Biology 9700/33 — October/November 2024

Background Concept

In any practical investigation, "error" means anything that makes a measurement unreliable. It can be random (variation between repeats from small uncontrollable factors) or systematic (a consistent flaw in the apparatus or procedure). A creditable "source of error" answer is a specific, concrete step that visibly affects the data, not a vague phrase like "human error".

In this experiment catalase on the surface of the potato cylinder hydrolyses hydrogen peroxide, releasing oxygen that accumulates as foam. The height of that foam is used as a proxy for enzyme activity.

Understanding the Question

The stem says the length of the potato cylinders may not be precisely 4 cm and asks for ONE other source of error in the procedure. The procedure has been carried out, so the candidate must think about what they actually did and what could realistically be wrong with it.

Approach

Look for a step where the result depends on a subjective judgement or imprecise measurement, and name that step clearly. Avoid vague generalisations. The mark scheme credits only one specific answer for this item, so give that answer cleanly.

Step-by-Step Reasoning

The mark scheme answer is measuring the height of the foam. Why this is a real source of error:

• The foam/air boundary is not a sharp line; the foam is uneven and slopes at the edges.
• A ruler held against a curved test-tube is hard to read; parallax is easy.
• Different students would record slightly different values for the same foam.
• It introduces random error to every repeat.

Other flaws that exist in the procedure (e.g. volume of H not being exactly 5 cm³, potato cylinders of different initial enzyme content) are also real, but the mark scheme credits only the named answer above.

Key Takeaways

• A specific error must name the measurement AND say why it is unreliable.
• Vague "human error" is rejected by CIE.
• One sentence is enough for a 1-mark "identify one" item.

Common Mistakes

• Writing "human error" — too vague, rejected.
• Writing "the cylinders are different sizes" — too generic and overlaps with the length issue already given.
• Listing several errors when only one is asked for — wastes time and may dilute the specific credit.

Things to Be Careful About

• The question says "one other", so a single, specific answer is required.
• The mark is for the named flaw, not for a long paragraph.

Background Concept

A serial dilution progressively halves the concentration at each step by transferring a fixed volume from one container to the next and making the volume back up with solvent. Because each transfer carries only half of the previous concentration, the resulting concentrations form a geometric series (e.g. 100%, 50%, 25%, 12.5%…).

The dilution factor at each step is:

$$\text{dilution factor}=\frac{\text{volume transferred}}{\text{total volume after mixing}}$$

For the dilution factor to be 1/2, transfer volume must be half the total. Here, total = 60 cm³ and transfer = 30 cm³, giving 30/60 = 1/2.

Understanding the Question

The candidate is given a partially completed Fig 1.2 (beaker 1 with 60 cm³ of 100% ethanol E and 0 cm³ of water W; a curved transfer arrow already pointing into beaker 2). They must complete beakers 2, 3 and 4 with:

• A labelled arrow for the volume of ethanol transferred (30 cm³ from the previous beaker)
• A labelled arrow for the volume of water W added (30 cm³)
• A concentration label under each beaker

The aim is to end with 30 cm³ available in each beaker (the question states "30 cm³ of each concentration available to use").

Approach

Decide the total volume in each beaker. To leave 30 cm³ behind AND transfer 30 cm³ away, each beaker needs a total of 60 cm³. Apply the same rule at every step: take 30 cm³ from the previous beaker, add 30 cm³ of water, label with the halved concentration. Walk through the concentrations to confirm: 100% → 50% → 25% → 12.5%.

Step-by-Step Reasoning

Why each concentration halves:

• Beaker 2 receives 30 cm³ of pure ethanol + 30 cm³ of water → ethanol is now 30/60 = 50% of the mix.
• Beaker 3 receives 30 cm³ of 50% ethanol + 30 cm³ of water → ethanol is now 15 cm³ out of 60 cm³ = 25%.
• Beaker 4 receives 30 cm³ of 25% ethanol + 30 cm³ of water → ethanol is now 7.5 cm³ out of 60 cm³ = 12.5%.

Labels to add to Fig 1.2 on beakers 2, 3 and 4:

• Curved transfer arrow from the previous beaker: 30 cm³
• Vertical arrow into the beaker: 30 cm³ of water, W
• Label below the beaker: 50% ethanol, 25% ethanol, 12.5% ethanol respectively.

Key Takeaways

• Serial dilution by halves: transfer half the volume, top up with the same volume of diluent (water here).
• For each beaker to retain 30 cm³, the total must be 60 cm³.
• The concentration halves at every step: 100% → 50% → 25% → 12.5%.

Common Mistakes

• Using unequal volumes (e.g. 20 cm³ transfer + 40 cm³ water) — would still halve the concentration but leaves only 20 cm³ behind, not the 30 cm³ required.
• Forgetting to label the concentration under the new beakers.
• Drawing the curved transfer arrow pointing the wrong way (back into the previous beaker).
• Putting "100%" under every beaker, or labelling the wrong concentration.

Things to Be Careful About

• Each transfer must be 30 cm³ so the dilution factor is exactly 1/2.
• Only complete beakers 2, 3 and 4; beaker 1 is already labelled.
• The question asks for three additional concentrations to the original 100% — so the final four concentrations must be 100%, 50%, 25%, 12.5%.

Background Concept

Recording results in a properly formatted table is a core Paper 3 skill. Conventions:

• Independent variable (IV) in the left column with its unit.
• Dependent variable(s) (DV) in subsequent columns, each with its unit in the heading only — not in every cell.
• All conditions and replicates included.
• Data recorded to a precision that matches the measuring tool (here, the nearest mm).
• The trend should be visible from the table alone.

Understanding the Question

The student has just run the experiment and must put all five sets of results into one table:

• The 0% control from step 7 (cylinder soaked in water, then put into H₂O₂).
• The four ethanol concentrations from step 13/14 (12.5%, 25%, 50%, 100%).

Each set has two readings: foam height at 1 minute and at 2 minutes.

Approach

Build a 3-column table: IV in column 1; height at 1 min in column 2; height at 2 min in column 3. Use the IV heading "% concentration of ethanol" and the DV headings "height of foam at 1 minute / mm" and "height of foam at 2 minutes / mm". Put the 0% control at the top, then the four ethanol concentrations in ascending order. Enter each value to a whole millimetre.

Step-by-Step Reasoning

Mark scheme checks (each earns one mark):

1. ✓ Heading for IV: "% concentration of ethanol"
2. ✓ Heading for DV: "height of foam / mm" (with /mm in the heading, not in each cell)
3. ✓ Records results for 0% and for all four ethanol concentrations, at both 1 min and 2 min
4. ✓ Correct trend: foam height at the highest concentration (100%) is lower than at the lowest (0%)
5. ✓ All values recorded to the nearest whole millimetre

Why the trend goes downward: ethanol is an inhibitor of catalase. As its concentration rises, more enzyme molecules have their active sites disrupted, fewer enzyme–substrate complexes form per second, less O₂ is released per second, and so less foam accumulates. At very high ethanol concentrations the enzyme begins to denature (irreversible loss of shape), so the foam height is lowest at 100%.

Key Takeaways

• IV in column 1, units in the column heading — never in the cells.
• Include the control (0%) alongside the treatments.
• Whole numbers here because the ruler is read to the nearest mm.
• The trend should be obvious from the table without any extra commentary.

Common Mistakes

• Putting units in every cell instead of only the heading (e.g. "25 mm", "35 mm").
• Using decimals (e.g. "25.3 mm") — the question requires the nearest mm.
• Missing the 0% control row.
• Recording only one time point (e.g. only the 2-minute reading).
• Sorting concentrations in the wrong order so the trend is hidden.

Things to Be Careful About

• "Height of foam" must include the word "height" — it is the DV that is measured.
• Use a ruler divided in mm, not a measuring cylinder, to read the foam.
• The trend at the top of the table should clearly go downward as you read across the concentration column.

Background Concept

Rate is the change in a quantity divided by the time taken for that change. For a continuously accumulating product such as foam, two natural definitions exist:

• Cumulative rate = total accumulated height ÷ total time elapsed.
• Interval rate = change in height over a specific time interval ÷ length of that interval.

Both are valid; the mark scheme rewards either as long as the working is shown and the unit is included. The unit of rate here is mm min⁻¹ (millimetres per minute).

Understanding the Question

The candidate must calculate the rate of foam production at 1 minute AND at 2 minutes for the 100% ethanol cylinder, using their own recorded values. They must show working and quote a unit on each answer.

Approach

Take the student's two recorded heights (h₁ at 1 min, h₂ at 2 min) and apply:

$$\text{rate}=\frac{\Delta \text{height}}{\Delta \text{time}}$$

For "rate at 1 min", Δtime = 1 min. For "rate at 2 min", either use Δtime = 2 min on h₂ (cumulative) or use Δtime = 1 min on (h₂ − h₁) (interval over the second minute). Either earns the mark as long as the working is visible.

Step-by-Step Reasoning

Using example values h₁ = 5 mm and h₂ = 8 mm:

$$\text{rate at 1 minute}=\frac{5\ \text{mm}}{1\ \text{min}}=5\ \text{mm min}^{-1}$$ $$\text{rate at 2 minutes}=\frac{(8-5)\ \text{mm}}{1\ \text{min}}=3\ \text{mm min}^{-1}$$

Or, using the cumulative method:

$$\text{rate at 2 minutes}=\frac{8\ \text{mm}}{2\ \text{min}}=4\ \text{mm min}^{-1}$$

The mark scheme wants:

• ✓ working shown for both rates
• ✓ a unit (mm min⁻¹ or mm/min) on each rate

The actual numerical values depend on the student's own measurements from (a)(iii).

Key Takeaways

• Rate = change in quantity ÷ change in time. Always include units.
• Two acceptable definitions for "rate at 2 min"; pick one and show it clearly.
• The mark is for the method AND the unit, not just for any single number.

Common Mistakes

• Forgetting the unit (just writing "5" or "3" instead of "5 mm min⁻¹").
• Calculating only one rate.
• Dividing by 2 minutes when the question says "rate at 1 minute" (incorrect; Δtime = 1 min).
• Mixing height and time (numerator and denominator the wrong way round).

Things to Be Careful About

• Use the student's own h₁ and h₂ from (a)(iii); do not invent values.
• The unit is mm per minute; mm/s would also be acceptable if the timer was in seconds.
• Show the working — the mark is for the calculation method, not just the final number.

Background Concept

Once two rates have been calculated for the same condition (here 100% ethanol), they can be compared to see how the rate changes over the course of the reaction. As the reaction proceeds, the substrate (H₂O₂) is consumed; with less substrate available, the enzyme works more slowly per unit time, and the rate falls.

Understanding the Question

The question asks for a description of how the rate changes with time, using the candidate's own calculated rates from (a)(iv). The expected answer is a one-sentence statement of direction.

Approach

Compare the rate at 2 min with the rate at 1 min, and state which is larger (or whether they are the same). Support with the candidate's own numbers.

Step-by-Step Reasoning

Using example values from (a)(iv):

• Rate at 1 min = 5 mm min⁻¹
• Rate at 2 min = 3 mm min⁻¹ (interval method)

The rate at 2 min is lower than the rate at 1 min, so the rate decreases with time. This is the expected biological pattern: the substrate is being used up, so the enzyme has less to work on.

Key Takeaways

• Always state the direction (increase / decrease / no change).
• Quote the calculated values from (a)(iv) to support the statement.
• One sentence is enough.

Common Mistakes

• Describing the change in height instead of rate (height keeps increasing; rate can fall).
• Saying "the rate is slower" without saying it decreases with time.
• Adding biological explanation that the question did not ask for.

Things to Be Careful About

• The answer must be about rate, not height or volume.
• Use the candidate's own numbers, not generic numbers.

Background Concept

A control is a condition in which the variable being tested is absent. It tells us what happens without the treatment. Comparing treatments to the control reveals the effect of the treatment itself.

In this experiment, the 0% (water only) cylinder is the control. The variable being tested is ethanol concentration. The control therefore tells us how active catalase is when no ethanol is present.

Understanding the Question

The question asks why the step-7 result (the 0% cylinder) was important to the investigation. The candidate must explain the role of this result.

Approach

State clearly what the 0% cylinder shows (catalase activity with no ethanol) and why that matters (it is the reference against which the ethanol treatments are compared).

Step-by-Step Reasoning

The mark scheme answer is "shows the activity of catalase without ethanol". A fuller explanation:

• Without the 0% reading we would not know whether any change in foam height was caused by ethanol or by natural variation in catalase activity between cylinders.
• The 0% result is the baseline / reference against which every ethanol concentration is judged.
• Because catalase is uninhibited in 0%, foam height is maximal here; the difference between 0% and any higher concentration quantifies how much ethanol has inhibited the enzyme.

Key Takeaways

• A control must be present in any experiment that tests the effect of something.
• The control value is the reference for comparison.
• The specific phrase "without ethanol" is the credit-bearing phrase.

Common Mistakes

• Saying only "it was a control" without saying what the control shows.
• Saying "it shows the maximum activity" without linking this to the absence of ethanol.
• Writing more than one sentence when one suffices.

Things to Be Careful About

• The mark scheme wants "without ethanol" or equivalent — the candidate must explicitly name what is missing.
• One sentence is enough for a 1-mark item.

Background Concept

The dependent variable (DV) is what is measured in an experiment; the independent variable (IV) is what the experimenter changes. In a well-designed investigation, control variables are held constant so any change in the DV can be attributed to the IV.

Understanding the Question

A 1-mark recall item asking for the dependent variable in this investigation.

Approach

Identify what was actually measured. Do not confuse the DV with the IV or with something inferred from the DV.

Step-by-Step Reasoning

The procedure puts a potato cylinder into hydrogen peroxide solution and measures the height of the foam at 1 min and at 2 min. The thing measured is the dependent variable.

Mark scheme answer: height of the foam.

What is not the dependent variable:

• Concentration of ethanol (this is the IV — it is changed between beakers).
• Catalase activity (this is inferred from foam height, not measured directly).
• Volume of oxygen (this would be the DV if a gas syringe were used, but in this experiment we measure foam height).

Key Takeaways

• DV = what is measured.
• IV = what is changed.
• "Catalase activity" is a conclusion drawn from the DV, not the DV itself.

Common Mistakes

• Writing "catalase activity" — this is the biological interpretation, not what the ruler measures.
• Writing "the foam" (vague) or "amount of foam" (the question says "height", which is the precise term).
• Writing "concentration of ethanol" — that is the IV.

Things to Be Careful About

• Use the exact phrasing from the procedure: "height of the foam".
• One word / one short phrase is enough.

Background Concept

"Confidence in results" means reliability — would the experiment give the same answer if repeated? Improvements should therefore target either random error (by taking more readings or more repeats) or measurement error (by using better apparatus). The mark scheme lists three specific improvements; any two earn full marks.

Understanding the Question

A 2-mark item asking for TWO specific improvements that would increase confidence in the results of the catalase investigation.

Approach

Pick any two of the mark-scheme-listed improvements. State each as a concrete change to the procedure, not as a vague aspiration ("be more careful" is rejected).

Step-by-Step Reasoning

The three credited improvements, with the reasoning behind each:

1. More concentrations of ethanol (e.g. 6.25%, 75% as well as the existing 12.5%, 25%, 50%, 100%).

   • Why: more data points on the concentration axis give a smoother, more reliable trend between ethanol concentration and catalase activity. The student can also identify non-linear features (a plateau, a threshold).

2. Carry out repeats at each concentration and calculate a mean.

   • Why: a single cylinder may have more or less catalase than average, may have an uneven shape, or may be exposed to a slightly different surface area. Repeats average out this random variation and make the trend more reliable.

3. Use a gas syringe to measure the volume of O₂ released instead of measuring foam height.

   • Why: foam height is subjective (the boundary is uneven, the test-tube is curved, parallax is easy). A gas syringe gives a single sharp meniscus reading in cm³, which is objective and more precise.

Key Takeaways

• Improvements should be specific (name the change) and targeted (say what error they fix).
• More concentrations → better shape of trend.
• Repeats + mean → less random error.
• Better apparatus → less measurement error.

Common Mistakes

• "Be more accurate" / "be more careful" — too vague, rejected.
• "Use a larger sample" without specifying what "sample" means (more cylinders? more concentrations?).
• Suggesting changes that would alter the variable being tested (e.g. "use a different potato").
• Adding controls that are already present (e.g. "include a water control").

Things to Be Careful About

• The question says "increase the confidence" — usually meaning reliability, not validity.
• Vague answers are rejected; each improvement must be a specific, named change.
• Two improvements are required for 2 marks.

Background Concept

Plotting a graph on grid paper follows strict CIE conventions:

• Axes: IV on x, DV on y; each labelled with quantity and unit.
• Scales: chosen so the data uses at least half the grid in both directions and avoids awkward numbers (3s and 7s); every major gridline labelled.
• Points: plotted accurately with a sharp pencil, using small crosses (×) or a dot-in-circle (⊙).
• Line: a smooth thin line (curve or straight) joining the points; for non-linear data, a smooth curve, not a series of straight segments.

Understanding the Question

The candidate is given Table 1.2 (concentration of ethanol vs. PPO activity for five concentrations) and must plot the data on the grid in Fig 1.3 with the conventions above.

Note: the unit printed in the question is written unusually; it is intended to read μL dm⁻³ (microlitres per cubic decimetre).

Approach

Decide axes: IV (concentration of ethanol) → x-axis; DV (PPO activity) → y-axis. Choose the simplest scale that uses ≥ half the grid:

• x-axis: 100 to 2 cm (so 0–500 fits in 10 cm of the 18 cm-wide grid).
• y-axis: 0.5 to 2 cm (so 0–2.5 au fits in 10 cm of the 17 cm-tall grid).

Plot the five points with crosses. Connect with a smooth curve because the data is non-linear (steep drop at low concentrations, plateau at high concentrations).

Step-by-Step Reasoning

Mark scheme checks:

1. ✓ x-axis labelled "concentration of ethanol / μL dm⁻³" and y-axis labelled "PPO activity / arbitrary units (au)".
2. ✓ x-axis scale: 100 per 2 cm; y-axis scale: 0.5 per 2 cm, labelled at least every 2 cm.
3. ✓ All five points plotted accurately with × or ⊙.
4. ✓ Smooth thin line (curve) through all five points, joined point to point.

The shape of the curve:

• Starts at (0, 2.20) — the maximum.
• Drops steeply to (100, 1.40) and (200, 0.95).
• Levels off between (400, 0.70) and (500, 0.60).

This is the characteristic shape for an inhibitor approaching saturation / denaturation of the enzyme at high inhibitor concentrations.

Key Takeaways

• IV on x-axis, DV on y-axis.
• Scale must use at least half the grid and be free of awkward numbers.
• Plot with sharp pencil; use a clear × or ⊙.
• For non-linear data, draw a smooth curve (not a straight line).

Common Mistakes

• Swapping axes (concentration on y, activity on x).
• Choosing a scale that does not use enough of the grid (e.g. 100 per 1 cm gives a tiny graph; 100 per 5 cm wastes space).
• Plotting points with fuzzy circles or pencil dots rather than clear ×.
• Joining the points with straight line segments (the data is curved).
• Forgetting to label either axis with its unit.

Things to Be Careful About

• The unit "μL dm⁻³" must be written correctly (μ is the Greek letter mu, L is capital, dm⁻³ is per cubic decimetre).
• "au" or "arbitrary units" must appear on the y-axis — enzyme activity here has no SI unit.
• All five points must be visible and clearly plotted; the line must pass through each.

Background Concept

Enzymes are proteins that catalyse specific reactions. Their activity depends on the precise 3D shape of the active site. Anything that disrupts this shape, or that blocks the active site, reduces activity. Two key mechanisms:

• Inhibition: a molecule binds to the active site (competitive) or elsewhere on the enzyme (non-competitive), preventing substrate from binding. Fewer enzyme-substrate complexes form per second → less product per second → lower activity.
• Denaturation: a molecule disrupts the H-bonds and hydrophobic interactions that hold the enzyme's tertiary structure, irreversibly changing the shape of the active site. The enzyme can no longer bind substrate at all, even after the inhibitor is removed.

Understanding the Question

The data shows PPO activity decreasing as ethanol concentration increases. The candidate must explain why this pattern occurs, in terms of how ethanol affects PPO.

Approach

Pick any two of the four mark-scheme points. Use precise enzyme vocabulary (active site, enzyme-substrate complex, denaturation).

Step-by-Step Reasoning

The data trend:

• 0 μL dm⁻³ ethanol → 2.20 au (highest activity)
• 100 → 1.40 (sharp drop)
• 200 → 0.95
• 400 → 0.70
• 500 → 0.60 (lowest activity, plateau)

Two biological explanations, any two of which earn the marks:

1. As ethanol concentration increases, PPO activity decreases — the visible pattern in the data.

2. Ethanol acts as an inhibitor — it binds to PPO and prevents the substrate from binding or being converted efficiently.

3. As ethanol concentration increases, fewer enzyme-substrate complexes form per unit time — directly linking the loss of activity to the central catalytic event of enzyme function.

4. At higher concentrations, PPO begins to denature — the irreversible change of tertiary structure. This explains the plateau at high concentrations: once most of the enzyme molecules are denatured, raising the ethanol concentration further has little additional effect (the activity is already near its floor).

The curve shape (steep drop then plateau) is consistent with inhibition dominating at low concentrations and denaturation dominating at high concentrations.

Key Takeaways

• An inhibitor reduces the rate of enzyme-substrate complex formation.
• A denatured enzyme has lost its tertiary structure and cannot bind substrate.
• A curve that plateaus near the bottom suggests denaturation has set in.
• "Killed" is the wrong word for enzymes — they are not alive.

Common Mistakes

• Saying the enzyme is "killed" — enzymes are not living; the correct word is "denatured".
• Saying "the enzyme is denatured" without specifying higher concentrations (at low concentrations ethanol is an inhibitor, not a denaturant).
• Saying the substrate is broken down faster — the data shows activity decreases.
• Confusing inhibition and denaturation — they are different mechanisms.
• Adding extra biology (e.g. about ethanol vapour, about PPO function) that the question did not ask for.

Things to Be Careful About

• The pattern is a decrease, not an increase.
• The trend is non-linear — describe it as a steep drop that plateaus.
• Use precise enzyme vocabulary: active site, enzyme-substrate complex, denaturation.
• Two creditable points are required for 2 marks.

Background Concept

A plan diagram (sometimes called a low-power outline drawing) is used to show the overall organisation of tissues in a specimen without the detail of individual cells. It is drawn at low magnification, using only clean, continuous outlines to mark the boundaries between different tissue layers. It is the standard way to record the gross anatomy of a plant or animal section on a Cambridge practical paper.

In a transverse section of a young dicot stem, the major tissue layers visible at low power, in order from the outside in, are:

• Epidermis — a single outer layer of cells, often with a cuticle.
• Cortex — a region of parenchyma cells just beneath the epidermis.
• Vascular bundles — discrete units of xylem and phloem, often arranged in a ring in dicots.
• Pith — central ground tissue (parenchyma), often with large cells.

The wedge indicated in Fig. 2.1 cuts from the outer surface inwards through the cortex and into a vascular bundle and the pith at its tip.

Understanding the Question

The shaded sector in Fig. 2.1 covers roughly one-eighth of the circular section. You are being asked to draw only that sector, large, at low magnification, with no cells shown, and to label the epidermis with one ruled line and label.

The command words are draw and label — so the answer is a piece of artwork on the paper, not a written description. Marks are awarded for the conventions of the plan diagram, not for biological content beyond showing the correct region.

Approach

1. Decide on the size — your drawing should fill most of the space provided in the answer booklet.
2. Sketch the outline of the wedge lightly first, then redraw with sharp pencil and continuous lines.
3. Mark the boundary between epidermis and cortex as a smooth line.
4. Mark the boundary of the vascular bundle(s) within the wedge.
5. Mark the pith boundary.
6. Draw one ruled label line from the word epidermis to the outermost layer.

Step-by-Step Reasoning

• Mark 1 — 'uses most of the available space and no shading': A plan diagram must fill most of the space; it must not be a tiny sketch. No shading anywhere — plan diagrams are line drawings only.
• Mark 2 — 'draws the correct region and no cells included': Only the wedge-shaped sector must be drawn, with the same proportions as in Fig. 2.1 (the wedge widens from the centre outwards to the surface). No individual cells drawn.
• Mark 3 — 'draws the layer of tissue beneath the epidermis': This is the cortex; show it as a continuous band of tissue between the epidermis and the next layer inwards.
• Mark 4 — 'draws at least one vascular bundle': Show the outline of at least one vascular bundle within the wedge. The wedge in the figure clearly passes through one.
• Mark 5 — 'label line and label to epidermis': One straight ruled line (drawn with a ruler) ending on the epidermis, with the label epidermis clearly written at the other end. Label lines must not have arrows or end in the middle of a layer.

Key Takeaways

• A plan diagram is a tissue-level outline drawing — no cells, no shading.
• Conventions: large, clean lines, correct proportions, ruler-straight label lines.
• Always show every relevant tissue layer visible in the region you are asked to draw.

Common Mistakes

• Drawing individual cells (this is a plan diagram, not a cell drawing).
• Using shading or stippling to differentiate tissues — lines only.
• Drawing the whole stem instead of the wedge.
• Label line ending in empty space or crossing other lines — the line must touch the structure labelled.

Things to Be Careful About

• The epidermis is a single cell layer; do not draw it as a thick band.
• The proportions of tissues must reflect those in the actual specimen — the cortex is wider than the epidermis and the vascular bundle is a discrete region.
• One label only — the mark scheme explicitly asks for one ruled label line.

Background Concept

Plant cells are bounded by a cell wall made primarily of cellulose. Under the light microscope, the wall appears as a thin line. Where two adjacent cells meet, the structure you see is the wall of cell A + middle lamella + wall of cell B, which is conventionally drawn as a single line shared between two cells.

In the centre (pith) of a young dicot stem, the cells are large, thin-walled parenchyma cells that fit together with no large intercellular spaces, giving them a polygonal (5- or 6-sided) shape in transverse section.

Understanding the Question

The question asks you to select a group of four adjacent cells from the central region of the stem on slide K1, where each cell touches at least two of the others, and to draw this group large.

The command word is make a large drawing — this is a high-power cell drawing, not a plan diagram and not a sketch of just outlines. The marks are for the conventions of biological drawing applied at cellular level.

Approach

1. Place the slide on the microscope and select the centre of the stem.
2. Identify four adjacent polygonal cells where each touches at least two of the others.
3. Decide on a size — large enough to fill most of the space.
4. Draw each cell wall as two thin parallel lines that meet at corners.
5. Where cells touch, only three lines appear (two for the outer walls of the pair, one for the shared middle wall).
6. Add the label cell wall by a single ruled line ending precisely on a wall.

Step-by-Step Reasoning

• Mark 1 — 'uses most of the available space and lines are continuous, thin and sharp': The drawing must be large, and the lines must be the kind of single, sharp pencil strokes you get with a hard pencil (HB or 2H) and a steady hand.
• Mark 2 — 'draws only four cells and each cell touches at least two other cells': Exactly four cells — not three, not five. The connectivity requirement rules out drawing a 'chain' of cells where the middle two only touch one neighbour each; the four must form a cluster with each cell in contact with at least two others.
• Mark 3 — 'two lines around each cell and three lines where cells touch': The cell wall is drawn as a double line. At a corner where three cells meet, the boundary is the two outer cell walls plus one wall between each pair of cells — drawn as the appropriate number of lines.
• Mark 4 — 'draws cells with 5 or 6 sides': Real parenchyma cells in transverse section are pentagons or hexagons (close-packed polygonal tiling). Drawing squares, circles or triangles loses this mark.
• Mark 5 — 'label line and label to cell wall of one cell': One ruled line ending on a cell wall, with the label cell wall written at the other end.

Key Takeaways

• A high-power cell drawing must show individual cells, not just outlines.
• Convention: cell walls = double lines; shared walls = one line between two cells (so three lines for the pair).
• Plant cells in section are polygonal (5–6 sides), never circular.

Common Mistakes

• Drawing single-line cell walls (this is for plan diagrams, not cell drawings).
• Drawing three lines at every cell–cell boundary (this is wrong; three lines only where two cells share a wall).
• Drawing the cells as round or oval — parenchyma cells in close contact are polygonal.
• Drawing more than four cells, or drawing four cells where one only touches one other (chain layout).

Things to Be Careful About

• The cell wall label must end on a wall line — not on the cytoplasm or in empty space.
• Only one label is required for this mark; additional labels are not penalised but waste time.
• No shading, no organelles (other than what is asked to be labelled) — this is a structural drawing.

Background Concept

Magnification is the ratio of image size to actual size:

$$M = \frac{\text{image size}}{\text{actual size}}$$

Rearranging gives:

$$\text{actual size} = \frac{\text{image size}}{M}$$

In Cambridge practical questions, the magnification of a printed image is given in the figure caption (here, ×12). To convert a length measured on the printed image into the true length of the specimen, divide by the magnification.

Understanding the Question

Five straight black lines labelled M1 to M5 cross the vascular bundles in Fig. 2.2. Each line is roughly the diameter of one vascular bundle. You must measure each line on the figure (in mm), calculate the mean, then convert that mean image length into the actual length of the vascular bundles using the stated ×12 magnification.

The question asks you to record five measurements and units, show the mean calculation, show the division by 12, and round the final answer to the nearest 0.5 mm.

Approach

1. Use a ruler to measure each of the five labelled lines M1–M5 on the photomicrograph in mm. Record each with its unit.
2. Sum the five measurements and divide by 5 to get the mean image length.
3. Divide the mean image length by the magnification (12) to obtain the mean actual length.
4. Round the answer to the nearest 0.5 mm (so possible answers: 0.0, 0.5, 1.0, 1.5, 2.0, 2.5 mm…).

Step-by-Step Reasoning

• Mark 1 — 'states five measurements for length of vascular bundles and units': Each of M1–M5 is given in mm (the unit is essential; numbers without units do not earn the mark). The image measurements depend on the ruler used; representative values are M1 = 18 mm, M2 = 24 mm, M3 = 18 mm, M4 = 12 mm, M5 = 18 mm.
• Mark 2 — 'shows five measurements added together and divided by 5': The mean is computed correctly: (18 + 24 + 18 + 12 + 18) ÷ 5 = 90 ÷ 5 = 18 mm.
• Mark 3 — 'shows value divided by 12': The actual length = image length ÷ magnification = 18 ÷ 12 = 1.5 mm.
• Mark 4 — 'answer given to nearest 0.5 mm': 1.5 mm already lies on a 0.5 mm gridline, so no further rounding is needed; the answer is 1.5 mm.

Key Takeaways

• actual size = image size ÷ magnification is the formula to remember.
• Always include units with measurements (mm here, not just numbers).
• The required precision is set by the mark scheme (here, nearest 0.5 mm) — give the answer to that precision, not to a tighter one.

Common Mistakes

• Forgetting units on the image measurements (loses the unit mark).
• Dividing by 12 first then averaging (works mathematically but does not match the mark scheme sequence — ecf is usually allowed but is less clear).
• Multiplying by 12 instead of dividing (treating image as actual).
• Giving the answer to a precision other than 0.5 mm, e.g. 1.4 mm or 1.50 mm — does not earn the final mark.

Things to Be Careful About

• The lines M1–M5 are not all the same length; some bundles are larger than others. Do not assume they are equal.
• The answer must be the mean actual length, not the mean image length. Many candidates stop at 'mean = 18 mm' and forget to divide by 12.
• The magnification (×12) is stated in the figure caption. Use that, not 10× or 40×.

Background Concept

A mean calculated from a small or non-representative sample may not be a reliable estimate of the whole population. To improve accuracy of a mean for an entire plant, you need either:

• a larger, more representative sample of measurements within the section you have, or
• samples from more locations/individuals to reduce the effect of local variation.

Understanding the Question

The student has calculated a mean vascular-bundle length from only five bundles in one section of one stem. They want a more accurate mean for the whole plant. You must describe two modifications to the method.

The mark scheme awards 1 mark for any two of three specific improvements.

Approach

Think about the three sources of error in the existing method:

1. Sample size within the section is small (only five bundles measured).
2. The magnification may be too low for accurate measurement of short bundles.
3. Only one section has been examined — there is biological variation along the length of the stem and between stems.

Then pick any two to write down.

Step-by-Step Reasoning

• Improvement 1 — measure all vascular bundles: instead of only five, measure every vascular bundle in the section. The mean is then based on the whole population within that section.
• Improvement 2 — use a larger magnification: a higher magnification enlarges each vascular bundle, making it easier to place the ruler accurately and reducing proportional error.
• Improvement 3 — take more sections: cut and examine additional sections from different parts of the same plant (or different plants of the same species) to capture biological variation.

Key Takeaways

• Reliability of a mean depends on sample size and representativeness.
• Improvements in practical biology usually fall into 'measure more', 'use better apparatus', or 'control variables better'.

Common Mistakes

• Saying 'repeat the experiment' without saying what specifically would be changed.
• Suggesting improvements unrelated to the measurement (e.g. staining, mounting) — these do not affect the accuracy of the length measurement.
• Repeating the same idea twice (e.g. 'measure more bundles' and 'measure all bundles' — only one mark available from that bullet).

Things to Be Careful About

• The question asks for two modifications to make the mean more accurate for the whole plant, not just more precise. Suggestions that improve precision but not representativeness (e.g. 'use a sharper pencil') do not earn marks.