Biology 9700/35 — October/November 2025

Background Concept

A serial dilution is a step-wise reduction in concentration by a fixed factor at each stage. For a 1:1 (halving) serial dilution, each solution in the series contains half the concentration of the previous one. The cleanest procedure is to transfer a fixed volume of the previous solution into the next beaker and top up with the same volume of diluent (water), so the final volume in each beaker stays constant while the solute is halved. This technique is widely used to produce a set of calibration standards of known concentration.

Understanding the Question

Fig. 1.3 already shows Beaker 1 containing 20 cm³ of 1.0% glucose solution (G) with 0 cm³ of water added. The candidate must complete Beakers 2–5 so that the glucose concentration halves at each step, giving four further concentrations in addition to the 1.0% starting solution. At the end of the dilution, 10 cm³ of each concentration must be available for use in the Benedict's tests.

Approach

Halve 1.0% four times: 1.0% → 0.5% → 0.25% → 0.125% → 0.0625%. Because each beaker must hold the same final volume (20 cm³), the most reliable way to halve the concentration is to transfer 10 cm³ of the previous beaker into the next and add 10 cm³ of distilled water (W).

Step-by-Step Reasoning

1. Beaker 1: 1.0% (already drawn — 20 cm³ of 1.0% glucose, 0 cm³ of W).
2. Beaker 2: 0.5%. Transfer 10 cm³ from Beaker 1 into Beaker 2, then add 10 cm³ of W. The 10 cm³ of 1.0% in 20 cm³ total = 0.5%.
3. Beaker 3: 0.25%. Transfer 10 cm³ from Beaker 2 into Beaker 3, then add 10 cm³ of W. The 10 cm³ of 0.5% in 20 cm³ total = 0.25%.
4. Beaker 4: 0.125%. Transfer 10 cm³ from Beaker 3 into Beaker 4, then add 10 cm³ of W. The 10 cm³ of 0.25% in 20 cm³ total = 0.125%.
5. Beaker 5: 0.0625%. Transfer 10 cm³ from Beaker 4 into Beaker 5, then add 10 cm³ of W. The 10 cm³ of 0.125% in 20 cm³ total = 0.0625%.

Each new beaker must therefore show:

• a transfer arrow labelled '10 cm³' from the previous beaker
• a water arrow labelled '10 cm³ of W'
• a label beneath the beaker giving the new concentration (with % sign)

Key Takeaways

• A halving serial dilution produces 1.0%, 0.5%, 0.25%, 0.125%, 0.0625%.
• Equal transfer volume and diluent volume (10 cm³ each) keeps the total volume at 20 cm³ while halving the concentration.
• The % sign must appear at least once on the completed diagram (in practice it should appear on all five beaker labels).

Common Mistakes

• Forgetting the % sign on any concentration label — the mark scheme requires % at least once.
• Adding 10 cm³ of water to the 20 cm³ already present in the beaker instead of transferring 10 cm³ first — this would not halve the concentration.
• Transferring a volume other than 10 cm³, which gives the wrong concentration.
• Drawing the transfer arrow in the wrong direction (e.g. from the new beaker back into the previous one).

Things to Be Careful About

• Both the volume transferred AND the volume of water added must be clearly labelled on every new beaker.
• Each new beaker needs its own pair of arrows.
• Keep the diagonal arrangement of the printed figure — the transfer arrows curve down from one beaker to the next.

Background Concept

The Benedict's test detects reducing sugars such as glucose. When heated with Benedict's solution, glucose reduces the blue copper(II) ions to a brick-red precipitate of copper(I) oxide. The colour sequence is blue → green → yellow → orange → brick-red. The time taken for the first colour change is inversely related to the concentration of glucose: a higher concentration reacts faster because more reducing sugar is available to react with the copper ions in a given time.

Understanding the Question

The candidate has prepared five glucose standards (1.0%, 0.5%, 0.25%, 0.125%, 0.0625%) by serial dilution. They perform a Benedict's test on each (1 cm³ of glucose + 1 cm³ of Benedict's, heated in a boiling water-bath) and record the time to the first colour change in a results table. If no colour change occurs within 120 s, the time is recorded as 'more than 120'.

Approach

Construct a table with two columns: the independent variable (glucose concentration) and the dependent variable (time to first colour change). The headings must include the unit (in the heading itself, not in the body of the table), and the independent variable heading must come before the dependent variable heading. Record whole-second times. Note the trend: higher concentration → shorter time.

Step-by-Step Reasoning

1. Independent variable column heading: 'percentage concentration of glucose (%)'. The unit % is in the heading.
2. Dependent variable column heading: 'time / s'. The unit s is in the heading.
3. Body of the table contains only numbers — no units are repeated in each cell.
4. Times are recorded to the nearest whole second (e.g. 25, 35, 50, 75, 110 — values are illustrative).
5. The trend must show that the highest concentration (1.0%) gives the shortest time, and the lowest concentration (0.0625%) gives the longest time (or 'more than 120').

Key Takeaways

• Headings must have a quantity AND a unit, with the independent variable heading first.
• No units in the body of the table.
• The expected trend: shorter time = higher glucose concentration.
• Times are whole seconds.

Common Mistakes

• Putting units in the body of the table (e.g. '25 s' in each cell) — the mark scheme rejects this.
• Reversing the columns (dependent variable before independent variable).
• Forgetting the unit in the heading.
• Recording 'more than 120' for the highest concentration — the trend must be that higher concentration → shorter time.

Things to Be Careful About

• Use only whole numbers for the times.
• Write the concentrations in the same order as prepared (1.0%, 0.5%, 0.25%, 0.125%, 0.0625%) for clarity.
• Note that the Benedict's test endpoint is the FIRST colour change — not necessarily the full brick-red precipitate.

Background Concept

A Benedict's test detects reducing sugars: the time to the first colour change is inversely related to the concentration. After 15 minutes of diffusion, the water surrounding each dialysis tubing will contain a small amount of glucose that has diffused out of the tubing. The Benedict's test on this water tells us how much glucose is present, by comparing the reaction time to the standard series in (a)(ii).

Understanding the Question

After 15 minutes of diffusion, the candidate uses a 1 cm³ syringe to take a 1 cm³ sample of the water surrounding the dialysis tubing containing R (20% glucose), and another 1 cm³ sample of the water surrounding the dialysis tubing containing S (10% glucose). These samples are labelled R1 and S1. A Benedict's test is then performed on each, and the time to the first colour change is recorded. If no change occurs within 120 s, the time is recorded as 'more than 120'.

Approach

Use the same technique as for the standards in (a)(ii): add 1 cm³ of Benedict's to 1 cm³ of sample, heat in the boiling water-bath, and time to the first colour change.

Step-by-Step Reasoning

1. R1: sample from water surrounding dialysis tubing containing 20% glucose (R). A higher initial concentration inside the tubing means more glucose will diffuse out, so R1 should have a higher glucose concentration than S1 → shorter time to colour change.
2. S1: sample from water surrounding dialysis tubing containing 10% glucose (S). A lower initial concentration inside the tubing means less glucose will diffuse out → longer time.
3. Both times are recorded in whole seconds (or 'more than 120' if no change).

Representative values: R1 ≈ 30 s, S1 ≈ 60 s (illustrative; actual values depend on the candidate's experiment).

Key Takeaways

• The Benedict's test on R1 and S1 gives an indirect measure of how much glucose diffused out of the tubing.
• R1 is expected to have a higher glucose concentration than S1, and therefore a shorter reaction time.
• This is the experimental data that will be used to estimate the concentration in (a)(iv).

Common Mistakes

• Mixing up which sample is R1 and which is S1.
• Forgetting to start timing as soon as the tube enters the boiling water-bath.
• Not stopping at the FIRST colour change (waiting for full brick red).
• Recording the time in minutes rather than seconds.

Things to Be Careful About

• The sample must be taken from the water surrounding the tubing (NOT from inside the tubing).
• The 1 cm³ volume is critical for fair comparison with the standards.
• The first colour change can be subtle — watch carefully for the blue → green transition.

Background Concept

This is a simple application of using a calibration (standard) series to estimate an unknown. The standards in (a)(ii) provide a relationship between glucose concentration and Benedict's reaction time. By matching the reaction time of an unknown sample to the standard with the closest (or interpolated) time, the unknown's concentration can be estimated. This is the principle behind colorimetric assays and many biosensors.

Understanding the Question

The candidate must use the times recorded in (a)(ii) for the standards and the times recorded in (a)(iii) for R1 and S1 to estimate the percentage concentrations of glucose in the unknown samples. The marks are awarded for an estimate consistent with the candidate's own data.

Approach

Look up the time recorded for R1 in the standard series. The concentration whose standard has the closest reaction time is the estimate. Repeat for S1. (Alternatively, draw a graph of time vs concentration and read off the values for R1 and S1.)

Step-by-Step Reasoning

1. From (a)(ii), the standard at 0.5% gives a time of (e.g.) 35 s. The standard at 0.25% gives (e.g.) 50 s.
2. From (a)(iii), R1 gave a time of (e.g.) 30 s. This is closest to the 0.5% standard.
3. S1 gave a time of (e.g.) 60 s. This is closest to the 0.25% standard.
4. So R1 ≈ 0.5% and S1 ≈ 0.25%.

The exact values depend on the candidate's data.

Key Takeaways

• A standard series provides a calibration curve (or table) for estimating unknowns.
• The closer the unknown's reaction time to a particular standard, the more confident the estimate.
• Estimates are limited by the spacing of the standards (which is why (a)(vii) asks for improvements).

Common Mistakes

• Using a textbook value instead of the candidate's own data.
• Giving a value that doesn't match the trend of the standards (e.g. R1 lower than S1).
• Reporting an exact value when only an estimate is possible (e.g. R1 = 0.46% rather than ≈ 0.5%).

Things to Be Careful About

• The estimate must be consistent with the candidate's own data in (a)(ii) and (a)(iii).
• Use '≈' or 'about' to indicate an estimate.
• Do not claim more precision than the standard series supports.

Background Concept

Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration, down a concentration gradient. Fick's law states that the rate of diffusion is proportional to the concentration gradient (and the surface area), and inversely proportional to the thickness of the membrane. A steeper gradient produces a faster rate of diffusion.

Understanding the Question

The candidate has estimated the glucose concentration in R1 (water around tubing R) and S1 (water around tubing S). R was 20% glucose inside; S was 10% glucose inside. The question asks why the two estimated concentrations differ.

Approach

Compare the two starting conditions: R had a higher internal concentration than S. Apply Fick's law: a steeper concentration gradient (R) produces a faster diffusion rate than a shallower gradient (S), so more glucose ends up in the surrounding water of R than of S in the same 15 minutes.

Step-by-Step Reasoning

1. R contained 20% glucose inside the tubing; the surrounding water was initially 0% glucose. Concentration gradient = 20%.
2. S contained 10% glucose inside the tubing; the surrounding water was initially 0% glucose. Concentration gradient = 10%.
3. R had a steeper concentration gradient than S.
4. By Fick's law, the rate of diffusion out of R was faster than out of S.
5. After 15 minutes, the water surrounding R (sample R1) contained more glucose than the water surrounding S (sample S1).

Key Takeaways

• Diffusion rate is proportional to the concentration gradient.
• A higher internal concentration produces a steeper gradient and a faster rate of diffusion.
• This is the same principle that explains why oxygen diffuses faster from alveoli into deoxygenated blood than into oxygenated blood.

Common Mistakes

• Saying 'there was more glucose in R so more diffused out' without mentioning the concentration gradient — the mark scheme requires reference to the gradient.
• Confusing concentration gradient with simple concentration.
• Saying the diffusion rate was the same in both (ignoring the difference in gradients).

Things to Be Careful About

• The reason must mention BOTH the steeper gradient AND the faster diffusion rate (two marking points).
• The gradient is between the INSIDE of the tubing and the OUTSIDE water — not just the internal concentration.

Background Concept

An average rate of change is the total change divided by the total time over which the change occurred. Here the change is the concentration of glucose that has appeared in the water outside the tubing, and the time is the 15 minutes of diffusion. This is a simple division, but the units and significant figures must be handled carefully.

Understanding the Question

The candidate must calculate the average rate at which the percentage concentration of glucose is increasing in the water surrounding the dialysis tubing containing R, using their own estimated concentration of R1 from (a)(iv) and the 15-minute diffusion time from step 8. The answer must be given to TWO significant figures.

Approach

Divide the concentration of R1 by 15 minutes to get % per minute. Round the result to two significant figures.

Step-by-Step Reasoning

1. Average rate = (concentration of R1) / (15 min).
2. Example: if R1 ≈ 0.5%, then rate = 0.5 / 15 = 0.0333...% per min.
3. Two significant figures → 0.033% per min (the leading zeros are not significant; the 3s are).
4. Other examples:
   • If R1 ≈ 0.4%, rate = 0.4 / 15 = 0.0266... ≈ 0.027% per min.
   • If R1 ≈ 0.3%, rate = 0.3 / 15 = 0.0200 ≈ 0.020% per min.
   • If R1 ≈ 0.6%, rate = 0.6 / 15 = 0.0400 ≈ 0.040% per min.
5. The unit must be 'per minute' (or '/ min' or 'min⁻¹') attached to the percentage.

Key Takeaways

• Average rate of change = change ÷ time.
• Two significant figures is the requirement — not two decimal places.
• The unit (% per min) must accompany the number.

Common Mistakes

• Giving the answer to too many or too few significant figures.
• Forgetting the unit (per minute).
• Using 5 min or 30 min instead of 15 min.
• Multiplying instead of dividing.

Things to Be Careful About

• 0.033 (to 2 sig figs) is correct; 0.03 is only 1 sig fig and would not earn the mark.
• The leading zero before the decimal is not significant; the 3s are.
• The unit should be written as '% per minute' or '%/min' — consistent with the question's wording.

Background Concept

A standard series provides a calibration for estimating unknowns, but the precision of the estimate is limited by the spacing between standards. Halving dilutions (1.0%, 0.5%, 0.25%, ...) give a broad range but widely spaced values. If the unknown's reaction time falls between two standards, the estimate is uncertain.

Understanding the Question

The candidate has estimated R1 to be (e.g.) 0.5%, but this estimate is based on widely spaced standards (the nearest are 0.25% and 1.0%). The question asks how to obtain a more accurate estimate of R1's concentration.

Approach

Reduce the spacing between standards around the estimated value of R1. Prepare standards both above and below the estimate so that the unknown can be matched against closely adjacent concentrations.

Step-by-Step Reasoning

1. The current standards are 1.0%, 0.5%, 0.25%, 0.125%, 0.0625% — each step is a halving, which is too coarse around the estimated value.
2. Identify the estimated R1 concentration from (a)(iv), e.g. ≈ 0.5%.
3. Prepare new standards with much smaller concentration differences, e.g. 0.4%, 0.45%, 0.5%, 0.55%, 0.6% — a set of five standards spanning the region around the estimate.
4. Each standard must be made with a precise transfer (e.g. 10 cm³ transfer + adjust water volume) so the concentrations are accurate.
5. Perform the Benedict's test on R1 alongside these new, finer standards; match the reaction time of R1 to the closest new standard.
6. The closer the spacing of the new standards, the more accurate the new estimate.

Key Takeaways

• The precision of an estimate from a standard series is limited by the spacing between standards.
• Narrowing the spacing around the estimated value gives a more precise estimate.
• The improvement is most valuable where the original standards are widely spaced (i.e., near the extremes of a halving dilution).

Common Mistakes

• Suggesting a vague improvement like 'repeat the experiment' or 'use better equipment' — these do not address the specific issue of calibration precision.
• Suggesting standards that are all on one side of the estimate (e.g. all higher than R1) — the mark scheme requires standards either side of the estimate.
• Not specifying the narrower intervals.

Things to Be Careful About

• The improvement must specifically target the precision of the concentration estimate.
• The improvement must mention BOTH narrower intervals AND standards on either side of the estimate (two marking points).
• It must be realistic — preparing solutions with concentration 0.43% or 0.567% is not practical; concentrations should be in simple, precise ratios or made with accurate pipettes.

Background Concept

Errors in measurement are classified as systematic (consistent bias in one direction, e.g. a miscalibrated instrument that always reads 5% low) or random (unpredictable variation around the true value, e.g. reading a meniscus slightly differently each time). The effect of a systematic error on a comparison depends on whether it affects both samples equally.

Understanding the Question

A 1 cm³ syringe in step 21 actually delivers 0.95 cm³ instead of 1 cm³ — a consistent 5% under-delivery. The candidate must identify this as systematic or random, and state its effect on the results.

Approach

A consistent error (every reading is 5% low) is systematic. The same syringe is used for both R1 and S1, so both samples are affected equally. The relative comparison between R1 and S1 is therefore unchanged.

Step-by-Step Reasoning

1. The error is consistent: every time the syringe is used, it delivers 0.95 cm³ instead of 1 cm³. This is a systematic error (not random variation).
2. The same syringe is used to sample the water around R (for R1) and the water around S (for S1).
3. Both R1 and S1 therefore contain 5% less water from the surrounding than intended.
4. The Benedict's test on each uses the same proportion of sample and reagent, so both R1 and S1 are affected by the same proportional error.
5. The relative difference between R1 and S1 is preserved — the absolute concentration is wrong, but the comparison is unaffected.
6. Hence: systematic error, no effect on the result of the comparison.

Key Takeaways

• A consistent bias is a systematic error; a fluctuating bias is a random error.
• If a systematic error affects two samples equally, the comparison between them is unaffected, even though the absolute values are wrong.
• This is why using the same instrument throughout an experiment is important.

Common Mistakes

• Classifying the error as random (because the candidate does not recognise the consistency).
• Saying the error affects the result, without specifying how or that the comparison is unaffected.
• Saying it would inflate the result (it would deflate the sampled volume, but the relative comparison is preserved).

Things to Be Careful About

• The mark scheme accepts 'systematic error' and 'no effect on the results' as the two required points.
• The candidate should think about whether the error affects the COMPARISON between R1 and S1, not just the absolute values.

Background Concept

A bar chart is used to compare discrete categories of data. A grouped (or clustered) bar chart is used when two sub-categories are being compared across the same categories — here, two fruits (apple and pineapple) across three sugars. Bars should be drawn with uniform width, with a small gap between bars in the same group and a larger gap between groups. Lines should be ruled sharply and meet axes precisely.

Understanding the Question

Table 1.3 gives the mass of three sugars (glucose, fructose, sucrose) in 100 g of two fruits (apple, pineapple). The candidate must draw a grouped bar chart on the grid in Fig. 1.4 to compare these six values.

Approach

Use the x-axis for the categorical variable (sugar type) with three groups (glucose, fructose, sucrose), each containing two bars (apple, pineapple). Use the y-axis for the numerical variable (mass of sugar), with a scale that uses at least half the grid and is labelled at regular intervals. Plot the six bars precisely, ruling all lines.

Step-by-Step Reasoning

1. x-axis label: 'type of sugar'. Three groups, evenly spaced: glucose, fructose, sucrose.
2. Within each group, two adjacent bars: one for apple (e.g. shaded one way), one for pineapple (e.g. shaded another). Even bar width.
3. y-axis label: 'mass of sugar / g per 100 g fruit'. Scale: 0 to 8, with labelled marks at 0, 2, 4, 6, 8 (i.e. every 2 cm if 1 cm = 1 g, or 2 g per 2 cm — the mark scheme specifies 2 g per 2 cm).
4. Bar heights:
   • Glucose: apple 2.3, pineapple 1.3
   • Fructose: apple 6.9, pineapple 2.3
   • Sucrose: apple 1.9, pineapple 5.2
5. All bars drawn with horizontal and vertical lines joined precisely.
6. Bars must be the same width, with a small gap within each group and a larger gap between groups.

Key Takeaways

• Grouped bar chart conventions: x-axis categorical, y-axis numerical, uniform bar widths, clear separation between groups.
• Scale must use at least half the grid, with even intervals.
• Lines must be ruled sharply and meet axes precisely.
• Each bar must be plotted accurately against the grid.

Common Mistakes

• Scale that does not use enough of the grid (e.g. 0 to 20 when max is 6.9).
• Uneven bar widths.
• Bars that do not align with the correct gridlines.
• Missing the y-axis units.
• Bars drawn freehand instead of with a sharp pencil and ruler.
• Confusing which bar is apple and which is pineapple.

Things to Be Careful About

• The mark scheme specifically requires: (1) axes labelled with both fruit names and sugar names AND y-axis units; (2) scale on y-axis as 2 g per 2 cm with labels every 2 cm; (3) all six bars plotted correctly; (4) lines joined precisely.
• The two bars within each group must be visually distinguishable (e.g. shaded differently, or labelled).

Background Concept

'Percentage difference' between two values can be calculated in two ways: (a) the difference as a percentage of the FIRST (reference) value, or (b) the difference as a percentage of the LARGER value. The mark scheme accepts both methods, so the candidate can use whichever is preferred.

Method A: (new − original) / original × 100, where 'new' is the value being compared to the original.
Method B: |A − B| / max(A, B) × 100.

Understanding the Question

Table 1.3 gives the mass of sucrose per 100 g of pineapple (5.2 g) and per 100 g of apple (1.9 g). The candidate must calculate the percentage difference in the mass of sucrose in pineapple compared to apple.

Approach

The mark scheme accepts two approaches:

• Approach 1 (apple as denominator): (5.2 − 1.9) / 1.9 × 100 = 173.7%, i.e. pineapple has ~174% more sucrose than apple.
• Approach 2 (pineapple as denominator): (5.2 − 1.9) / 5.2 × 100 = 63.5%, i.e. apple has ~64% less sucrose than pineapple (or pineapple has ~64% more sucrose than apple relative to its own value).

Step-by-Step Reasoning

1. Identify the two values: sucrose in pineapple = 5.2 g per 100 g; sucrose in apple = 1.9 g per 100 g.
2. Difference = 5.2 − 1.9 = 3.3 g per 100 g.
3. Approach 1: percentage difference = 3.3 / 1.9 × 100 = 173.7% ≈ 174%. (Here apple is the reference, since the question says 'pineapple compared to apple'.)
4. Approach 2: percentage difference = 3.3 / 5.2 × 100 = 63.5% ≈ 64%. (Here pineapple is the reference, giving the relative difference as a proportion of the larger value.)

The mark scheme accepts either answer — both earn the mark.

Key Takeaways

• Percentage difference = (difference between two values) / (reference value) × 100.
• The choice of reference value matters and gives different numerical answers.
• The mark scheme for this question accepts both interpretations.

Common Mistakes

• Dividing by the wrong reference value (e.g. using 5.2 when the question implies apple).
• Forgetting to multiply by 100.
• Using absolute difference in the numerator (the negative sign should not appear if the larger value is on top).
• Not showing working clearly.

Things to Be Careful About

• The mark scheme accepts BOTH 174% and 64% (or 173.7% and 63.5% if not rounded).
• The candidate should show their working with clear substitution.

Background Concept

A plan diagram is a low-power, outline-only drawing of a specimen. It shows the arrangement and proportion of tissues — not individual cells. Conventions for plan diagrams are heavily rewarded in CIE Paper 3 marking:

• Use sharp, single, continuous lines (no sketchy/fuzzy outlines).
• Do not shade any region.
• Do not draw individual cells anywhere — only the boundaries between tissues.
• Make the drawing large (it should fill most of the space provided).
• Keep proportions accurate — the cortex should not be drawn thicker than the vascular tissue if it isn't, etc.
• Labels use a ruled line that ends precisely on the structure; no arrowheads.

A plant stem in transverse section typically shows (outside → inside): epidermis → cortex → vascular tissue (xylem on the inside, phloem on the outside) → pith. The way these tissues are arranged (continuous ring vs separate bundles, number of rings, presence/absence of pith) is what distinguishes a young dicot stem, a monocot stem, a root, etc.

Understanding the Question

The question asks the candidate to draw a plan diagram of a specific wedge-shaped region of the stem section M1 indicated in Fig. 2.1. The diagram must show the tissues correctly, follow plan-diagram conventions, and include a labelled xylem. The marks reward size, shading rules, proportions, shape of the outline and vascular tissue, and the label.

The marking scheme tells us that M1 has continuous vascular tissue (one ring, cells in the centre). This is the key feature distinguishing M1 from Fig. 2.2, which has separate vascular bundles.

Approach

The strategy is to:

1. Identify the tissues in M1 — look down the microscope and decide where the epidermis, cortex, vascular tissue, and pith lie in the wedge.
2. Sketch lightly first to get the proportions right.
3. Trace the final outlines in sharp, single, continuous pencil lines.
4. Add the one required label (xylem) using a ruler; the label line must touch the structure being labelled and the label written clearly at the end of the line.
5. Check there is no shading and no cells drawn anywhere.

Step-by-Step Reasoning

• Appropriate size and no shading — fill most of the answer space; leave all interior regions blank (no stippling, hatching, or pencil shading). This single mark covers both points.
• Correct section of stem — the outline should match the curved outer boundary of the stem in the wedge, with the inner tissues drawn in proportion. The wedge should not look like a rectangle or circle.
• At least three vascular bundles, correct proportion, no cells — the vascular tissue crosses the wedge as a band; the candidate must show at least three recognisable bundle 'bulges' in that band. Critically, do not draw the individual cells of the cortex or pith — only their boundaries.
• Correct outline shape and continuous vascular tissue — the epidermis must be drawn as a single smooth curve, and the vascular tissue must be drawn as a continuous ring, not as separate islands. This matches the biology of M1 (continuous ring of vascular tissue).
• Label line to xylem — a single straight line drawn with a ruler, ending on the xylem tissue (the inner part of the vascular band), labelled 'xylem'. Do not label any other structures.

Key Takeaways

• A plan diagram is about tissue arrangement, not cellular detail.
• Conventions (sharp lines, no shading, no cells, large size, ruled label lines) are themselves worth marks — they are easy to lose and easy to earn with care.
• Recognising whether the vascular tissue forms a continuous ring or separate bundles is a key diagnostic feature of stem anatomy.

Common Mistakes

• Drawing individual cells inside tissues — this loses the 'no cells' mark and the diagram becomes a high-power drawing rather than a plan.
• Shading any region (e.g. filling in the pith with pencil). Plan diagrams must be unshaded.
• Using unruled, wobbly label lines, arrowheads, or labels that don't end on the structure.
• Drawing the vascular tissue as separate bundles when in fact it is continuous in M1 (this is the diagnostic feature being tested).

Things to Be Careful About

• The label line must end on the xylem, not on the phloem or cortex.
• The diagram should be of the shaded wedge in Fig. 2.1, not of the whole section.
• Make the vascular tissue a continuous ring (M1's biology) and show its inner edge clearly so the pith is distinct.

Background Concept

Xylem vessel elements are dead, hollow cells with thickened, lignified walls. They are joined end-to-end and the end walls dissolve to form continuous xylem vessels for water transport. Each element therefore has a thick wall around a hollow lumen with no cytoplasm, nucleus or organelles. Under the microscope they appear as large empty polygonal or circular cells with prominent walls.

In a high-power biological drawing, the double-line convention represents the cell wall — because the cell wall itself has thickness and is built of layers. Where two cells touch, each contributes its own two lines, giving three lines in total (one line from each cell's wall plus the shared middle line representing the middle lamella / shared wall). This is one of the most heavily tested conventions in CIE Paper 3.

Understanding the Question

The candidate must select four adjacent xylem vessel elements on M1 such that each element touches at least one other. The high-power drawing must follow biological-drawing conventions: sharp lines, double walls, three lines where elements touch, empty lumens, no shading, and one label to the wall of one element.

The marks reward: (1) appropriate size and sharp continuous lines; (2) only four elements drawn, each touching at least one other; (3) two-line walls and three lines where elements touch; (4) correct shape; (5) label line to the wall.

Approach

The strategy is to:

1. Move to high power on the microscope.
2. Identify a region where you can clearly see four adjacent xylem vessel elements with each touching at least one other — usually where vessels are densely packed.
3. Note the actual shape (round, polygonal, oval) and the relative sizes of the elements before drawing.
4. Sketch the cluster lightly, then trace with sharp single lines, drawing two lines around each element.
5. Where elements touch, draw three lines (one shared boundary between the two walls).
6. Leave lumens empty (no dots, no nucleus, no shading).
7. Add one ruled label line to the wall of one element, labelled 'wall'.

Step-by-Step Reasoning

• Appropriate size, sharp continuous lines — draw large (fill most of the space); lines must be single, crisp pencil strokes with no fuzzy gaps.
• Only four xylem vessel elements, each touching at least one other — pick exactly four. A common error is to draw five because an extra element creeps in. Arrangements that satisfy 'each touches at least one other': a row of four in contact, a 2 × 2 cluster, or a 'T' shape.
• Two lines around each, three lines where elements touch — for every element draw its outer wall as two parallel lines. Where two elements share a wall, the shared boundary is drawn as three lines (two walls pressed together).
• Correct shape — match the observed shape: roughly circular if vessels were round in T.S., polygonal if angular. Sizes should reflect relative sizes seen.
• Label line to the wall — one ruler-drawn line ending exactly on one wall (the pair of lines making up that wall), labelled 'wall'. Do not label the lumen or any other feature.

Key Takeaways

• High-power drawings of plant cells must use the double-line wall convention.
• Three lines where two cells touch is a signature mark — forgetting it costs marks.
• Xylem vessel elements are dead at maturity, so the lumen is empty in a mature section — do not draw cytoplasm, nucleus or organelles.

Common Mistakes

• Drawing only one line around each element (looks like an animal cell or a plan diagram outline).
• Drawing five or six elements instead of exactly four.
• Drawing cytoplasm, a nucleus or organelles inside the lumen.
• Labelling the lumen instead of the wall, or labelling with an arrowhead or unruled line.
• Drawing elements that don't actually touch, failing the 'each touches at least one other' rule.

Things to Be Careful About

• The two lines around each element should be parallel and close together — not so far apart that they look like a single thick wall with an internal space.
• The lumen must be truly empty (no shading, no stippling, no nucleus dot).
• Only one label is required; adding extra labels (lumen, xylem, vessel, etc.) is not wrong but is unnecessary.

Background Concept

An eyepiece graticule is a tiny scale (usually 100 divisions) etched into a glass disc that sits inside the microscope eyepiece. Its divisions look the same size regardless of the magnification of the objective lens — but the actual length that each division represents depends on which objective lens is in use.

A stage micrometer is a slide with a precisely known scale (each division = $0.1\ \text{mm}$ or $1.0\ \text{mm}$ typically). To calibrate the eyepiece graticule for a particular objective lens, the two scales are superimposed and aligned. The number of eyepiece graticule divisions that line up with a known length on the stage micrometer is then counted. From this:

$\text{length of one eyepiece graticule division} = \frac{\text{known length on stage micrometer}}{\text{number of eyepiece divisions}}$

Once calibrated, the eyepiece graticule can be used to measure any specimen on that microscope at that objective lens simply by counting divisions.

Understanding the Question

The candidate is given Fig. 2.3 showing the stage micrometer and eyepiece graticule superimposed. They are told that one division on the stage micrometer is $1.0\ \text{mm}$. They must read the alignment and work out the actual length (in µm) of one eyepiece graticule division, to three significant figures.

From Fig. 2.3, 100 eyepiece graticule divisions span 10 stage micrometer divisions — the graticule's full 0 to 100 range lines up with 10 mm of stage micrometer scale. This is the alignment the markscheme requires.

Approach

1. Read the number of eyepiece graticule divisions that align with one stage micrometer division (or vice versa).
2. Convert the stage micrometer length to µm.
3. Divide to find the actual length of one eyepiece graticule division.
4. Express the answer to three significant figures (a specific marking requirement).

Step-by-Step Reasoning

• Counting the alignment — 100 eyepiece graticule units span 10 stage micrometer divisions. So one stage micrometer division (1.0 mm) = 10 eyepiece graticule units.
• Unit conversion — $1.0\ \text{mm} = 1000\ \text{µm}$. Therefore 10 eyepiece graticule units = $1000\ \text{µm}$.
• Divide — 1 eyepiece graticule unit = $\frac{1000}{10}\ \text{µm} = 100\ \text{µm}$.
• Three significant figures — write the answer as $100\ \text{µm}$ (not $1 \times 10^{2}$ or $0.1\ \text{mm}$ etc.).

Equivalently: 100 eyepiece graticule units = $10\ \text{mm}$ = $10\,000\ \text{µm}$, so 1 eyepiece graticule unit = $100\ \text{µm}$.

Key Takeaways

• An eyepiece graticule must be calibrated for each objective lens because its actual length per division changes with magnification.
• The conversion $1\ \text{mm} = 1000\ \text{µm}$ is essential in microscope work.
• Always quote answers to the number of significant figures the question asks for — here, three.

Common Mistakes

• Forgetting to convert mm to µm (writing the answer as $0.1\ \text{mm}$ instead of $100\ \text{µm}$).
• Quoting an answer to the wrong number of significant figures.
• Reading the alignment incorrectly (e.g. counting the wrong number of eyepiece divisions across one stage division).
• Forgetting to show the working — the mark scheme requires both division and unit conversion steps.

Things to Be Careful About

• The question explicitly says three significant figures — $100$ already has three significant figures, but be careful not to round a different value to fewer.
• Make sure the units in the final answer match the units requested (µm, not mm).

Background Concept

Once the eyepiece graticule has been calibrated (part (c)(i)), any specimen on the same microscope with the same objective lens can be measured simply by counting graticule divisions across the feature of interest and multiplying by the calibration factor.

The relationship is:

$\text{actual length} = \text{number of eyepiece graticule divisions} \times \text{length per division (calibration)}$

Units must be consistent — if calibration is in µm, the result is in µm; convert to mm if a more convenient unit is wanted.

Understanding the Question

Fig. 2.4 is the same photomicrograph as Fig. 2.2, taken with the same microscope and same lenses as Fig. 2.3 (the calibration image). An eyepiece graticule has been placed across a vascular bundle. The candidate must:

1. Read the number of eyepiece graticule divisions spanned by the vascular bundle.
2. Multiply by the calibration from (c)(i) ($100\ \text{µm}$ per division) to obtain the actual length.
3. Show the working and use appropriate units (µm or mm).

Approach

1. Identify the two ends of the vascular bundle on the graticule and read the number of divisions spanned (e.g. 70).
2. Multiply by the calibration from (c)(i).
3. Convert to a sensible unit (µm or mm) and present the final answer.

Step-by-Step Reasoning

• Reading the graticule — in Fig. 2.4 the graticule runs diagonally across the vascular bundle. The bundle starts at approximately graticule mark 5 and ends at approximately mark 75, so the bundle spans about 70 eyepiece graticule divisions.
• Applying the calibration — 70 divisions × $100\ \text{µm/division}$ = $7000\ \text{µm}$.
• Unit conversion — the question says 'use appropriate units'. $7000\ \text{µm}$ is awkward; converting gives $7.0\ \text{mm}$. Either is acceptable, but converting makes the size of the vascular bundle easier to interpret.
• Error carried forward (ecf) — if the candidate's calibration in (c)(i) was wrong but they correctly multiplied the graticule reading by their calibration value, full marks are still awarded here.

Key Takeaways

• The calibration is specific to the objective lens in use — change the lens and you must recalibrate.
• 'Appropriate units' means a unit that makes the size of the specimen sensible — µm for cells, mm for whole organs. Either is fine if stated correctly.
• Error carried forward allows the candidate to recover from a wrong calibration in (c)(i) provided the multiplication in (c)(ii) is performed correctly.

Common Mistakes

• Forgetting to show the working — the mark scheme explicitly requires the multiplication step to be visible.
• Reading the bundle length as the total graticule span (e.g. 100) rather than the actual span covered by the bundle.
• Mixing units — writing the calibration in mm but applying it in µm, or vice versa.
• Not converting to appropriate units (giving an answer in pm, m, or with no unit).

Things to Be Careful About

• The reading of the bundle length is the only measurement-dependent step — different candidates will read it slightly differently (e.g. 65 vs 70 vs 75 divisions). Any reasonable reading is creditable, and the working carries the mark.
• Always state the unit with the answer — bare numbers without units are not credited in CIE marking.