Biology 9700/36 — October/November 2025

Background Concept

The Benedict's test detects reducing sugars. A positive result is shown by a colour change from blue through green, yellow and orange to a brick-red precipitate of copper(I) oxide. The reaction requires reducing sugar in excess of the copper(II) sulfate present, so the time to the first appearance of any colour change is inversely related to the concentration of reducing sugar in the sample: more reducing sugar → faster colour change → shorter time.

In this investigation, sucrose is hydrolysed by yeast enzymes into glucose and fructose, both of which are reducing sugars. As the reaction proceeds, the concentration of reducing sugars in the surrounding solution rises, and the Benedict's test becomes positive in a shorter time.

Two preparations of yeast are compared:

• Immobilised yeast in alginate beads — yeast cells trapped inside calcium-alginate gel beads formed when drops of a yeast-alginate mixture fall into calcium chloride solution.
• Free yeast cells — the same yeast suspension used directly, without immobilisation.

Immobilisation may slow the apparent reaction because substrate (sucrose) has to diffuse into the bead and product (reducing sugars) has to diffuse out, whereas free yeast enzymes are in direct contact with the sucrose.

Understanding the Question

You have carried out the practical and now need to record what you observed. From each of two beakers (B with beads, F with free yeast) you took a 1 cm³ sample every minute for 5 minutes and tested it with Benedict's solution. For each sample you timed, in seconds, how long it took to see the first colour change; if no change appeared within 60 s, you recorded 'more than 60'.

Approach

The table must compare two conditions at five sampling times. The independent variable (sampling time) goes down the left. The two conditions are the column headings placed to the left of the dependent variable (time to first colour change), which appears with the unit '/ s' in each column heading. Units must not appear in the body of the table. Times are recorded to the nearest whole second.

Step-by-Step Reasoning

1. Independent variable first: the time of sampling (1, 2, 3, 4, 5 min) is the independent variable and forms the leftmost column.
2. Two conditions as columns: yeast in beads (B) and free yeast cells (F) are the two conditions. The mark scheme requires them to be placed to the left of the dependent variable — achieved by making them the two right-most column headings while the dependent variable (time to colour change / s) is included in each heading.
3. Heading conventions: each column heading has both the quantity and the unit ('/ s'); units are not repeated in the body.
4. Five results per condition: one result per sampling time per condition, so ten cells in total.
5. Trend check: as sampling time increases, time to colour change must decrease in both columns because reducing sugars accumulate.
6. Comparison check: at every sampling time, free yeast must give a shorter time than beads, because direct enzyme–substrate contact is faster than diffusion through a bead.
7. Whole seconds: the instructions and the mark scheme require times to the nearest whole second, with 'more than 60' written if no colour change occurred within the 60-second window.

Key Takeaways

• Two conditions × five sampling times = ten data points in a 5 × 2 table.
• Heading format: quantity + unit; no unit in the body.
• Independent variable on the left; conditions placed to the left of the dependent variable.
• Whole-second recordings; use 'more than 60' for non-reactors within the timing window.
• The expected trend is decreasing times for both conditions, with free yeast always faster than beads.

Common Mistakes

• Forgetting to record 'more than 60' for the bead sample at 1 min — losing a mark because a value is missing where the instructions tell you how to record it.
• Repeating the unit 's' in every cell of the body — only the column heading should carry it.
• Putting the two conditions in rows and the time in columns — the mark scheme requires conditions to the left of the dependent variable.
• Recording times to one decimal place — the instructions and mark scheme require whole seconds.
• Missing out a time point so the table only has four results per condition.

Things to Be Careful About

• Judge the first colour change (often a faint green) rather than waiting for the brick-red end-point.
• Stir immediately before sampling so the value represents the bulk solution, not a local pocket of substrate or product.
• Match each test-tube (1–5) to its correct sampling time when recording — easy to muddle under exam pressure.

Background Concept

A trend is a general pattern of change in the dependent variable as the independent variable changes. Comparing trends between two conditions means describing both individually and then stating how they differ.

The dependent variable here is the time to first colour change, which is inversely related to reducing-sugar concentration. A shorter time therefore means more reducing sugar is present in the sample at the moment it was taken.

Understanding the Question

You need to do three things:

1. Describe how the time to colour change changed over the 5-minute sampling period for the beads.
2. Describe how it changed over the 5-minute sampling period for the free yeast.
3. Compare the two — say which condition gave shorter times and quote data to back it up.

The command word 'compare' requires a direct comparison, not just two separate descriptions.

Approach

Look down each column of your own results table. In both columns the times fall as the sampling time rises — that is the trend. Compare the two columns cell-by-cell: free yeast is always faster (shorter time) than beads. Pick one time point and quote both values as evidence.

Step-by-Step Reasoning

1. Beads trend: read down the B column. Times fall from 1 min to 5 min. State that as time increases, time to colour change decreases.
2. Free-yeast trend: read down the F column. Times also fall from 1 min to 5 min. State the same decreasing pattern.
3. Comparison: read across each row. Free yeast is always shorter than beads. State that comparison explicitly.
4. Data quote: choose one row (typically the clearest — here 5 min) and quote both numbers to give numerical support.

Key Takeaways

• 'Describe and compare' demands a direct comparison statement, not just two parallel descriptions.
• A data quote (one pair of numbers with units) earns the supporting-evidence mark.
• Decreasing time = increasing reducing-sugar concentration = faster enzyme activity.

Common Mistakes

• Writing 'the time decreases' without saying with what it decreases (i.e. with sampling time).
• Describing the two conditions separately without ever comparing them — only the description mark is earned, not the comparison mark.
• Vague comparison ('free is faster') without numbers — the data-quote mark is lost.
• Mixing up which condition is faster.

Things to Be Careful About

• Use the candidate's own numbers from (a)(i); do not invent values.
• Quote the numbers exactly as recorded, with the unit 's'.
• The comparison should reference the trend, not just state which is faster.

Background Concept

A control is a parallel experiment in which the variable being tested is removed or held constant, so that any observed effect can be attributed to that variable alone and not to something else in the procedure.

In this investigation the variable being tested is the activity of yeast enzymes on sucrose. A negative control therefore uses a preparation in which those enzymes are absent or inactive.

Understanding the Question

You need a control that proves the Benedict's colour change is caused by the yeast enzymes (and not, for example, by sucrose spontaneously reducing copper ions, or by impurities in the alginate beads).

Approach

The cleanest negative control is to remove the active enzyme. There are two ways:

• Boil the yeast (or boiled beads) — heat denatures the enzymes irreversibly, so any colour change observed cannot be due to enzyme activity.
• Replace the yeast with distilled water — there is no enzyme at all, so no reaction should occur.

Step-by-Step Reasoning

1. The Benedict's test is positive because enzymes in the yeast hydrolyse sucrose → reducing sugars → reduce Cu²⁺ to Cu⁺.
2. To show that the colour change depends on the enzymes, repeat the experiment with the enzymes removed.
3. Boiling denatures proteins, so boiled yeast contains no active enzyme. Boiled beads therefore give no hydrolysis and Benedict's remains blue (or 'more than 60').
4. Alternatively, use distilled water in place of the yeast suspension — same outcome.

Key Takeaways

• A control isolates the variable being tested.
• For an enzyme experiment, the standard control is boiled enzyme (denatured) or no enzyme (water).
• The control must be run through exactly the same procedure as the test, with only the enzyme variable changed.

Common Mistakes

• 'Use Benedict's on its own' — this does not control for the sucrose; it only checks that Benedict's does not change on its own.
• 'Use a different sugar' — that is an additional experimental variable, not a control.
• 'Repeat the experiment' — that is replication, not a control.

Things to Be Careful About

• The control should test the absence of the active variable, not a different concentration of it.
• Be explicit about what is being controlled and why.

Background Concept

An enzyme-catalysed reaction proceeds as long as substrate, enzyme and suitable conditions are present. Removing a sample from the reaction mixture does not stop the reaction unless the enzyme is also removed or inactivated.

The Benedict's test measures the concentration of reducing sugar at the moment the sample is heated with copper(II) sulfate. The time to first colour change is inversely related to that concentration.

Understanding the Question

In the F beaker the yeast cells are in suspension. When you take 1 cm³ into Benedict's, those yeast cells are still alive and still catalysing the hydrolysis of any sucrose that came with them. In the B beaker the beads are removed (the beads are bigger than the cells and stay in the funnel), so the sample does not contain active yeast.

The procedure specifies that F samples must be tested immediately, but B samples can be batched and tested together at the end. Why?

Approach

Compare what is in the F sample vs the B sample at the moment it is taken:

• B sample: sucrose solution surrounding the beads. The beads (and therefore the yeast enzymes) are left behind in the funnel, so the sample contains no active enzyme. The reducing-sugar concentration is now 'frozen' and the samples can be tested in batch.
• F sample: sucrose solution containing free yeast cells. The yeast enzymes are still present and will keep hydrolysing any remaining sucrose. The reducing-sugar concentration therefore rises during any delay.

Step-by-Step Reasoning

1. The reaction in F is sucrose → reducing sugars, catalysed by yeast enzymes.
2. As long as yeast cells, sucrose and water are all in the same tube, the reaction continues.
3. Removing 1 cm³ from beaker F does not stop the reaction — the cells and sucrose come with the sample.
4. While waiting, more sucrose is hydrolysed, so the reducing-sugar concentration rises.
5. A delayed Benedict's test therefore gives a shorter time to colour change than the true value at the sampling instant — the data point is biased high.
6. To avoid this bias, the F sample must be put into the boiling Benedict's immediately, which both stops the enzyme and starts the test.

Key Takeaways

• The Benedict's test must capture the reducing-sugar concentration at a defined moment.
• Free enzyme in a sample keeps working; immobilised enzyme is removed with the apparatus.
• The asymmetry in the procedure (batch test for B, immediate test for F) reflects this difference.

Common Mistakes

• Saying 'because Benedict's cools down' or 'to save time' — neither captures the enzymatic reason.
• Saying 'to avoid contamination' — irrelevant here.
• Saying 'because the samples would evaporate' — irrelevant.

Things to Be Careful About

• The free yeast cells are small enough to come through the syringe and remain in the sample; that is why the timing matters.
• The beads in B are large and stay in the beaker, so the sample is enzyme-free.

Background Concept

A suspension of cells in solution is not, at the microscopic scale, homogeneous. Cells settle, droplets form local regions of higher concentration, and any small sample taken without mixing may over- or under-represent the bulk composition. Stirring (or shaking) before each sample is taken equalises the composition so the sample reflects the bulk.

Accuracy, in this context, means that the value recorded is close to the true value for the bulk at that moment. Stirring improves accuracy by removing sampling bias.

Understanding the Question

You stirred the yeast-alginate mixture, the beads in beaker B, and the free yeast in beaker F. The question asks you to explain two distinct ways that stirring increased the accuracy of the readings.

Approach

Think of the contents of the beaker at the moment of sampling:

• Yeast cells (the enzyme source).
• Sucrose (the substrate).
• Reducing sugars (the accumulating product).

Stirring helps each of these be uniformly distributed. Pick two distinct benefits that translate into a more accurate reading.

Step-by-Step Reasoning

1. Yeast-cell uniformity — without stirring, yeast cells settle or form clumps. Some 1 cm³ samples would contain more cells (and therefore more enzyme activity) than others. Stirring suspends the cells evenly so every sample has the same number of cells in it; the time recorded then reflects a true rate rather than a sampling artefact.
2. Substrate uniformity — sucrose can form local concentration gradients as it is consumed near the cells. Stirring equalises the sucrose concentration so every sample sees the same bulk substrate level.
3. Product uniformity — reducing sugars produced by the reaction also need to be evenly distributed before a sample is taken, otherwise some samples reflect local accumulation rather than the bulk value.
4. Even temperature distribution (less relevant here, but a valid general point) — stirring avoids hot or cold spots, particularly relevant if the beaker sits over a heat source.

Any two of these accuracy benefits earns the marks.

Key Takeaways

• Stirring in a heterogeneous mixture is essential to make every small sample representative of the bulk.
• Accuracy = closeness to the true value; stirring removes sampling bias introduced by local heterogeneity.
• For this experiment, the three components that must be evenly distributed are yeast cells, sucrose and reducing sugars.

Common Mistakes

• Vague answers such as 'it mixes things up' or 'so the reaction happens properly' — too imprecise to earn credit.
• 'Stops the reaction' — stirring does not stop the reaction; it merely mixes the contents.
• 'Increases the rate' — stirring may give a small kinetic benefit, but the question asks about accuracy, not rate.

Things to Be Careful About

• The question asks specifically about accuracy, not rate or yield.
• Each answer must name a specific component (cells, substrate, product) that becomes evenly distributed.

Background Concept

In the free-yeast part of the investigation, step 26 (taking a sample) and step 27 (timing the Benedict's test) are supposed to be done back-to-back, but in practice the candidate has to:

1. Pick up the syringe.
2. Draw up 1 cm³ from F.
3. Transfer it to test-tube 1.
4. Start the Benedict's timer immediately.

This is hard to do smoothly, and the moment the colour change appears is easily missed because the candidate's attention is split between the sample transfer and the colour judgement.

A source of error in this context is anything that makes the recorded value different from the true value.

Understanding the Question

You must identify one source of error specific to the free yeast part of the procedure (where the sample has to be tested immediately) and propose one improvement that would make the timing more accurate.

Approach

The error must be something the procedure forces the candidate to do, not a vague comment about 'human error'. The improvement must directly address that error.

Look at steps 25–32: at each one-minute mark the candidate both takes the sample and begins the Benedict's timing. That is the moment of difficulty.

Step-by-Step Reasoning

1. Source of error: the candidate has to simultaneously take a sample and start timing the Benedict's colour change. This is a difficult multitasking task and the first appearance of colour is easily overlooked.
2. Improvement 1: set up five separate beakers, each started one minute after the previous one, so the Benedict's timer for beaker 1 is started at the same time as the addition of yeast to beaker 1, and so on — only one task at a time.
3. Improvement 2: take the sample, then immediately add it to Benedict's that is already at boiling, so the candidate can focus purely on judging the colour change without taking a sample at the same time.
4. Improvement 3: stop the enzyme reaction as soon as the sample is taken, e.g. by adding the sample to Benedict's that contains a denaturant, or by adding the sample to a hot tube — this removes the timing pressure from the Benedict's step.

Any of these earns the improvement mark, provided the error and improvement are clearly linked.

Key Takeaways

• An 'error' must be specific to the procedure, not generic ('human error').
• The improvement must directly solve the error identified.
• In multitasking procedures, separating the tasks into separate steps almost always improves accuracy.

Common Mistakes

• 'Human error' — too vague, never credited.
• 'Not accurate measurements' — not a specific source of error.
• Improvement that does not match the error (e.g. error about temperature, improvement about stirring).
• Improvement that simply repeats the procedure ('do it again', 'be more careful') — these are not changes to the procedure.

Things to Be Careful About

• The error and improvement are specific to the free yeast part, where multitasking is forced.
• A valid improvement must be a real change to the procedure, not just a vague intention.

Background Concept

A line graph is the correct choice when both variables are continuous (numerical) and the aim is to show how one changes with the other. Two conditions on the same axes allow the reader to compare them directly.

Standard conventions:

• The independent variable goes on the x-axis (horizontal).
• The dependent variable goes on the y-axis (vertical).
• Each axis is labelled with the quantity and the unit.
• The scale uses at least half the grid in both directions and is not awkward (e.g. avoid 30 or 70 per 2 cm).
• Each axis is labelled at least every 2 cm.
• Points are plotted precisely with a small symbol (dot in circle or cross).
• Points are joined with a thin line — straight between successive points, no 'dot-to-dot' embellishment.
• Each line is clearly labelled so the reader can tell which is which.

Understanding the Question

Table 1.2 gives the activity of immobilised and free catalase at five pH values (5, 6, 7, 8, 9). You need to plot both sets of data on Fig. 1.4 so the reader can see how activity varies with pH for each preparation.

Approach

• Choose axes: pH is the independent variable (set by the scientist); catalase activity is the dependent variable (measured).
• Choose scales: pH from 5 to 9 fits easily on the grid at 1 cm per unit. Activity from 0 to 100 fits at 20 per 2 cm (5 × 2 cm = 10 cm on the y-axis).
• Plot the two sets of points using distinguishable symbols.
• Join each set with a thin straight line through all its points.
• Label each line.

Step-by-Step Reasoning

1. Axes:
   • x-axis label: 'pH'.
   • y-axis label: 'activity of catalase / arbitrary units (au)'.
2. Scale on x-axis: 1 cm per unit. Labels at 5, 7, 9 (every 2 cm). All five pH values fit within the grid.
3. Scale on y-axis: 20 au per 2 cm. Labels at 0, 20, 40, 60, 80, 100 (every 2 cm). 100 au fits comfortably on the y-axis.
4. Plotting:
   • Immobilised: (5, 68), (6, 88), (7, 99), (8, 98), (9, 94). Use small dot in circle (or cross) symbols.
   • Free: (5, 50), (6, 62), (7, 96), (8, 65), (9, 48). Use a different symbol (e.g. plain cross) so the two are distinguishable.
5. Lines: thin, straight, passing through every plotted point in each series.
6. Labels: write 'immobilised' near the right-hand end of the immobilised line and 'free' near the right-hand end of the free line (or use a key/legend).

Key Takeaways

• Line graph for two continuous variables.
• x-axis = independent, y-axis = dependent; both labelled with quantity and unit.
• Scales that use at least half the grid, labelled every 2 cm.
• Small precise point symbols; thin straight lines through all points; clear labels.
• Two conditions can be shown on one graph using distinguishable symbols.

Common Mistakes

• Reversing axes (activity on x, pH on y).
• Plotting pH as a categorical axis (only the five values, with no scale).
• Using awkward scales (e.g. 30 or 70 au per 2 cm).
• Drawing smooth curves rather than straight lines between the points.
• Thick, fuzzy lines or large blobs as point symbols.
• Joining the two series with a single line by mistake.
• Failing to label the two lines.

Things to Be Careful About

• Use a sharp pencil so the points and lines are precise.
• Read each y-value from Table 1.2 carefully — easy to misread 68 as 86.
• The two lines cross each other between pH 6 and pH 7; make sure the symbols and lines are clearly distinguishable at the crossing point.

Background Concept

The optimum pH of an enzyme is the pH at which its activity is highest. Many enzymes work over only a narrow pH range and lose activity quickly on either side of the optimum.

Immobilisation can change the apparent properties of an enzyme in two ways:

• It may physically protect the enzyme from extremes of pH, denaturation, or temperature.
• It may constrain the enzyme's conformation, sometimes broadening the range of conditions over which it remains active.

A wider activity range is a useful practical property because it means the immobilised enzyme is more tolerant of varying conditions in a real reactor.

Understanding the Question

You have a graph with two lines: immobilised catalase activity and free catalase activity, both plotted against pH. From this graph, state two conclusions about how pH affects the activity of the immobilised catalase compared to the free catalase.

Approach

Look at the shape of each curve:

• Where is the peak (highest point) of each? Both peaks are at pH 7, so the optimum pH is the same.
• How steeply does each curve fall away from the peak on either side? The immobilised curve falls much more gently than the free curve, so the immobilised enzyme retains more activity across the range tested.

Step-by-Step Reasoning

1. Optimum pH: read the highest point of each line. The immobilised line peaks at (7, 99) and the free line peaks at (7, 96). Both optima are at pH 7.
2. Range of activity: compare the slopes away from the optimum. The free line falls steeply on either side of pH 7 (from 96 at pH 7 down to 50 at pH 5 and 48 at pH 9). The immobilised line falls only gently (from 99 at pH 7 down to 68 at pH 5 and 94 at pH 9).
3. Therefore, the immobilised catalase has a higher activity over a wider range of pH than the free catalase.
4. Quote values to support the second conclusion: e.g. at pH 9, immobilised = 94 au but free = 48 au; at pH 5, immobilised = 68 au but free = 50 au.

Key Takeaways

• The optimum pH is the pH at which activity is highest; it can be read directly from the peak of a graph.
• The range of activity over which an enzyme remains effective is often as important as the optimum in real applications.
• Immobilisation can broaden the effective pH range of an enzyme.

Common Mistakes

• Saying the optimum pH is different for the two preparations — both peak at pH 7.
• Saying immobilised catalase is always more active than free catalase — at pH 7 they are very similar (99 vs 96), but elsewhere the immobilised line is consistently higher.
• Stating the conclusion without reference to data (no pH or activity values quoted).
• Confusing the optimum with the range — these are two distinct features of the graph.

Things to Be Careful About

• A 'conclusion' must be a complete sentence stating a relationship, not a single value.
• The two conclusions should be distinct; 'same optimum' and 'wider range' are the two the data clearly support.

Background Concept

A plan diagram (sometimes called a low-power plan) is a simplified outline drawing of a specimen as seen down the microscope. The conventions exist so that information about the distribution of tissues, and the shape and proportion of the whole organ, is recorded clearly and consistently.

Key conventions for a plan diagram:

• No cells are drawn — only the boundaries of tissues.
• Continuous, thin, sharp lines (drawn with a sharp pencil).
• No shading anywhere.
• The epidermis (where present) is drawn as two lines drawn close together to show it is a distinct, narrow tissue layer.
• The drawing should fill most of the available space and should be in correct proportion to the specimen.
• Labels are written in pencil on the page outside the drawing, joined to the structure by a straight, horizontal ruler line ending precisely on the structure (the line must touch the structure and the label).

The transverse section of a young dicotyledonous stem (such as a sunflower, Helianthus, or buttercup, Ranunculus, stem — typical Cambridge A-Level specimens) shows, from outside in:

• Epidermis (single layer of cells, often with a thin cuticle).
• Cortex (parenchyma below the epidermis).
• A ring of vascular bundles, each containing xylem (towards the inside of the section) and phloem (towards the outside of the section), separated by a cambium in young stems.
• Pith (parenchyma in the centre).

Within each bundle the phloem is the tissue on the outer side, the xylem is the tissue on the inner side, and the small-celled cambium lies between them.

Understanding the Question

The candidate has been given slide N1, a stained transverse section (TS) through a plant organ. They must produce a large plan diagram of the WHOLE section. They must also use one ruled label line to identify the phloem. Five marks are available, all for plan-diagram conventions and the phloem label.

Approach

First, look at the section under the lowest power of the microscope to see the whole section. Identify the outline shape and the position of every vascular bundle. Choose a clear region of the section to draw (avoid torn or folded areas) and draw the outline of the epidermis, then sketch in the cortex, vascular bundle ring, and pith as smooth continuous outlines. Add the epidermis as a second line just inside the outer line. Finally, draw a ruler line from the phloem in one bundle out to one side and write phloem.

Step-by-Step Reasoning

• Mark 1 — size and correct number of tissues: The plan must be large (use most of the available space) and must show the correct number of tissue layers visible in the section. For a young dicot stem these are: epidermis, cortex, vascular bundles, pith.
• Mark 2 — whole section, no cells: The whole outline of the section must be included. No individual cells should be drawn — this is what makes the diagram a "plan" rather than a cell drawing.
• Mark 3 — epidermis as 2 lines close together: A second line is drawn just inside the outer outline to represent the inner edge of the epidermis.
• Mark 4 — correct number of vascular bundles: The candidate must count and draw the correct number of bundles in their correct positions (a ring near the outside of the section, inside the cortex).
• Mark 5 — phloem label: A single straight ruler line is drawn from the phloem (the small-celled outer part of a bundle) out to the side of the drawing, with the label phloem written at the end. The line must touch the phloem and the line must be horizontal.

Key Takeaways

• A plan diagram is a tissue-level drawing, never a cell-level drawing.
• The epidermis, when present, is always drawn as a double line.
• The phloem is the outer component of each vascular bundle, the xylem is the inner component, and the cambium is the small-celled band between them.
• Labels must be on a horizontal ruler line that ends precisely on the structure.

Common Mistakes

• Drawing individual cells inside the section (this makes it a low-power drawing, not a plan) — loses the "no cells" mark.
• Drawing the epidermis as a single line.
• Drawing too few or too many vascular bundles.
• Labelling xylem where phloem should be (xylem is the inner, larger-celled part of a bundle; phloem is the outer, smaller-celled part).
• Using a label line that does not touch the structure, or that has an arrow, or that is not horizontal.

Things to Be Careful About

• The phloem in a TS through a dicot stem is the part of the bundle nearer the outside of the section.
• The line must be drawn with a ruler, and the line must end on the phloem, not in empty space.
• The drawing must be in the correct proportion to the section — a very small drawing cannot earn the "uses most of the available space" mark.

Background Concept

The cortex in a young dicot stem is the parenchymatous tissue that lies immediately inside the epidermis and outside the ring of vascular bundles. Its cells are:

• Roughly isodiametric (similar length and width) or polygonal.
• Have thin cellulose cell walls (drawn as two lines close together because the cell wall plus middle lamella sits between the two adjacent protoplasts).
• Often have small intercellular air spaces at the corners where three or more cells meet.

A drawing of cells under high power is a biological drawing, not a plan diagram. Conventions for cell drawings:

• Continuous, thin, sharp, single lines drawn with a sharp HB pencil.
• No shading anywhere.
• Each cell wall is drawn as two lines close together.
• The relative sizes, shapes, and positions of the cells must be as observed under the microscope — not an idealised textbook version.
• No labels inside the cells (other than, for example, a nucleus if visible and required).
• One horizontal ruler line from the structure to be labelled, ending precisely on the structure, with the label written in pencil at the outer end.

Understanding the Question

The candidate has to look at slide N1, find the cortex (the tissue immediately below the epidermis), and choose four adjacent cortex cells in which each cell touches at least two of the others. They then make a large drawing of those four cells, using one ruled label line to identify the cell wall. Five marks are available.

Approach

First locate the cortex (just under the epidermis). Switch to a higher power objective so the cell walls are clearly visible. Look for a small group of four cortex cells in a tight cluster where each cell shares a wall with at least two of the others — a 2 × 2 block is the simplest. Draw the outlines carefully with a sharp pencil, then add the second line of each cell wall. Finally, draw one horizontal ruler line from the cell wall of one of the cells out to the side, and label it cell wall.

Step-by-Step Reasoning

• Mark 1 — quality of lines and no shading: The drawing must be done with a sharp pencil in continuous thin lines. No broken, fuzzy, or shaded lines anywhere.
• Mark 2 — exactly four cells, each touching at least two others: A 2 × 2 arrangement satisfies this. If a cell only touches one other, the mark is lost.
• Mark 3 — cell walls as two lines close together: Each shared wall is drawn as a pair of lines (the wall of one cell plus the wall of its neighbour).
• Mark 4 — correct shapes: The cells must look like cortex parenchyma — polygonal / isodiametric with rounded corners — and the relative sizes must match what is seen.
• Mark 5 — cell wall label: A single horizontal ruler line ending on the cell wall (between the two lines) with the words cell wall written at the outer end.

Key Takeaways

• Cell drawings show cells, plan diagrams show tissue outlines.
• Cell walls are always drawn as two lines close together.
• The "four adjacent cells" rule forces the candidate to think about which cells share a wall — this is good practice for interpreting the tissue.
• Labels must use a horizontal ruler line that ends on the structure.

Common Mistakes

• Drawing the cell wall as a single line.
• Drawing five or three cells, or a single cell.
• Drawing cells that touch at only one point (corner contact) rather than sharing a wall.
• Using shading or coloured pencil.
• Label line that does not end on the cell wall, or that is drawn with a freehand line instead of a ruler.

Things to Be Careful About

• "Touches" in the Cambridge mark scheme means shares a wall, not merely corner-to-corner contact.
• The drawing must be large (use most of the available space).
• Cortex cells in young dicot stems are typically parenchyma; if the section is from a different organ (e.g. a root), the shapes will be different — use what is actually seen.

Background Concept

To find a representative size of an irregularly shaped specimen (such as a roughly circular TS of a stem), it is good practice to take several measurements through the centre and then calculate a mean. This averages out the effect of any irregularity in the outline and reduces the effect of a single inaccurate measurement.

For a roughly circular section, the diameters should be measured across the section passing through the centre (or as close to the centre as possible). The black dot in Fig. 2.2 has been placed at the centre specifically so that the candidate can align each diameter line through it.

Understanding the Question

Fig. 2.2 is the same photomicrograph as Fig. 2.1 with a black dot at the centre of the section. The candidate is asked to determine the mean diameter of the section as it appears on the page. The marks are for: taking at least three diameter measurements, and recording a mean value with appropriate units.

Approach

Use a ruler. Measure the distance from one side of the section to the other, passing through the black dot. Take at least three such measurements, ideally at different angles (e.g. horizontal, and two diagonals). Record each one with the correct unit, then calculate the mean.

Step-by-Step Reasoning

• Mark 1 — at least three measurements: The mark scheme requires at least three diameter measurements through (or near) the centre.
• Mark 2 — values within range and units: The measurements must be sensible (not 5 mm, not 500 mm) and must include the unit (mm on a printed figure, µm if measuring on the microscope).

For Fig. 2.2 as printed, the candidate's actual numbers will depend on the size of the figure in their exam paper. The example above (95 mm) is a representative value consistent with a printed figure of around 8–10 cm in diameter with a scale bar of around 4 mm.

Key Takeaways

• Multiple measurements through the centre, then a mean, give a more reliable value than a single measurement.
• Always include the unit with a measurement.
• The black dot in Fig. 2.2 is provided specifically as a reference for the centre of the section.

Common Mistakes

• Taking only one measurement.
• Recording a measurement without a unit.
• Measuring the radius (from centre to edge) and forgetting to double it.
• Measuring diagonally across the page rather than through the centre of the section.

Things to Be Careful About

• The diameter must pass through the centre — otherwise the measurement is a chord, not a diameter.
• Record the unit with every measurement and with the mean.

Background Concept

A scale bar on a photomicrograph is a printed line of known actual length. By measuring how long the scale bar is on the page (its image length) and how long the structure of interest is on the page, the actual size of the structure can be calculated by proportion:

This is a direct application of similar triangles: the printed image is a scaled version of the actual specimen, so the ratio of image to actual is the same for any feature on the image.

Understanding the Question

The scale bar on Fig. 2.2 represents $310\ \text{µm}$. The candidate is asked to use this scale bar to convert the mean image diameter measured in (c)(i) into a mean actual diameter of the section.

Approach

Measure the scale bar with a ruler. Substitute the mean image diameter (from (c)(i)) and the scale-bar image length into the proportion above. Multiply by $310\ \text{µm}$.

Step-by-Step Reasoning

• Mark 1 — measure the scale bar with units: Measure the printed length of the scale bar to the nearest mm and record the unit.

• Mark 2 — correct substitution:

  $$\frac{\text{mean diameter}}{\text{scale bar image length}} \times 310\ \text{µm}$$

  must be shown explicitly in the working.

Using the representative values of $95\ \text{mm}$ (mean image diameter) and $4.0\ \text{mm}$ (scale bar image length):

This rounds to $7400\ \text{µm}$ (or $7.4\ \text{mm}$). The candidate's answer will vary with their own measurements; the mark scheme credits any answer obtained by this method.

Key Takeaways

• A scale bar allows you to convert any measured image length to actual size by simple proportion.
• The actual size of the structure is larger than its image if the magnification is greater than 1, and smaller if the magnification is less than 1.
• Always include the unit of the scale bar in your working.

Common Mistakes

• Dividing the wrong way round (dividing the scale bar by the diameter instead of the diameter by the scale bar).
• Forgetting to multiply by the scale-bar value ($310\ \text{µm}$).
• Quoting the answer in mm instead of µm, or vice versa, without converting.
• Using the wrong measurement from (c)(i) (for example, the radius instead of the diameter).

Things to Be Careful About

• The scale bar unit is µm; the mean image diameter is in mm. The proportion is dimensionless, so the units cancel cleanly and the answer is in µm.
• Keep at least 2 significant figures in the final answer.

Background Concept

Magnification is defined as

Both sizes must be expressed in the same unit before the ratio is taken, or the answer will be wrong by a factor of 1000 (typical mm to µm conversion error). The answer is dimensionless — it is just a number — and is conventionally written with a multiplication sign in front, e.g. $\times 13$.

The magnification of a photomicrograph is the factor by which the printed image is larger than the actual specimen. A young dicotyledonous stem of a few mm diameter, when photographed and printed to a few cm, will have a magnification of around $\times 10$ to $\times 30$, depending on the print size.

Understanding the Question

The candidate is asked to use the mean actual diameter calculated in (c)(ii) and the mean image diameter from (c)(i) to calculate the magnification of Fig. 2.2, giving the answer to the nearest whole number.

Approach

Convert the image size into the same unit as the actual size (µm in this case: $1\ \text{mm} = 1000\ \text{µm}$). Divide the image size by the actual size. Round to the nearest whole number.

Step-by-Step Reasoning

• Mark 1 — correct substitution: The mean image diameter is divided by the answer from (c)(ii).
• Mark 2 — correct answer to the nearest whole number.

Using the representative values:

• Mean image diameter $= 95\ \text{mm} = 95000\ \text{µm}$.
• Mean actual diameter $= 7362.5\ \text{µm}$.

So the magnification is $\times 13$ to the nearest whole number.

The candidate's own answer will depend on their measurements in (c)(i) and (c)(ii); the method is the same.

Key Takeaways

• $M = \dfrac{\text{image}}{\text{actual}}$.
• Always convert to the same unit before dividing.
• A photomicrograph typically has a magnification of $\times 10$ to $\times 1000$ depending on the objective and the print size.

Common Mistakes

• Dividing the actual size by the image size (giving a number less than 1, which cannot be a magnification greater than 1).
• Forgetting to convert mm to µm (off by a factor of 1000).
• Not rounding to the nearest whole number.
• Quoting the answer without the multiplication sign, e.g. "13" rather than "$\times 13$".

Things to Be Careful About

• Use the mean actual diameter from (c)(ii), not the actual diameter of the scale bar ($310\ \text{µm}$).
• Magnification has no units — it is just a number.
• Write the answer as $\times 13$ (or $13$ if you cannot use a multiplication sign).