Biology 9700/33 — October/November 2025

Background Concept

A dilution reduces the concentration of a solute by adding more solvent. The relationship between the concentration and volumes of stock and diluted solution is governed by the equation

where $C_1$ and $V_1$ are the initial concentration and volume, and $C_2$ and $V_2$ are the final concentration and volume after dilution.

Understanding the Question

The question provides $1\ \text{cm}^3$ of 20.0% glucose solution G as the stock, and asks how much distilled water W must be added to give a total of $20\ \text{cm}^3$ at 1.0% glucose. The answer is one number, written in the empty cell of Table 1.2.

Approach

Apply $C_1 V_1 = C_2 V_2$ to find the total final volume needed, then subtract the volume of stock added to obtain the volume of water.

Step-by-Step Reasoning

1. $C_1 = 20.0\%$, $V_1 = 1\ \text{cm}^3$ (stock); $C_2 = 1.0\%$, $V_2 = ?$ (final).
2. $V_2 = \dfrac{C_1 V_1}{C_2} = \dfrac{20.0 \times 1}{1.0} = 20\ \text{cm}^3$.
3. The water to add is the final volume minus the stock volume: $20 - 1 = 19\ \text{cm}^3$.

Key Takeaways

• The volume of diluent is final volume minus the volume of stock solution added.
• In CIE practical papers, units must always be quoted with numerical answers.

Common Mistakes

• Forgetting to subtract $1\ \text{cm}^3$ and writing $20\ \text{cm}^3$ (this is the final volume, not the water volume).
• Using $1.0$ instead of $1$ in $V_1$ — the unit is $\text{cm}^3$ so the number is dimensionless for the equation.

Things to Be Careful About

Units must be consistent on both sides of the equation; percentages cancel and only volumes remain.

Background Concept

A serial dilution halves the concentration at every step. Starting from 1.0%, successive halvings give 0.5%, 0.25%, 0.125% and 0.0625%. Each step transfers a fixed volume into a beaker, then makes it back up to the same total volume with water.

Understanding the Question

Fig. 1.3 shows five beakers already labelled with the target concentrations. The candidate must add two labelled arrows to each of the four transfer steps: one showing the volume moved between beakers, and one showing the volume of water added to make the concentration correct.

Approach

Because each step halves the concentration, the cleanest scheme is to transfer $10\ \text{cm}^3$ and add $10\ \text{cm}^3$ of water (1:1 mix). This keeps every beaker at the same working volume and is straightforward to draw.

Step-by-Step Reasoning

1. From the 1.0% beaker, transfer $10\ \text{cm}^3$ into the 0.5% beaker, then add $10\ \text{cm}^3$ W to give a total of $20\ \text{cm}^3$ at 0.5%.
2. Mix the 0.5% beaker; transfer $10\ \text{cm}^3$ into the 0.25% beaker, then add $10\ \text{cm}^3$ W.
3. Repeat for 0.25% → 0.125% (add $10\ \text{cm}^3$ W) and 0.125% → 0.0625% (add $10\ \text{cm}^3$ W).
4. Each step halves the concentration: 1.0% → 0.5% → 0.25% → 0.125% → 0.0625%.

Key Takeaways

• A serial dilution with a 1:1 transfer-to-water ratio halves the concentration each time.
• The volume of water added at each step is the same as the volume transferred.

Common Mistakes

• Forgetting to add the water arrow to every beaker (the mark scheme requires an arrow for W at every step).
• Using unequal volumes, which would give the wrong final concentration.

Things to Be Careful About

Both arrows (transfer AND water) must be shown for each of the four transfer steps; missing either loses a mark.

Background Concept

The Benedict's test detects reducing sugars. Benedict's solution (copper(II) sulfate in alkaline citrate) is blue; when heated with a reducing sugar it forms a brick-red precipitate of copper(I) oxide. For the colour change to develop reliably, the Benedict's solution must be present in equal or excess amount relative to the sugar being tested.

Understanding the Question

The candidate is asked to choose a specific volume of Benedict's solution to add to each $1\ \text{cm}^3$ glucose sample and to justify the choice in one or two sentences. The mark scheme credits any answer of $2\ \text{cm}^3$ or more provided the reason given is 'equal or excess Benedict's so all the sugar reacts'.

Approach

The most efficient use of the $40\ \text{cm}^3$ provided is $2\ \text{cm}^3$ per test (giving enough for $5 + 3 = 8$ tests comfortably). The reason is that Benedict's must be in excess to ensure the reaction is not limited by reagent, so that the colour change depends on the sugar concentration, not the amount of Benedict's.

Step-by-Step Reasoning

1. A typical Benedict's test uses approximately $2\ \text{cm}^3$ Benedict's to $1\ \text{cm}^3$ sugar solution.
2. A 1:1 or greater ratio guarantees that Benedict's is in excess, so the limiting reagent is the glucose, and the colour developed reflects the glucose concentration.
3. With $40\ \text{cm}^3$ provided, $2\ \text{cm}^3 \times 8$ tests fits within the volume available.

Key Takeaways

• Quantitative biochemical tests need the reagent in excess so that the analyte is the limiting factor.
• A typical school/college Benedict's test uses about $2\ \text{cm}^3$ of Benedict's per sample.

Common Mistakes

• Choosing $1\ \text{cm}^3$ (equal to the sample volume is acceptable only if stated as 'equal'; the safer mark-scheme answer is $2\ \text{cm}^3$ and 'excess').
• Giving a reason about boiling rather than about reagent excess — the question asks why the volume is chosen.

Things to Be Careful About

The mark scheme credits the volume AND the explanation; both must be present.

Background Concept

The Benedict's test relies on reducing sugars reducing the blue $\text{Cu}^{2+}$ ions in Benedict's solution to brick-red $\text{Cu}_2\text{O}$. The reaction is concentration-dependent: the more reducing sugar present, the faster the colour change appears (and the more intense the final colour). For very low concentrations the colour change can take several minutes; for high concentrations it appears within seconds.

Understanding the Question

Table 1.3 has five known concentrations of glucose (1.0%, 0.5%, 0.25%, 0.125%, 0.0625%) and a blank column for the candidate to record what they observed during the Benedict's test in step 14. The mark is for showing the correct trend.

Approach

Run the Benedict's test on each of the five standards in turn (each freshly prepared, so all at the same starting temperature, with the same volume of Benedict's). Time how long each takes to show its first colour change. The trend should be: time-to-first-colour-change decreases as concentration increases.

Step-by-Step Reasoning

1. Heat each standard with $2\ \text{cm}^3$ Benedict's in a boiling water bath (do not reuse tubes — cross-contamination would obscure the trend).
2. Record the time at which the first permanent colour change becomes visible in each tube.
3. Sort the times: 1.0% shortest, 0.0625% longest, in order of decreasing concentration.
4. The required observation is therefore a strictly decreasing series of times as concentration rises.

Key Takeaways

• A standard series of known concentrations is essential for estimating the concentration of an unknown.
• The Benedict's reaction rate is proportional to the reducing-sugar concentration under controlled conditions.

Common Mistakes

• Recording final colour rather than the trend (e.g. 'red', 'green', 'blue') — the mark is for the time/trend, not the colour.
• Inverting the trend — slower change at higher concentration is incorrect.

Things to Be Careful About

The exact times are student-dependent; only the trend is marked.

Background Concept

In CIE practical papers, results tables must follow strict conventions: the independent variable (here, the sample time) goes on the left; the dependent variable (the time to first colour change) goes on the right; column headings carry a quantity and a unit separated by a solidus, with no units in the body of the table; and numerical data are recorded to a sensible, consistent precision.

Understanding the Question

Part (v) asks the candidate to record the times taken for the Benedict's test on the three water samples (taken at 5, 10, 15 minutes) to show their first colour change. These are the candidate's own measurements, so the marks are for the layout and conventions rather than for any particular value.

Approach

Set up the table with the IV (sample time) on the left and the DV (time to colour change) on the right, with units included only in the headings. Time the colour change in each tube to the nearest whole second, and record consistently. The expected trend is shorter time to colour change at longer sampling times because more glucose has diffused out into the surrounding water.

Step-by-Step Reasoning

1. Identify the independent variable: sample time (5, 10, 15 min).
2. Identify the dependent variable: time from heating to first permanent colour change.
3. Place the IV heading on the left ('sample time / min') — this is a marking-point.
4. Place the DV heading on the right ('time to first appearance of a colour change / s').
5. Use whole seconds, with no units in the body of the table.
6. Record the three measured times; the trend should show time decreasing as sample time increases (because glucose concentration in the water rises).

Key Takeaways

• Tables must have headings with a quantity and a unit, and units must NOT appear in the body of the table.
• The independent variable is conventionally on the left.
• Whole-second precision is appropriate for human-reaction-time measurements.

Common Mistakes

• Putting the unit in the body of the table (e.g. '60 s') — the mark is lost for not keeping the unit in the heading.
• Putting IV on the right and DV on the left.
• Mixing precision (some values to the nearest second, some to 0.1 s).

Things to Be Careful About

The four marks break down as: IV heading correctly placed (with unit); DV heading correctly placed (with unit); three results recorded; correct precision (whole seconds).

Background Concept

A 'source of error' in an experiment is a feature of the procedure that introduces uncertainty into the result. Subjective judgements — the end-point of a colour change, the start of a colour change, the exact moment a precipitate appears — are classic sources of error because they depend on the observer's perception and reaction time, and the change itself is gradual rather than instantaneous.

Understanding the Question

Part (vi) asks for one specific source of error from the Benedict's test. The mark scheme specifically credits the answer 'time to first colour change is subjective'.

Approach

Identify the step in the Benedict's test where human judgement most affects the result. The colour change happens gradually (blue → green → yellow → orange → brick red), and the moment the first permanent change is detected is decided by the person watching the tube.

Step-by-Step Reasoning

1. Look at the test procedure: heat the sample, watch for the first appearance of a colour change, record the time.
2. The 'first appearance' is not a single discrete event — it is a gradual transition that the observer judges.
3. Different observers (or the same observer on different days) will give different times, even on the same sample.
4. This makes the time measurement subjective, which is a valid source of error.

Key Takeaways

• 'Subjective' is the key word the mark scheme looks for.
• The source of error must be specific to the procedure, not a generic 'human error'.

Common Mistakes

• 'Human error' is too vague and is rejected by the mark scheme.
• 'The Benedict's solution might not be in excess' is a different kind of error and does not match the mark scheme's accepted answer.
• 'The temperature of the water bath fluctuates' is an error of a different type and not the one credited here.

Things to Be Careful About

The mark is for a source of error specifically in the Benedict's test, not in the diffusion step or the dilution step.

Background Concept

A calibration series (a set of standards of known concentration) is used to estimate the concentration of an unknown. The unknown's response (here, the time to first colour change in the Benedict's test) is compared with the response of the standards: the unknown is taken to have the concentration of the standard that gives the closest response.

Understanding the Question

Part (vii) asks the candidate to take their results from (a)(iv) (the five known concentrations) and (a)(v) (the three water samples) and read off an estimated concentration for each of samples 5, 10 and 15. The mark is for using the results correctly — the actual numbers will vary with the candidate's own data, but the direction of change (rising concentration with longer sampling time) is fixed.

Approach

For each sample, find the standard with the closest time-to-first-colour-change. That standard's concentration is the estimate.

Step-by-Step Reasoning

1. In (a)(iv) the standard at 1.0% gave the shortest time, the one at 0.0625% the longest, and the times rise as concentration falls.
2. In (a)(v) the 5-minute sample gave a long time (low glucose), the 10-minute sample a shorter time, the 15-minute sample the shortest time.
3. Match: 5-min sample ≈ concentration of the standard with a similar long time (low %, e.g. ~0.125%); 10-min ≈ ~0.25%; 15-min ≈ ~0.5% (these are illustrative — the actual read depends on the candidate's data).
4. The expected pattern is increasing concentration at later sampling times.

Key Takeaways

• A calibration series turns a quantitative measurement (time) into a concentration estimate.
• A well-designed calibration should bracket the unknown — the unknown should fall between two standards for a reliable estimate.

Common Mistakes

• Estimating the concentration by eye from the test-tube colour alone rather than from the timing data.
• Giving the same concentration for all three samples.
• Reversing the trend (claiming concentration falls with time).

Things to Be Careful About

The exact figures depend on the candidate's own timings; the mark is for the procedure of using the calibration correctly, not for one specific value.

Background Concept

Diffusion is the net movement of molecules from a region of higher concentration to a region of lower concentration, down a concentration gradient. The rate of diffusion is greatest when the gradient is steepest, but the total amount transferred increases with time.

Understanding the Question

Part (viii) asks the candidate to explain why the estimated glucose concentration was higher in the 15-minute sample than in the 5-minute sample. The mark scheme credits the explanation 'glucose had more time to diffuse out of the dialysis tubing into the water'.

Approach

Recognise that diffusion is a continuous process, and that more time means a greater cumulative quantity of glucose crossing the membrane.

Step-by-Step Reasoning

1. The dialysis tubing contains 20% glucose; the surrounding water contains 0% glucose.
2. Glucose molecules continually cross the membrane down this concentration gradient.
3. Over 5 minutes, only a small amount has crossed.
4. Over 10 minutes, more has crossed; over 15 minutes, more still.
5. Therefore the concentration of glucose in the surrounding water rises with time.

Key Takeaways

• Diffusion is a time-dependent process; cumulative transfer increases with elapsed time.
• The rate may slow as the gradient lessens, but the total transferred keeps rising.

Common Mistakes

• Saying 'diffusion happened faster' rather than 'more time elapsed' — the rate may not be constant, but the amount transferred must be greater at later times.
• Confusing the direction (claiming glucose moves out of the water into the tubing — wrong, the gradient is from tubing to water).

Things to Be Careful About

The question asks for a reason, not a restatement of the observation; the explanation must refer to the time allowed for diffusion.

Background Concept

A line graph is used to show how a continuous variable (concentration) changes with another continuous variable (time). CIE conventions require: clear, ruled axes with full quantity-and-unit headings; linear scales that use at least half the grid in both directions; neat, accurate plotting with small crosses or dots in circles; and thin lines joining the points (or a curve of best fit).

Understanding the Question

The candidate is given a data table (Table 1.4) with five time points and two concentrations (P and Q) and must plot a line graph on the grid in Fig. 1.5. Both data series should be plotted on the same axes and clearly distinguished.

Approach

Choose sensible scales (5 min → 1 cm or 2 cm on x-axis; 50 units → 1 cm or 2 cm on y-axis) so that the data fill at least half the grid in each direction. Plot each point with a small, neat cross or dot in a circle. Join each set of points with a thin line. Add a key or label the lines so that P and Q can be distinguished.

Step-by-Step Reasoning

1. x-axis range needed: 0 to 20 min. y-axis range needed: 0 to 400 (just above the highest value of 400).
2. A scale of 5 min to 2 cm and 50 units to 2 cm fits comfortably on the supplied grid and uses most of it.
3. Plot the five points for P: (0,200), (5,205), (10,200), (15,195), (20,200). The points should lie almost on a horizontal line at 200.
4. Plot the five points for Q: (0,400), (5,280), (10,150), (15,90), (20,75). These describe a curve that falls steeply then levels off.
5. Mark each point with a small cross or dot in a circle; join the points within each series with a thin straight line.
6. Label each line (P and Q) or provide a key.

Key Takeaways

• CIE graph marks are split: axes (1), scales (1), plotting (1), joining (1).
• The independent variable (time) goes on the x-axis; the dependent variable (concentration) on the y-axis.
• Two data series on one graph need a key or direct labels.

Common Mistakes

• Forgetting to label which line is P and which is Q (the marker 'x' or 'o' alone is not enough).
• Using awkward scales (e.g. 7 min to 1 cm).
• Plotting dots without joining them, or joining them with a thick wobbly line.

Things to Be Careful About

The grid provided in Fig. 1.5 has a specific aspect ratio; pick a scale that fills it without running off the edge.

Background Concept

Percentage decrease is calculated as

The 'initial' value here is the concentration of Q at 5 minutes (280 arbitrary units) and the 'final' value is the concentration at 20 minutes (75 arbitrary units). The initial and final values are subtracted, divided by the initial, and multiplied by 100 to give a percentage.

Understanding the Question

The question asks for the percentage decrease in concentration of Q between 5 and 20 minutes, to two significant figures, with working shown.

Approach

Apply the percentage-change formula. Take care to use the value at 5 minutes (not at 0 minutes) as the starting value, because the question specifies 'between 5 and 20 minutes'.

Step-by-Step Reasoning

1. Initial value (at 5 min) = 280 arbitrary units.
2. Final value (at 20 min) = 75 arbitrary units.
3. Decrease = $280 - 75 = 205$ arbitrary units.
4. $\dfrac{205}{280} \times 100 = 73.214\ldots\%$.
5. To 2 significant figures: $73\%$.

Key Takeaways

• Always quote the answer to the precision asked for in the question.
• The denominator of a percentage change is the initial (earlier) value, not the final one.

Common Mistakes

• Dividing by 75 instead of 280 (gives 273%, wrong).
• Using the value at 0 minutes (400) instead of 5 minutes (280), because the question specifies 'between 5 and 20 minutes'.
• Reporting 73.2% (3 sig figs) instead of 73% (2 sig figs).

Things to Be Careful About

The 2-sig-fig rounding is part of the question's mark scheme; the mark for 'correct answer to 2 significant figures' depends on this.

Background Concept

The rate of diffusion across a membrane is proportional to the concentration gradient across that membrane (Fick's law). A steep gradient gives a high rate; a shallow gradient gives a low rate. As diffusion proceeds, the gradient is gradually reduced as the high side becomes depleted and the low side becomes enriched, so the rate slows over time.

Understanding the Question

The question provides a table showing the concentration of Q in the blood falling steeply between 0 and 10 minutes (400 → 280 → 150) and then more gradually between 10 and 20 minutes (150 → 90 → 75). The candidate must suggest a reason for the two different rates.

Approach

Identify the driving force for diffusion (the concentration gradient) and explain how it changes over the two intervals. The gradient is large initially because the blood is full of Q and the dialysate is fresh; as time passes, the blood Q falls and the dialysate Q rises, so the gradient shrinks.

Step-by-Step Reasoning

1. 0–10 minutes: Q in blood is high (400 → 150) and Q in dialysate is still low (the dialysate is being refreshed continuously, or its Q is still small). The gradient is therefore steep, and the rate of diffusion is high.
2. 10–20 minutes: Q in blood is now much lower (150 → 75) while Q in the dialysate is rising. The gradient between blood and dialysate is therefore less steep, and the rate of diffusion is correspondingly lower.
3. The flat line for P (which stays at ~200 throughout) shows that P does not cross the membrane — its concentration in the blood does not change — confirming that the falling Q is due to diffusion, not to dilution by the blood sampling.

Key Takeaways

• A changing rate of diffusion reflects a changing concentration gradient.
• The reference to 'less steep' gradient in the second interval is the mark-scheme wording.

Common Mistakes

• Saying 'the concentration in the blood was lower' without saying why that matters (the gradient is what drives diffusion, not the absolute concentration).
• Confusing gradient and absolute concentration — the answer must refer to the gradient between blood and dialysate.

Things to Be Careful About

The mark scheme credits 'less steep concentration gradient' rather than 'smaller concentration in the blood'; the gradient is the contrast between blood and dialysate, not the blood level alone.

Background Concept

A plan diagram is a low-power, simplified map of a specimen that records the distribution of tissues but never shows individual cells. It is used to capture the overall organisation of a specimen without the detail of a cell drawing. The conventions CIE expects are: (1) continuous, thin, sharp pencil lines; (2) no shading; (3) no individual cells drawn; (4) correct proportions between tissue layers; (5) labels on ruled lines that end exactly at the chosen letter.

In a transverse section of a young dicot root the tissues, from outside to centre, are:

• Epidermis — a single outer cell layer that may carry root hairs (trichomes).
• Cortex — a wide zone of parenchyma used for starch storage and for the apoplastic movement of water and minerals toward the stele.
• Endodermis — a single ring of cells with a Casparian strip; on a plan diagram it shows as a clear boundary, and is reinforced as two close parallel lines marking the inner edge of the cortex and the outer edge of the stele.
• Pericycle — a layer just inside the endodermis from which lateral roots arise.
• Vascular tissue (stele) — xylem (water and mineral transport, often arranged in a star/X-shape with the largest vessels toward the centre of the section) and phloem (assimilate transport) in the gaps between the xylem arms.

The xylem is the main tissue that transports water and mineral ions upward throughout the plant; the phloem transports organic solutes. Either is acceptable as the tissue to label T, but xylem is the most conspicuous structure in a root cross-section and is the easiest to point to with a label line.

Understanding the Question

You are given a slide K1 of a stained transverse section through a root. Fig. 2.1 shades one sector (about a quarter) of the circular section and tells you to draw that region as a plan diagram. The sector is one half of the root (the other half is the unshaded bottom half — only the shaded part is to be drawn, but it represents half of the root, not a single quarter wedge). You must use plan-diagram conventions and add a single ruled label line ending in T that identifies a transport tissue.

The command word is draw. Marks are awarded for using the space, drawing only half the root, using plan-diagram conventions, drawing the endodermis correctly, getting the proportions right, and labelling correctly.

Approach

1. Rotate the view in your head so the flat side of the half-disc lies along the bottom of the page, giving a half-root plan.
2. Lay out the regions in order: epidermis (thin outer curve), cortex (wide band), endodermis (two close lines), vascular cylinder with xylem shapes inside.
3. Draw with continuous thin sharp pencil lines, no cells anywhere, no shading.
4. Make the drawing large — at least two-thirds of the available width — and keep the proportions: cortex should be the widest band, vascular cylinder roughly the central third.
5. Add the single label T to a xylem vessel (or to a phloem region).

Step-by-Step Reasoning

1. Use most of the available space and include the correct number of tissues (epidermis, cortex, endodermis, vascular tissue) — a half-disc at least two-thirds of the drawing area.
2. Half the root, no cells — outline the curved outer edge (epidermis) and a straight diameter line at the bottom; do not draw any individual cells inside any tissue.
3. Endodermis as two close lines — inside the cortex, draw two thin lines very close together marking the outer boundary of the vascular tissue.
4. Correct proportions of the vascular tissue — the central region containing xylem shapes should be in roughly the right proportion to the cortex width.
5. Label T — one ruled label line from a large open xylem vessel (the transport tissue) ending in the letter T.

Key Takeaways

• A plan diagram shows tissue distribution only, never cells.
• The endodermis is drawn as two close lines because it is the clearest visible boundary of the stele.
• Use the xylem (or phloem) as the labelled transport tissue.
• Always rule the label line cleanly and end it exactly at the chosen letter.

Common Mistakes

• Drawing cells inside the tissues (e.g. dotting the cortex or the vascular cylinder).
• Drawing the whole root, or only a quarter-wedge, instead of half.
• Drawing the endodermis as a single line (not credited).
• Labelling cortex or epidermis as T — these are not transport tissues.
• Adding shading, broken lines, or using a blunt pencil.

Things to Be Careful About

• Keep the xylem vessels as simple open outlines — do not fill them in.
• The vascular cylinder width should be noticeably less than the cortex width to reflect correct proportions in a typical young root.
• The label line must end exactly at the letter T, not on the tissue outline.

Background Concept

A high-power cell drawing is used to record the structure of a small number of cells. Unlike a plan diagram, the candidate does draw cells here, but only a few. The conventions CIE expects are:

• Continuous, thin, sharp lines drawn with a sharp pencil.
• No shading, no stippling, no colouring — structure is shown through line work alone.
• Cell walls drawn as two lines where the wall has visible thickness, to show that the wall is a discrete structure rather than just a line.
• Correct shapes of the cells — only draw what is actually visible.
• Correct relative sizes — do not make all the cells identical if they are visibly different.
• Only what is asked — exactly the number of cells requested, no extra cells peeking in at the edges.

In the centre of a young dicot root, parenchyma cells of the stele and xylem parenchyma are typically isodiametric polygonal cells with relatively thin, straight walls. Mature xylem vessels are larger open cells with thicker walls.

Understanding the Question

You are told to switch to high power on slide K1, look at the cells in the centre of the root, and draw four adjacent cells so that each cell touches at least one other. The cells must be drawn large (high-power drawing), with the conventions above. The marks reward: lines and shading conventions; number and arrangement of cells; double-line walls; and correct shapes/sizes.

The command word is make a large drawing. Marks are awarded for the drawing itself, not for any written description.

Approach

1. Switch to high power and focus on the central stele of the root.
2. Pick four cells that form a small cluster in which each cell shares at least one wall with another.
3. Sketch the cluster at large size on the answer sheet.
4. Use a sharp pencil, draw each wall with two close parallel lines, and avoid any shading.
5. After drawing, double-check that you have exactly four cells, all correctly proportioned and touching.

Step-by-Step Reasoning

1. Lines and shading — continuous thin sharp lines; no shading anywhere.
2. Only four cells, each touching at least one other — draw exactly four cells; choose cells so that the cluster is connected, not four isolated cells.
3. Cell walls as two lines — each wall should be a double line, indicating wall thickness.
4. Correct shapes and relative sizes — draw the cells with the actual polygonal shapes and sizes seen down the microscope; do not regularise them into identical rectangles.

Key Takeaways

• Cell drawings use double lines for walls and no shading.
• Draw only the requested number of cells, no extras.
• Cells in the central stele of a young root are roughly polygonal and isodiametric.
• Choose cells that touch each other — a cluster, not four isolated cells.

Common Mistakes

• Drawing more than four cells (extra cells on the edges are not credited and may lose the only four cells mark).
• Drawing four cells that do not touch.
• Using a single line for the cell wall instead of two — loses the cell wall drawn as two lines mark.
• Adding shading or colouring inside the cells.
• Drawing cells as perfectly identical — the cells should reflect the variation seen.

Things to Be Careful About

• The double-line wall must be two close parallel lines, not two lines far apart (which would look like an extra cell between them).
• Each cell must share at least one wall with at least one other cell — a corner-touching is not enough.
• Keep the drawing large — examiners need clear detail to award marks.

Background Concept

A printed photomicrograph shows a magnified specimen. To find the actual size of a structure in the specimen, divide the image size (what you measure on the print) by the magnification stated on the figure. The vascular cylinder in a young dicot root cross-section is roughly circular, so its area is estimated using the formula

where $r$ is half of the diameter $A$–$B$. The magnification is given on the figure (×35), so all that is needed is a careful measurement of $A$–$B$ on the printed page, conversion to actual size, halving for the radius, and substitution into the area formula.

Understanding the Question

The student must measure line A–B (the diameter of the vascular tissue) on the printed Fig. 2.3, convert that measurement to the actual size in the plant, and use $\pi r^2$ to find the area. The answer must be in mm².

The four marks are for:

1. Recording the measurement of A–B.
2. Dividing by 35 to get the actual length.
3. Showing $r = (\text{actual length})/2$ and substituting into $\pi r^2$.
4. The final correct numerical answer.

Approach

1. Use a ruler to measure the length of the printed line A–B on Fig. 2.3 in mm (this is the image size).
2. Divide by the magnification (×35) to obtain the actual diameter in mm.
3. Halve the actual diameter to obtain the radius.
4. Square the radius, multiply by $\pi$, and quote the answer in mm² with an appropriate number of significant figures (typically 2 sig figs, matching the precision of the measurement).

Step-by-Step Reasoning

For the representative measurement above:

1. Image length of A–B: read off the printed figure — here $\approx 56\ \text{mm}$.
2. Actual length: $\dfrac{56}{35} = 1.6\ \text{mm}$.
3. Radius: $r = \dfrac{1.6}{2} = 0.8\ \text{mm}$, then $\pi r^2 = 3.14 \times (0.8)^2 = 3.14 \times 0.64 = 2.01\ \text{mm}^2$.
4. Final answer (to 2 sig figs): $\approx 2.0\ \text{mm}^2$.

If the candidate's own measurement of A–B differs (e.g. $50\ \text{mm}$ → $1.43\ \text{mm}$ diameter → $0.715\ \text{mm}$ radius → $\pi \times 0.715^2 = 1.6\ \text{mm}^2$; or $60\ \text{mm}$ → $1.71\ \text{mm}$ diameter → $\pi \times 0.857^2 = 2.3\ \text{mm}^2$), the same four steps apply and full marks are awarded provided each step is shown clearly.

Key Takeaways

• Actual size = image size ÷ magnification.
• Use $\pi = 3.14$ (as the question states).
• Quote the answer in the unit requested (mm² here, not µm²).
• Show every step — marks are split across the working, not just the final answer.

Common Mistakes

• Forgetting to divide by the magnification (using the image measurement directly).
• Forgetting to halve the diameter before squaring (using the diameter as $r$ — gives an answer 4× too large).
• Quoting the answer in the wrong units (e.g. µm² instead of mm²).
• Omitting working and giving only a final number — the four marks require the steps to be visible.

Things to Be Careful About

• The mark scheme accepts error carried forward: if your measurement is slightly different from the examiner's, you still get the working marks provided each manipulation is correct.
• Use the magnification stated on Fig. 2.3 (×35), not any other figure.
• Keep at least 2 significant figures in the final answer to reflect the precision of the measurement.

Background Concept

A hypothesis that "vascular tissue area changes as the plant grows" implies that plant age is the independent variable. To test it, the candidate must compare plants of different ages while keeping everything else constant (the species, the growing conditions, the sectioning method, the measuring method). This is the standard structure for an experiment with one independent variable and several controls.

The original method only used one plant of one age, so it cannot address change over time at all — the modification must add age as an independent variable and replicate the measurement across several ages.

Understanding the Question

You are asked to suggest modifications to the original method in part (c) so that it can test whether vascular tissue area changes with plant age. The marks are for: varying the plant's age, keeping the species the same, and keeping the growing variables the same.

The command word is suggest — any reasonable modification that genuinely tests the hypothesis is acceptable, but the mark scheme expects three specific points.

Approach

1. Identify the independent variable that needs to vary: plant age.
2. Identify the controlled variables that must be held constant: species, growing conditions.
3. State the modification as a clear instruction.

Step-by-Step Reasoning

1. Plant age — grow plants for several different lengths of time and take a section from each, so that vascular area can be plotted against age.
2. Species — use the same species in every age group; otherwise differences might be due to species, not age.
3. Growing variables — keep soil, water, light, temperature, pot size, and any other environmental variable identical between age groups.

Key Takeaways

• Vary the independent variable (plant age), control everything else.
• Replicate across several ages to show a trend.
• Keep species and growing conditions constant.

Common Mistakes

• Suggesting only one age (does not test change with age).
• Forgetting to specify that the species must be kept the same.
• Naming only one or two growing variables to control — the mark scheme treats "keep the variables for growing the plants the same" as one mark; you do not have to list every variable individually.

Things to Be Careful About

• "Take sections at different ages" is the heart of the modification — without it, the hypothesis cannot be tested.
• The mark scheme treats "keep the variables for growing the plants the same" as one mark, but the candidate should give at least one example (e.g. same light, same water) to show understanding.