available.

Approach

Recall the three temperature stages and the molecular event at each, plus the reason the cycle is repeated. Present them in the order they occur in the thermocycler.

Step-by-Step Reasoning

• Denaturation (90–98 °C). High temperature breaks the hydrogen bonds between complementary base pairs, separating the double-stranded DNA into two single strands. This temperature is needed to overcome the energy of those hydrogen bonds; without denaturation the primers could not access their target sequences.
• Annealing (50–65 °C). Cooling allows the short synthetic primers to base-pair (anneal) with their complementary sequences at the two flanks of the target region on each single strand. The lower temperature is needed so that primers (which are short) form stable base pairs without the DNA strands simply re-annealing to each other.
• Extension (68–75 °C). Raising the temperature to the optimum for Taq polymerase allows it to add free nucleotides to the 3′ end of each primer, synthesising a new complementary DNA strand. Taq is used because it remains active at the high denaturation temperature; ordinary DNA polymerase would be denatured.
• Repetition. Each new strand acts as a template in the next cycle, so the number of target copies roughly doubles per cycle. Repeating the cycle ~25–35 times produces millions of copies from a single starting molecule — enough DNA to detect.

Key Takeaways

• The three temperatures correspond to denaturation, annealing and extension.
• Taq polymerase is essential because it survives the high denaturation temperature.
• Each cycle roughly doubles the amount of target DNA — amplification is exponential, not linear.

Common Mistakes

• Omitting the temperatures — the mark scheme requires them.
• Saying 'DNA is added' or 'primers are made' — primers are added once at the start.
• Confusing the order (e.g. extension before annealing) — PCR runs hot–cool–warm each cycle.
• Describing 'helicase unwinds the DNA' — wrong; PCR uses heat, not helicase.
• Forgetting to explain WHY each temperature is used (the question says 'outline AND explain').

Things to Be Careful About

• Quote temperatures as ranges (the mark scheme gives 90–98, 50–65, 68–75).
• The mark scheme uses 'denatures / strands separate' for step 1 — 'separate' alone is acceptable.
• The primer-annealing step is sometimes called 'binding' — both are accepted.

Background Concept

PCR primers are short (~18–25 nt) single-stranded oligonucleotides that must base-pair with the DNA flanking the target region. To design them, you need to know the sequence of the target DNA so that you can choose stretches complementary to the two primer-binding sites. Modern bioinformatics tools compare, annotate and search published DNA sequences in databases such as NCBI GenBank, Ensembl, EMBL-EBI and UniProt, then software such as Primer3 designs suitable primer pairs.

Understanding the Question

Part (b)(ii) asks how the researchers obtained suitable primer sequences for A. pallipes. The parent stem notes that the primers were 'specific to A. pallipes DNA', meaning they had to be designed from prior knowledge of the species' DNA sequence.

Approach

Identify the standard source of primer sequences: an online DNA sequence database accessed via bioinformatics tools.

Step-by-Step Reasoning

• The A. pallipes genome (or at least the relevant gene region) has already been sequenced and deposited in a public database.
• Researchers used bioinformatics (e.g. BLAST searches, Primer3) to locate the sequence and identify short flanking regions that would serve as primer-binding sites.
• The primer sequences were then synthesised commercially.
  The mark scheme accepts 'a (named) database / bioinformatics' as the answer.

Key Takeaways

• Primer design depends on prior knowledge of the target DNA sequence.
• Bioinformatics and sequence databases (GenBank, Ensembl, EMBL) underpin modern molecular biology.

Common Mistakes

• Saying 'they read the DNA' — too vague; specify a database or bioinformatics.
• Confusing this with reverse genetics or RNA-seq — irrelevant here.

Things to Be Careful About

• A named database (e.g. GenBank) is acceptable but not required; 'a database / bioinformatics' is sufficient.

Background Concept

In quantitative PCR (qPCR) a fluorescent dye (commonly SYBR Green) binds to double-stranded DNA and the fluorescence is monitored after each cycle. The fluorescence curve shows a flat baseline phase (signal too low to detect), an exponential phase (the dye-bound DNA doubles each cycle) and a plateau (reagents become limiting). The cycle at which the fluorescence first crosses a fixed 'threshold' is the Ct (cycle threshold). Because each cycle doubles the amount of DNA, the sample that started with the highest concentration reaches the threshold in the fewest cycles.

Understanding the Question

Fig. 4.1 shows four standard curves (A, B, C, D) of known DNA concentrations, plus a dotted line for the A. pallipes sample. The question asks for the letter of the curve with the highest starting concentration. The further LEFT a curve crosses the threshold, the more starting DNA it had.

Approach

Read off which standard curve first reaches the horizontal threshold line at fluorescence $\approx 0.2$. From the figure, curve A crosses the threshold at the lowest cycle number (around cycle 12–13), so curve A had the highest starting concentration.

Step-by-Step Reasoning

• Looking at Fig. 4.1, the four standard curves cross the threshold in the order A (earliest) → B → C → D (latest). The dotted sample line crosses the threshold later still (around cycle 28).
• The earlier a curve crosses the threshold, the more starting DNA was present — because fewer doubling cycles were needed to reach the detectable fluorescence.
• Therefore curve A has the highest starting concentration of DNA.

Key Takeaways

• In qPCR, the cycle threshold (Ct) is inversely related to the starting DNA concentration.
• A standard curve of serial dilutions lets you compare an unknown sample's Ct to known standards and back-calculate the original concentration.

Common Mistakes

• Picking the curve with the highest fluorescence plateau — wrong; the plateau is set by reagent limitation, not starting amount.
• Picking the rightmost curve assuming it has 'more' DNA — it actually started with LESS and needed more cycles to be detected.

Things to Be Careful About

• The fluorescence axis is logarithmic — don't be misled by the shape.
• All curves share the same plateau height by the end, because reagents run out at similar absolute amounts of DNA.

Background Concept

In PCR (and qPCR) the amount of target DNA doubles with each cycle, so after $n$ cycles the amount is $2^n$ times the starting amount. If two samples reach the same fluorescence threshold but one took $n$ more cycles than the other, then the first sample must have started with $2^n$ times MORE DNA than the second (equivalently, the second has $2^n$ times less).

Understanding the Question

You are told the A. pallipes sample reaches the threshold at 28 cycles and curve C reaches it at 25 cycles. You must calculate the relative difference in starting concentrations. The key relationship is that each PCR cycle doubles the DNA.

Approach

Find the difference in cycle number, then raise 2 to that power. The answer is the fold-difference.

Step-by-Step Reasoning

• Cycle difference: $28 - 25 = 3$ cycles.
• DNA doubles per cycle, so the sample that reached the threshold 3 cycles earlier must have started with $2^3$ times more DNA.
• Curve C started with $2^3 = 8$ times more DNA than the A. pallipes sample.
• Equivalently, the A. pallipes sample started with 8 times less DNA than C.

Key Takeaways

• PCR is exponential: $2^n$ over $n$ cycles.
• The threshold-cycle difference directly gives the fold-difference in starting DNA.

Common Mistakes

• Forgetting to take the difference in cycles first.
• Multiplying rather than raising 2 to the power (e.g. saying '3 × 2 = 6').
• Inverting the answer — stating '8 times more' instead of '8 times less' (or vice versa) without making clear which sample is the reference.
• Saying 'the sample has 8 times the DNA' without specifying relative to what.

Things to Be Careful About

• The mark scheme accepts either '8' alone or '$2^3 = 8$' as the working.
• The question is about RELATIVE difference, not absolute concentration.
• Choose one direction ('times less' or 'times more') and state it clearly; either is correct as long as the direction is unambiguous.

Background Concept

Bacteria do not naturally take up whole chromosomes from other species, but they can be persuaded to take up small, circular, self-replicating DNA molecules called plasmids. By inserting a target gene (along with a promoter the bacterium's RNA polymerase can recognise) into a plasmid, the bacterium's own transcription and translation machinery expresses the gene as a protein. The plasmid thus acts as a cloning and expression vector. A selectable marker (often an antibiotic-resistance gene) is usually included so that only transformed bacteria grow on selective agar.

Understanding the Question

Fig. 4.2 shows normal (non-fluorescent) bacteria on the left and genetically modified, fluorescent bacteria on the right of an agar plate. (d)(i) asks: what component do normal bacteria take up to become the GM strain, and what must that component contain for the bacteria to make green fluorescent protein (GFP)?

Approach

Identify the vector (a plasmid), then name the two essential features inside it: the GFP gene itself and a promoter that the bacterium can recognise.

Step-by-Step Reasoning

• The component is a plasmid (vector). Bacteria readily take up plasmids, especially after treatments such as CaCl₂ plus heat-shock or electroporation. The plasmid replicates independently inside the bacterium, copying the inserted gene along with its own DNA.
• It contains the gene (DNA) for green fluorescent protein. The GFP gene originally comes from the jellyfish Aequorea victoria. It codes for a protein that fluoresces green under UV / blue light. Without this gene, the bacterium cannot make GFP.
• It contains a (prokaryotic) promoter. A promoter is the DNA sequence that RNA polymerase binds to in order to start transcription. Bacteria will only transcribe genes downstream of a promoter they recognise. The plasmid must therefore carry a bacterial (prokaryotic) promoter upstream of the GFP gene so that the host's RNA polymerase initiates transcription.
  Additional features such as a selectable marker gene (e.g. antibiotic resistance) and restriction sites for inserting the GFP gene are typically also present, but the question only asks what is needed 'to allow the bacteria to make green fluorescent protein'.

Key Takeaways

• A plasmid is a small, circular DNA molecule used as a vector to introduce new genes into bacteria.
• For a bacterium to express a foreign gene, the plasmid must carry the gene itself AND a promoter the bacterium's RNA polymerase can recognise.
• GFP from Aequorea victoria is widely used as a visual marker.

Common Mistakes

• Saying 'a gene' without specifying that it is the GFP gene.
• Forgetting the promoter — without it the bacterium cannot transcribe the gene.
• Confusing 'gene' with 'allele' or 'protein'.
• Saying the bacteria take up 'DNA' generically — too vague; specify a plasmid / vector.

Things to Be Careful About

• The mark scheme requires three explicit points: (1) plasmid / vector, (2) GFP gene, (3) prokaryotic promoter.
• 'Promoter for bacteria' or 'bacterial promoter' is acceptable for the third point.

Background Concept

A marker (or reporter) gene is a gene whose phenotype is easily detected and which is co-transferred with the gene of interest. Common markers include antibiotic-resistance genes (selection on selective agar) and fluorescent-protein genes (visual identification under UV / blue light). If the marker is expressed, the cell must contain the construct — so the marker tells you whether transformation was successful.

Understanding the Question

Part (d)(ii) asks why the GFP gene is sometimes transferred IN ADDITION to the desired gene. The implication is that GFP is not itself the gene of interest but a tool.

Approach

Recognise that GFP is a marker / reporter. Its purpose is to identify cells that have taken up the plasmid — those cells will glow under UV light.

Step-by-Step Reasoning

• When a plasmid carrying both the gene of interest AND the GFP gene is taken up by a bacterium, the bacterium will express both proteins.
• Under UV / blue light, only successfully transformed bacteria fluoresce green — non-transformed bacteria do not.
• This makes it easy to visually identify and select GM bacteria (or any GM organism) that carry the desired gene.
• It also confirms that the construct is being expressed, not merely carried.

Key Takeaways

• Marker / reporter genes allow rapid identification of successfully transformed cells.
• GFP is a visual marker; antibiotic-resistance genes are selectable markers.
• Co-transferring a marker is a standard element of most genetic-engineering constructs.

Common Mistakes

• Saying 'to make the bacteria glow for fun' — misses the purpose.
• Saying 'so the bacteria produce GFP' without explaining the identification role.
• Confusing a marker gene with a selectable marker (they overlap, but 'marker' is the broader term the mark scheme uses).

Things to Be Careful About

• The mark scheme accepts three phrasings: 'marker / reporter gene', 'identify recombinant organisms', or 'show expression of another transferred gene' — any one is sufficient for the single mark.

Background Concept

A chloroplast has two internal aqueous regions separated by the thylakoid membrane. The stroma is the fluid matrix that surrounds the thylakoids and contains the enzymes of the Calvin cycle (including rubisco). The thylakoid space (also called the thylakoid lumen) is the interior of each thylakoid disc, enclosed by the thylakoid membrane. The thylakoid membrane carries the photosystems, electron carriers, and ATP synthase used in the light-dependent stage.

Understanding the Question

The figure shows the thylakoid membrane in cross-section with PSII and PSI embedded in it. Area A lies outside the thylakoid (on the same side as NADP being reduced to product D and ATP being made from ADP + Pi), while area B lies inside the thylakoid (where H⁺ accumulates and from which H⁺ flows back through ATP synthase). The task is to name these two aqueous compartments.

Approach

Match each labelled region to the correct compartment using its position relative to the thylakoid membrane and the events happening on each side: reduction of NADP and ATP synthesis occur on the stromal side; proton accumulation occurs on the luminal side.

Step-by-Step Reasoning

• A is on the side of the thylakoid membrane where NADP is reduced to D (reduced NADP) and where ADP + Pi is converted to ATP by ATP synthase. Both of these reactions occur in the stroma, so A = stroma.
• B is enclosed by the thylakoid membrane and is the region into which H⁺ ions are pumped during electron transport. This is the thylakoid space (also accepted as thylakoid lumen).

Key Takeaways

The stroma and the thylakoid lumen are chemically distinct compartments; the proton gradient that drives ATP synthesis is built up across the thylakoid membrane separating them.

Common Mistakes

Writing "grana" instead of stroma for A (grana are stacks of thylakoids, not the surrounding fluid); calling B "the thylakoid membrane" rather than the lumen/space inside it.

Things to Be Careful About

The mark scheme accepts "thylakoid lumen", "thylakoid space" or simply "thylakoid" (with lumen/space implied) for B — but rejects "inside the thylakoid membrane". Use one of the precise terms.

Background Concept

In the light-dependent stage of photosynthesis there are two routes by which electrons can flow. In non-cyclic photophosphorylation electrons pass from water (via PSII) through an electron-transport chain to PSI and finally to NADP, producing both reduced NADP and ATP. In cyclic photophosphorylation only PSI is used: an excited electron leaves PSI, passes through a chain of electron carriers (including an electron-transport complex analogous to the one between PSII and PSI), and returns to PSI. Only ATP is produced — no reduced NADP, no O₂, no photolysis of water. The energy released as electrons flow through the chain is used to pump H⁺ into the thylakoid lumen; the resulting proton gradient drives ATP synthesis as H⁺ flows back through ATP synthase (an enzyme that uses chemiosmotic coupling to phosphorylate ADP).

Understanding the Question

The figure shows PSI, the protein complex C on the stromal side of PSI (which receives electrons from PSI and passes them back), and protein E spanning the membrane with H⁺ flowing through it to make ATP. Note that no electrons are shown leaving to reduce NADP in this loop and PSII is not part of the cycle — this is the diagnostic feature of the cyclic pathway.

Approach

Identify which photosystem(s) the electrons are cycling through and what the membrane protein E (with H⁺ passing through it to make ATP) must be.

Step-by-Step Reasoning

• Electrons leave PSI, pass through complex C, and return to PSI — only PSI is involved, so this is cyclic photophosphorylation (mark 1).
• Protein E sits in the thylakoid membrane, has a channel through which H⁺ flows down their electrochemical gradient from the lumen to the stroma, and is shown making ATP from ADP + Pi as H⁺ passes through. This is the structure and function of ATP synthase (mark 2).

Key Takeaways

Cyclic photophosphorylation is an ATP-only pathway that uses PSI and an electron-transport complex. ATP synthase is the enzyme that uses the proton gradient to phosphorylate ADP.

Common Mistakes

Naming the process "non-cyclic photophosphorylation" (only PSI is used here, and no NADP reduction is shown). Calling protein E "a channel protein" or "a proton pump" — it is specifically ATP synthase; pumps pump H⁺ against the gradient, ATP synthase lets H⁺ flow down the gradient.

Things to Be Careful About

The mark scheme accepts "ATP synth(et)ase" so either spelling is fine. The process mark requires "cyclic photophosphorylation" — do not abbreviate to "cyclic phosphorylation" or describe it in words.

Background Concept

The two stages of photosynthesis are linked by ATP and reduced NADP produced in the light-dependent stage. In the light-dependent stage water is split at PSII, electrons flow through the electron-transport chain to PSI, and NADP is reduced to reduced NADP (NADPH) on the stromal side of the thylakoid membrane. Reduced NADP then carries that reducing power into the Calvin cycle, where it is used in the reduction step: CO₂ is first fixed by rubisco onto RuBP to give glycerate 3-phosphate (GP), and GP is then reduced to triose phosphate (TP) using both ATP (to phosphorylate GP) and reduced NADP (to supply the hydrogen). TP can then be used to regenerate RuBP or to build carbohydrates.

Understanding the Question

The figure shows NADP being converted into product D on the stromal side of the thylakoid. The question asks for the identity of D and for its specific role in the Calvin cycle. Three marks are available: one for identification and two for the role.

Approach

Identify D as the reduced form of NADP (NADPH), then state which substrate it reduces and what that substrate becomes. Use the exact Calvin-cycle abbreviations GP and TP that the mark scheme expects.

Step-by-Step Reasoning

• NADP gains electrons and H⁺ in the light-dependent stage to become reduced NADP, so D = reduced NADP (mark 1).
• In the Calvin cycle, the carbon-fixing enzyme rubisco produces GP from CO₂ and RuBP (mark scheme abbreviations).
• Reduced NADP supplies hydrogen (and electrons) to reduce GP, forming TP (mark 2 for reducing GP; mark 3 for forming TP).

Key Takeaways

Reduced NADP and ATP carry the chemical products of the light-dependent stage into the Calvin cycle. Reduced NADP is specifically used in the reduction of GP to TP; ATP is used in the same step and to regenerate RuBP. Together they couple the two stages of photosynthesis.

Common Mistakes

Writing "NADP" without "reduced" — NADP itself is the oxidised form and cannot reduce anything. Saying reduced NADP "provides energy" or "is used in respiration" — it supplies reducing power (hydrogen), not energy directly. Calling the product "glucose" or "G3P" — these are downstream, not the immediate product of this reduction step.

Things to Be Careful About

Use the abbreviations GP and TP exactly as the mark scheme does, and spell "glycerate (3-)phosphate" and "triose phosphate" in full at least once. State both halves of the reduction: GP is reduced TO TP. Don't say reduced NADP "produces" TP — it is a cofactor that reduces GP.

Background Concept

The rate of photosynthesis is limited by whichever environmental factor is in shortest supply relative to demand. The three classical limiting factors are light intensity, carbon dioxide concentration and temperature. Light intensity limits the light-dependent reactions directly; temperature affects the activity of Calvin-cycle enzymes (notably rubisco) and so limits the light-independent reactions; CO₂ concentration limits carbon fixation. When a factor is in short supply, raising the others does not increase the rate.

In real ecosystems, plants experience a combination of all three; the factor that is most different from the optimum controls the local rate. Tropical forests are warm, wet and brightly lit for most of the year; snow forests are cold, with a short growing season and low sun angles for much of that time.

Understanding the Question

Fig. 5.2 shows a bar chart of rate of energy transfer (in $\text{MJ m}^{-2}\ \text{yr}^{-1}$) for six ecosystems. The three forest ecosystems form a clear descending order — tropical forest highest, then temperate forest, then snow forest — and the question asks you to (1) describe that pattern with a paired data quote (i.e. read two values and compare them, including units) and (2) suggest the biological reason in terms of limiting factors.

Approach

First read the chart carefully: tropical ≈ 37, temperate ≈ 24, snow ≈ 15 (units $\text{MJ m}^{-2}\ \text{yr}^{-1}$). Then link each difference to an environmental variable that is biologically limiting — temperature and light intensity are the two variables the mark scheme rewards.

Step-by-Step Reasoning

• D1: rate of energy transfer is highest for tropical forest and lowest for snow forest, with temperate forest in between (mark 1).
• D2: quote paired values with units, e.g. tropical ≈ $37\ \text{MJ m}^{-2}\ \text{yr}^{-1}$ vs snow ≈ $15\ \text{MJ m}^{-2}\ \text{yr}^{-1}$ (mark 2).
• S3: tropical forest has the highest temperature of the three; cold temperatures slow enzyme-controlled reactions (mark 3 — S3 if D2 was used; otherwise substitute).
• S4: tropical forest has the highest light intensity and longest growing season, providing more photons per unit area per year (mark 4).
• S5: therefore temperature and/or light intensity are acting as limiting factors on photosynthesis in the cooler forests (mark 5).

Key Takeaways

A bar chart can demonstrate a limiting-factor effect if the only systematic environmental difference along the trend matches a known limiting factor. Reading off paired values with units is an essential skill on these questions, and the conclusion must name the factor.

Common Mistakes

Stating the trend without quoting numbers and units (no mark for D2). Saying "more rainfall" or "more biodiversity" as the cause — neither is a photosynthesis limiting factor. Saying "more plants" in tropical forest — the chart shows per-area energy transfer, so the rate per unit area, not total biomass, is what differs.

Things to Be Careful About

Always quote the units at least once in the answer (the mark scheme requires "with units at least once"). Limit your suggested factors to the three classical limiting factors (light, CO₂, temperature) — extra "explanations" that are not factors will not score.

Background Concept

Two environmental realities dominate grasslands and desert: water scarcity and open, less productive vegetation. Plants in dry, hot or windy habitats lose water rapidly through their stomata. To conserve water they close their stomata, but in doing so they also restrict the inward diffusion of CO₂ — so the Calvin cycle runs out of carbon substrate and CO₂ becomes a limiting factor even when light and temperature are favourable. In addition, deserts and grasslands support fewer individual plants per unit area, and those plants are often smaller (grasses, scattered shrubs) than forest trees. The total leaf area index is therefore much lower, so the proportion of incoming light actually intercepted by photosynthesis is much smaller. Forest canopies, by contrast, present multiple layers of leaves that together absorb almost all incoming light.

Understanding the Question

The question extends the comparison in Fig. 5.2 from forest-to-forest (5b i) to forests versus the open ecosystems. You must suggest why the open ecosystems have a lower rate of energy transfer by photosynthesis. Three marks are available, taken from a list that includes transpiration, stomatal closure, CO₂ limitation, plant cover, leaf area and light absorption.

Approach

Two parallel arguments both work here:

1. Water-loss / CO₂-limitation pathway: open, dry habitats → high transpiration → stomatal closure → CO₂ entry falls → CO₂ becomes limiting → photosynthesis rate falls.
2. Plant-cover / light-limitation pathway: open habitats have fewer, smaller plants → less total leaf area per m² → less light absorbed per m² → lower energy transfer per m².

Either pathway alone can score all three marks; combining them strengthens the answer.

Step-by-Step Reasoning

• Mark 1: grasslands/desert have more transpiration / water loss from leaves.
• Mark 2: stomata close, reducing CO₂ entry / gas exchange.
• Mark 3: therefore CO₂ (concentration) becomes a limiting factor.
• Alternative Mark 1: grasslands/desert have fewer plants / smaller plants / less plant cover.
• Alternative Mark 2: this gives a smaller total leaf surface area.
• Alternative Mark 3: therefore less light is absorbed.

Key Takeaways

The rate of photosynthesis per unit area depends both on how efficiently each leaf photosynthesises (limited by CO₂, light, temperature) and on how much leaf area is present per unit ground area. Open ecosystems suffer on both counts: water stress limits the CO₂ supply to each leaf, and sparse vegetation limits the total light intercepted.

Common Mistakes

Saying "because deserts are dry" without spelling out the mechanism (stomatal closure → CO₂ entry falls). Confusing plant biomass with leaf area index — the chart is per m² of ground, not per plant. Saying "less chlorophyll" — chlorophyll per leaf is not the issue; it is total leaf area and stomatal aperture.

Things to Be Careful About

The mark scheme accepts any three from the listed points; an AVP (additional valid point) can be credited. Make sure each sentence carries one distinct idea: don't combine "stomata close" with "CO₂ limiting" into one sentence — the mark scheme treats them as two separate points.

Background Concept

A homologous pair of chromosomes carries the same genes at the same loci — one chromosome inherited from the mother, the other from the father. As well as having the same loci, homologous chromosomes share a number of structural features that allow them to pair up precisely during prophase I of meiosis and to undergo crossing over. The features that define a homologous pair are: same length, same shape, centromere in the same position, and — when chromosomes are stained for karyotyping — the same banding pattern.

Understanding the Question

The stem has already told the candidate that homologous chromosomes share the same genes at the same loci. The question asks for ONE OTHER shared feature. This is a single-mark, single-fact recall item.

Approach

Recall the structural features that define a homologous pair and pick any one that is not already mentioned in the stem. The mark scheme accepts any one of: same size / length, same shape / position of centromere, or same banding pattern.

Step-by-Step Reasoning

The mark scheme offers three acceptable answers; the candidate needs only one. A complete sentence is not required — naming the feature is enough to score the mark. If pressed for the most reliable answer, 'same length' is the textbook definition.

Key Takeaways

• A homologous pair shares: the same genes, the same loci, the same length, the same shape, the same centromere position and the same banding pattern.
• These shared structural features are what allow synapsis and crossing over to occur accurately during meiosis I.

Common Mistakes

• Restating that they have 'the same genes at the same loci' — this is given in the question and gains no credit.
• Writing 'they have the same alleles' — this is incorrect; homologues may carry different alleles of the same gene, which is precisely why they are not identical.

Things to Be Careful About

• The question says 'one other feature', so do not pad the answer with extra features; one mark requires one correct feature.

Background Concept

The TYR gene codes for the enzyme tyrosinase, which is made inside melanocytes (pigment-producing cells). Tyrosinase catalyses the first step in the pathway that converts the amino acid tyrosine into melanin — the dark pigment that colours skin, hair and the iris. A recessive mutation in TYR usually alters the active site, producing a non-functional enzyme. Because albinism is recessive, a person only shows the phenotype if BOTH copies of the gene carry the mutant allele; one functional copy is enough to make sufficient tyrosinase for normal pigmentation.

Understanding the Question

The stem gives the inheritance pattern (recessive) and asks for an explanation of the MOLECULAR and CELLULAR basis of the phenotype. The command word is 'explain', worth three marks, so the answer must build a chain of reasoning: allele → protein → biochemical product → cell function → tissue phenotype.

Approach

Work outwards from the gene: identify what the protein is, what happens when it is faulty, which cells are affected, which biochemical product is missing, and which tissues show the visible phenotype. The mark scheme rewards any three of the following four points:

1. recessive (mutant) allele = faulty / inactive / no tyrosinase
2. (melanocytes) do not produce melanin / pigment
3. reference to skin / hair / iris
4. any valid extra point (AVP) — e.g. tyrosinase converts tyrosine to melanin; both alleles are non-functional so no working enzyme is produced.

Step-by-Step Reasoning

• A homozygous recessive individual has two mutant TYR alleles, so every tyrosinase molecule made is non-functional.
• Without working tyrosinase, melanocytes cannot convert tyrosine into the intermediates that polymerise to form melanin.
• The absence of melanin in skin, hair and iris produces the visible albino phenotype (very pale skin, white hair, pink/red iris because the blood vessels of the eye show through).

The mark scheme accepts any three of the four creditable points; the strongest answer names all of them.

Key Takeaways

• Genes code for proteins; a recessive mutation usually means the protein is non-functional.
• The phenotype (albinism) is the direct consequence of a missing enzyme in a specific cell type (melanocytes).
• This question links genotype → protein → cell biochemistry → tissue phenotype — the classic 'gene expression' chain.

Common Mistakes

• Saying 'the gene is not expressed' — too vague; the mark scheme requires the answer to specify that the protein (tyrosinase) is faulty or absent.
• Saying the person 'cannot make the right colour' — the precise wording is 'cannot make melanin / pigment'.
• Only describing the phenotype ('white skin and hair') without explaining the underlying biochemistry — this loses marks because the question asks to explain.

Things to Be Careful About

• Use the word 'tyrosinase' (the protein the gene codes for); do not substitute vague terms such as 'colour-making protein'.
• Name the cell type (melanocyte) and the tissues affected (skin, hair and iris) to access all three marks.
• 'AVP' on the mark scheme means any other valid biological point — for example, the two recessive alleles mean the heterozygote 'escape' (one working copy) does not apply, so no working enzyme is produced at all.

Background Concept

A mutation rate is the frequency at which new changes in DNA sequence arise in a germline, usually expressed per base pair per year. Mean lifespan is the average number of years an individual of a species lives. When two variables are measured across a number of species and plotted against each other, the resulting pattern is a correlation — a relationship, not a proof of cause. A negative (inverse) correlation means that as one variable increases, the other decreases.

Understanding the Question

Table 7.1 lists five mammal species with two measurements each: a mutation rate (per million base pairs per year) and a mean lifespan (years). The command word is describe, which only requires the candidate to state the trend shown in the data — not to explain why the trend exists.

Approach

Rank the species by mutation rate and see what happens to lifespan:

• human: 0.02 → 75 years
• horse: 0.05 → 30 years
• cattle: 0.08 → 20 years
• ferret: 0.20 → 8 years
• mouse: 0.28 → 3.6 years

As mutation rate rises (0.02 → 0.28), lifespan falls (75 → 3.6). One variable goes up while the other goes down: a negative correlation.

Step-by-Step Reasoning

• The species with the lowest mutation rate (human, 0.02) has the longest lifespan (75 years).
• The species with the highest mutation rate (mouse, 0.28) has the shortest lifespan (3.6 years).
• The intermediate species (horse, cattle, ferret) lie between these extremes in both columns, in the same order.
• Therefore: as mutation rate increases, lifespan decreases. This is the mark-worthy description.

Key Takeaways

• A 'describe' command word only needs the trend and its direction; no causal explanation is required.
• Negative correlation is the correct technical term, and 'as X increases, Y decreases' is the safest phrasing.

Common Mistakes

• Stating only 'they are related' without specifying the direction of the relationship — this does not earn the mark.
• Calling the relationship 'positive' — wrong: as one variable rises the other falls.
• Offering a causal explanation ('mutation causes ageing') when the question only asks to describe the pattern.

Things to Be Careful About

• Describe means what the data show, not why. Keep the answer to the observed pattern.
• 'Inverse' and 'negative' are both acceptable; 'indirect' is not standard and should be avoided.

Background Concept

Evolution by natural selection needs two ingredients: (1) genetic variation in the population, so that some individuals carry alleles better suited to a new environment, and (2) time — measured in generations, since each generation is one round of selection. Mutation is the ultimate source of new alleles; without it, selection has nothing new to act on. The number of generations per unit time is the inverse of the mean generation time, and for mammals this scales closely with lifespan.

Understanding the Question

The data in Table 7.1 show two features that both vary across the five species. The command word is explain, so the answer must connect these features to a biological outcome — the rate at which each species could adapt if the environment changed. The mark scheme offers five acceptable points and three marks are available.

Approach

• Ask: what does each of the two variables (mutation rate, lifespan) contribute to adaptation?
  • mutation rate → new alleles / new genetic variation
  • lifespan → generation time → number of generations per year
• Combine: a species high in both adapts fastest because it both generates more variation and tests it more often.
• Then apply to the named species in the table.

Step-by-Step Reasoning

• Point 1 (mutation → variation). A higher mutation rate per base pair per year means more new DNA changes arise in the germline each generation. This raises the standing genetic variation in the population and increases the chance that a beneficial allele — one conferring, say, resistance to a new pathogen or a new food source — will appear. Without such an allele, selection has nothing to favour and adaptation cannot occur.
• Point 2 (lifespan → generations). A shorter mean lifespan (e.g. mouse at 3.6 years) means a shorter generation time and therefore more generations per year than a long-lived species (e.g. human at 75 years). Each generation is one cycle of selection: the better-adapted individuals leave more offspring. The more cycles per unit time, the faster allele frequencies can change in response to a new selection pressure.
• Point 3 (combining the two). Mouse and ferret have both the highest mutation rates and the shortest lifespans, so they have the most new variation and the most generations in which selection can act on it. They are predicted to adapt fastest. Cattle, horse and human have low mutation rates and long lifespans, so they should adapt more slowly. The two effects compound.
• An additional valid point (AVP) would be to note that small effective population sizes can sometimes reduce the benefit of high mutation rates (drift can eliminate new variants), but this is not required.

Key Takeaways

• Rate of adaptation ≈ (supply of new beneficial alleles) × (number of generations per unit time) that selection can act on.
• Mutation rate supplies the alleles; lifespan determines the number of selection cycles per year.
• Mouse/ferret should evolve faster than cattle/horse/human when environments change.

Common Mistakes

• Only mentioning mutation rate or only mentioning lifespan, without linking both to adaptation.
• Saying 'they have more mutations' without stating that this gives more genetic variation / more alleles / more chance of a beneficial allele.
• Failing to make the final link to 'adapt / evolve faster'.
• Naming only 'mouse' when both mouse and ferret satisfy the criteria.
• Confusing the direction: saying species with high mutation rates adapt slower (they do not — the mark scheme explicitly says the opposite).

Things to Be Careful About

• Three distinct ideas are needed for three marks: variation, generation time, and the combined prediction of faster adaptation.
• 'Genetic variation', 'alleles' and 'beneficial allele' are the precise terms the mark scheme credits.

Background Concept

Selective breeding is the traditional method of changing the genetic makeup of a population: the breeder identifies individuals with the desired trait, allows only those individuals to reproduce, and repeats the process generation after generation. Each generation produces only a small shift in allele frequency because inheritance is halved at every cross and the chosen parents may not be homozygous. Genetic engineering inserts a chosen allele directly into fertilised eggs (or early embryos), which are then transferred into surrogate mothers. The engineered individuals carry the allele in the heterozygous state at minimum, and breeding from them in a single subsequent generation can fix it.

Understanding the Question

The stem defines the task precisely: compare the timescales needed for selective breeding and for genetic engineering to raise the frequency of a rare allele in cattle, and give a reason for the comparison. Two marks are available — one for the comparison and one for the reason. The candidate must therefore make an explicit comparative statement and then justify it.

Approach

• Recall the reproductive biology of cattle: one calf per pregnancy, ~9-month gestation, puberty at ~1 year. So a generation is at least a year, and increasing an allele from rare to common by selecting parents typically needs many generations.
• Recall the GE route: an allele can be inserted into many embryos at once, and every cow in the herd can be implanted with an engineered embryo.
• The timescale gap is therefore many generations (selective breeding) versus one generation (GE), i.e. decades versus months.

Step-by-Step Reasoning

• C1 — comparison. Genetic engineering is faster than selective breeding at increasing the frequency of a rare allele in cattle.
• R2 — generation-time reason. Selective breeding acts on the existing allele frequency generation by generation; in cattle, a generation is at least one year, so even under intense selection it takes many years (often decades) to shift a rare allele to a useful frequency.
• R3 — surrogate-mother reason (alternative). With genetic engineering, a single allele can be inserted into many embryos in the laboratory and all of the cows in the herd can be used as surrogate mothers, each carrying an engineered calf to term. This raises the frequency of the allele in the next generation dramatically — potentially from rare to majority in one step.
• Either R2 or R3 (or both) earns the second mark. R3 is the more striking reason and is the one the mark scheme highlights first.

Key Takeaways

• Selective breeding is slow because each generation produces only a small, statistical change in allele frequency.
• Genetic engineering is fast because it bypasses the need for natural or selected reproduction — surrogate mothers multiply the engineered embryos in parallel.
• The comparison and a reason are both required for full marks.

Common Mistakes

• Stating only that 'GE is faster' with no justification — the comparison earns one mark but the reason mark is lost.
• Giving the wrong direction ('selective breeding is faster') — the rare allele would have to be concentrated by slow, gradual selection, which is the very thing GE avoids.
• Confusing the timescale: in cattle a generation is roughly a year, so 'many generations' is many years, not many weeks.
• Saying 'GE is faster because it is more accurate' — accuracy is not what makes it faster; it is the parallel use of surrogates / avoidance of generations.

Things to Be Careful About

• Two distinct ideas for two marks: comparison, then reason.
• A valid reason must be biological (generations, surrogates), not vague ('because it is modern' is rejected).

Background Concept

Genetic engineering of farmed animals aims to introduce alleles that improve commercially important traits. For dairy cattle the most economically valuable outputs are the milk itself (its composition and the volume produced per cow per lactation) and the health and longevity of the cow (which affect how many lactations she completes in her lifetime). Disease resistance is a major target because infection (especially mastitis — inflammation of the udder) is one of the largest costs in dairy farming.

Understanding the Question

The stem names the context: genetic engineering of dairy cattle. The command word is state, so the candidate simply needs to name two features — no explanation is required. Two marks are available.

Approach

Ask 'what makes a dairy cow profitable to a farmer?' The answer falls into three broad buckets:

1. What comes out of her — quantity and quality of milk.
2. What keeps her healthy — resistance to common, costly diseases.
3. How efficiently she converts feed into milk — growth rate and feed conversion efficiency.

Any two of these (or a more specific sub-feature) earn the marks.

Step-by-Step Reasoning

• Milk quality. This covers the composition of the milk — higher protein or butterfat content, altered casein profile, reduced lactose for low-lactose milk, or the presence of added beneficial compounds. The mark scheme credits 'quality of milk' (or quality of the animal) without requiring a specific example, but a specific example strengthens the answer.
• Milk quantity / yield. Volume of milk per lactation is the most direct measure of dairy productivity. Engineering cows to produce more milk per day or per lactation is a well-established target.
• Disease resistance (AVP). Mastitis resistance is the most common example, but resistance to bovine tuberculosis, tick-borne diseases or other endemic infections are all valid.
• Growth rate / feed conversion efficiency (AVP). Faster growth or more efficient use of feed reduces the cost per litre of milk and shortens the time to first lactation.
• Any two distinct, plausible features earn both marks.

Key Takeaways

• Dairy cattle improvement targets either the product (milk) or the producer (cow health/efficiency).
• 'Quality' and 'quantity' are the two simplest, most general terms the mark scheme accepts.
• A specific example is not required but is rewarded (an AVP).

Common Mistakes

• Naming traits that apply to beef cattle (e.g. 'meat quality', 'carcase conformation') rather than dairy cattle — the question specifies dairy.
• Vague answers such as 'make them better' or 'improve them' — these do not name a feature and score zero.
• Repeating the same idea in two forms (e.g. 'more milk' and 'higher milk yield' are the same feature and would earn only one mark).
• Naming a feature that genetic engineering could not realistically target in cattle (e.g. 'intelligence' is not a commercial target).

Things to Be Careful About

• Two distinct features are needed for two marks.
• Keep the answer focused on dairy, not beef, traits.
• 'State' means a short, named feature — no explanation is required or expected.

Background Concept

Simpson's Index of Diversity is one of several diversity indices used in ecology to summarise how many different species are present in a sample and how evenly their individuals are distributed. It ranges from 0 (lowest diversity — all individuals belong to a single species) up to a maximum approaching 1 (very high diversity, with many evenly represented species).

The formula is:

where $n$ is the number of individuals of one species and $N$ is the total number of individuals across all species. Subtracting the sum from 1 converts a "dominance" measure (large when one species dominates) into a "diversity" measure (large when no species dominates).

Understanding the Question

Part (c)(i) is a 3-mark calculation. The candidate is given a partially completed table for the disturbed population, with one row (Gaultheria hapalotricha) and the total $\sum (n/N)^2$ still to complete, and must finish the calculation to obtain $D$. The undisturbed population is given later for comparison.

Approach

Three steps, each worth a mark:

1. Compute $\frac{n}{N}$ for Gaultheria hapalotricha (and its square).
2. Sum all the $(n/N)^2$ values.
3. Subtract the sum from 1 to give $D$.

Show working clearly so each step is visible to the examiner.

Step-by-Step Reasoning

Step 1 — proportions for Gaultheria hapalotricha
With $n = 12$ and $N = 62$:

Squared:

Step 2 — sum of $(n/N)^2$
Adding the seven squared proportions:

Step 3 — Simpson's Index

Sanity check — the disturbed population is dominated by Puya atra (28 of 62 individuals ≈ 45%), so $D$ should be well below 1; 0.728 is consistent with moderate-to-low diversity. By contrast, the undisturbed population has $D = 0.833$, indicating more even representation across species — consistent with the biological expectation that undisturbed habitat supports more diverse feeding.

Key Takeaways

• Simpson's Index uses individuals (not just species counts) — so a community with one super-abundant species scores lower than a community with several evenly represented species, even if both have the same species richness.
• $D$ is dimensionless and bounded between 0 and 1.
• The arithmetic is straightforward but each step (proportion, square, sum, subtract) must be shown explicitly to earn the marks.

Common Mistakes

• Using the species richness (number of species = 7) instead of the formula — this is not Simpson's Index and scores 0.
• Forgetting to square the proportion, or summing the proportions instead of the squared proportions.
• Rounding too early — squaring 0.19 instead of 0.194, which gives a slightly different total.
• Misreading $N$ — the total is 62, not 60 (60 bears were sampled but the total individuals across all food species is 62 because some faeces contained more than one prey item, or some contained none — the mark scheme uses 62).
• Reporting $D$ as a percentage, or with units — $D$ has no units and lies between 0 and 1.

Things to Be Careful About

• Match the precision given in the question (the other squared proportions are given to 3 decimal places), so report $(n/N)^2 = 0.037$ to match the mark scheme.
• Round only the final displayed value, not intermediate workings.
• The total in the final row is the sum of the $(n/N)^2$ column only — do not add the $n/N$ column.

Background Concept

Simpson's Index of Diversity increases as both the number of species and the evenness of their representation increase. A higher $D$ therefore corresponds to greater diversity in the community being sampled.

Understanding the Question

The candidate is told that the undisturbed population has $D = 0.833$ (calculated in part (c)(i) as $D = 0.728$ for the disturbed population). The 1-mark question asks what the difference between these two $D$ values shows about the two populations.

Approach

Identify which population has the higher $D$, then state what a higher $D$ means biologically. One clear sentence is enough.

Step-by-Step Reasoning

• $D_{\text{undisturbed}} = 0.833$ and $D_{\text{disturbed}} = 0.728$.
• $D_{\text{undisturbed}} > D_{\text{disturbed}}$.
• A higher $D$ means more diversity (more species, or the same number of species more evenly distributed).
• Therefore the undisturbed population eats a greater variety of food species than the disturbed population.

This matches the mark scheme: "undisturbed / higher D, has, larger / higher / more, (food), diversity / variation".

Key Takeaways

• $D$ is a comparative index — always interpret it in relation to another $D$, or against a benchmark.
• A higher $D$ reflects greater species diversity AND greater evenness of species representation, not simply more species.

Common Mistakes

• Saying "more species" without using the word "diversity" or "variation" — the mark scheme specifically rewards the term "diversity".
• Stating that the disturbed population has more diversity because it has fewer species — the direction of the comparison is the opposite.
• Vague answers such as "the populations eat different food" — the question asks specifically what the difference in $D$ shows.

Things to Be Careful About

• The marks reward the term "diversity" or a clear synonym ("variation"); avoid hedging with "number of animals".
• Keep the comparison explicit: state which population has the higher $D$ and what that means.