Biology 9700/33 — February/March 2025

Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Analysis, Conclusions and Evaluation · Manipulation, Measurement and Observation · Presentation of Data and Observations · Use of the Light Microscope

Q1MediumManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

Agar cubes that have been stained with a blue indicator called DCPIP can be used to investigate diffusion.

When ascorbic acid diffuses into an agar cube stained blue with DCPIP, it causes the DCPIP to decolourise (the blue colour disappears). The end-point is reached when the agar cube has completely decolourised all the way through to the centre.

You will investigate the effect of temperature on the time taken to reach the end-point.

You are provided with the materials shown in Table 1.1.

Table 1.1

labelled	materials	hazard	volume / $\text{cm}^3$
A	ascorbic acid solution in a beaker	low	100
B	agar block stained blue with DCPIP in a Petri dish	low	—

If A or B comes into contact with your skin, wash the affected area under cold water.

It is recommended that you wear suitable eye protection and disposable gloves.

You will need to:

cut agar block B into cubes of equal size
incubate the agar cubes in the ascorbic acid solution A at different temperatures
record the time taken for each cube to reach the end-point.

The cubes will all be cut to a size of $5\text{mm} \times 5\text{mm} \times 5\text{mm}$ , as shown in Fig. 1.1.

Fig. 1.1

(a)

You will use five different temperatures. The lowest temperature will be the temperature of the water in the beaker labelled water-bath before heating. The highest temperature will be $60\,^\circ\text{C}$ . You will need to decide on the three other temperatures that you will use.

13M

(i)

Measure the temperature of the water in the beaker labelled water-bath.

Decide on the three other temperatures that you will use.

Complete Table 1.2 to show the temperature of the water in the water-bath and the three other temperatures that you have decided to use. The maximum temperature is already included.

Table 1.2

temperature / $^\circ\text{C}$
______	______	______	______	60
water-bath				maximum

1M Similar

DifficultyEasy

Worked solution

Answer

Table 1.2 completed (representative values, with a water-bath measured at $20\,^\circ\text{C}$ ):

temperature / $^\circ\text{C}$	20	30	40	50	60
	water-bath				maximum

Final answer

Water-bath temperature recorded (e.g. 20 °C) and three intermediate temperatures of e.g. 30, 40 and 50 °C

Detailed explanation

Background Concept

This investigation is a classic rates-of-diffusion practical. Ascorbic acid diffuses from the surrounding solution into a stained agar cube and reacts with the indicator DCPIP, bleaching it. The time taken for the colour to disappear completely is an indirect measure of the rate of diffusion into the cube. Rate of diffusion is affected by temperature because particles gain kinetic energy.

Understanding the Question

The first sub-part of the practical asks the candidate to (1) measure the starting temperature of the water in the water-bath and (2) choose the three intermediate temperatures to use, given that the maximum is fixed at $60\,^\circ\text{C}$ . The values must (a) all lie below the $60\,^\circ\text{C}$ maximum, (b) include the starting water-bath temperature as the lowest, and (c) be spaced at least $5\,^\circ\text{C}$ apart.

Approach

Read the thermometer in the beaker to obtain the lowest temperature. Pick three values between this reading and $60\,^\circ\text{C}$ , evenly spaced, that are sensible for the practical. The mark scheme explicitly requires the three chosen values to be below $60\,^\circ\text{C}$ and at least $5\,^\circ\text{C}$ apart from one another and from the maximum.

Step-by-Step Reasoning

The water-bath reading is taken first, for example $20\,^\circ\text{C}$ (this will vary between sittings and between candidates).
With a maximum of $60\,^\circ\text{C}$ , three additional temperatures evenly spaced at $10\,^\circ\text{C}$ intervals gives 30, 40, 50 °C; these are below the maximum and at least $5\,^\circ\text{C}$ apart.
Other valid combinations include 25, 35, 45 °C or 20, 35, 50 °C provided each gap is $\geq 5\,^\circ\text{C}$ and all values are $\leq 60\,^\circ\text{C}$ .

Key Takeaways

Identify the independent variable and select a sensible range with regular intervals.
A minimum of five values of the IV is required for a trend to be visible.
The range should span from a low/room temperature to a high but safe maximum.

Common Mistakes

Choosing values that exceed $60\,^\circ\text{C}$ (rejected).
Spacing values less than $5\,^\circ\text{C}$ apart (rejected by the mark scheme).
Omitting the actual water-bath reading instead of measuring it.

Things to Be Careful About

Always use the actual thermometer reading for the lowest value; do not just assume a value such as 20 °C. The gap rule is at least 5 °C — equal spacing is conventional but not required.

Techniques used

choose five temperatures spanning a suitable rangespace intermediate temperatures at equal intervals of at least 5 °C

(ii)

Carry out step 1 to step 8.

step 1 On the tile provided, cut 5 agar cubes to the size shown in Fig. 1.1.

Put any waste pieces of agar into the container labelled For waste.

step 2 Put $10\,\text{cm}^3$ of A into a large test-tube.

step 3 Put the large test-tube into the water-bath and wait for 2 minutes.

Explain why the test-tube is left in the water-bath for 2 minutes in step 3.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

So that the ascorbic acid solution reaches the temperature of the water-bath (equilibrates) before the agar cube is added, ensuring the cube is tested at the stated temperature.

Final answer

So that the ascorbic acid reaches the temperature of the water-bath before the agar cube is added

Detailed explanation

Background Concept

A water-bath is used to hold a liquid at a constant, known temperature. When a small volume of liquid is placed in a much larger volume of water at a different temperature, heat flows between them until the smaller volume reaches the same temperature as the bath.

Understanding the Question

In step 3 the test-tube containing $10\,\text{cm}^3$ of ascorbic acid solution is placed in the water-bath and left for 2 minutes before the agar cube is added in step 4. The candidate must explain the purpose of this wait.

Approach

Think about what state the ascorbic acid needs to be in when timing begins. If the ascorbic acid is not at the bath temperature when the cube is added, the cube will not be at the intended experimental temperature, and the time recorded will not be a fair test of that temperature. The 2-minute wait allows heat transfer to occur.

Step-by-Step Reasoning

Heat moves from the water-bath to the test-tube wall and then to the ascorbic acid solution.
After ~2 minutes the ascorbic acid has reached the same temperature as the bath.
When the agar cube is then added, both the cube and the ascorbic acid are at the stated temperature, so the only variable affecting the rate of diffusion is temperature itself.
Starting timing at this point gives a fair measurement of the time taken at that temperature.

Key Takeaways

A 'wait' or equilibration step is a control-of-variables technique.
Standardising the temperature of the ascorbic acid before adding the agar cube is essential for a valid comparison across temperatures.

Common Mistakes

Saying 'to heat up the cube' — the cube is not in the bath during step 3.
Saying 'so the DCPIP reacts' — the timing has not yet started.
Vague answers such as 'to make it work properly' — the mark scheme requires a specific link between temperature equilibration and a fair test.

Things to Be Careful About

The mark is awarded for stating that the ascorbic acid (or its contents) reaches the water-bath temperature. Avoid referencing the cube or DCPIP in this answer.

Techniques used

explain the need to equilibrate temperature before starting the timer

(iii)

step 4 After 2 minutes, put one of the agar cubes into the large test-tube and immediately start timing.

step 5 Measure the time taken for the agar cube to reach the end-point. Record this time in (a)(iii).

The end-point is when the blue colour disappears from the whole agar cube.

If the end-point has not been reached after 300 seconds, stop timing and record the result as ‘more than 300’.

step 6 Remove the large test-tube from the water-bath and place it in the test-tube rack.

step 7 Increase the temperature of the water-bath to the next temperature stated in Table 1.2 and maintain this temperature.

step 8 Repeat step 2 to step 7 until all of the temperatures stated in Table 1.2 have been tested.

Record your results in an appropriate table.

5M Similar

DifficultyMedium

Worked solution

Answer

Representative results table (values are student-dependent; the trend is what the mark scheme requires):

temperature / $^\circ\text{C}$	time / s
20	285
30	175
40	95
50	55
60	25

Final answer

A table with two column headings (temperature / °C and time / s), five time values recorded to the nearest whole second, showing a clear decrease in time as temperature increases.

Detailed explanation

Background Concept

Paper 3 marks for recording data in a table reward the table's conventions rather than any single set of numbers. Conventions include: a heading for the independent variable, a heading for the dependent variable, each heading carrying a unit, the data going down the page in a consistent direction, and the use of whole seconds because the stopwatch is read in seconds.

Understanding the Question

The candidate carries out steps 1–8 for each of the five temperatures in Table 1.2 and records the time taken for the agar cube to decolourise in their own results table. The marking scheme awards five separate marks: (1) a heading for the independent variable with its unit, (2) a heading for the dependent variable with its unit, (3) a time recorded for each temperature, (4) the correct trend, and (5) times recorded as whole seconds.

Approach

Before the experiment, sketch a table with two columns: 'temperature / $^\circ\text{C}$ ' and 'time / s'. As each cube is timed, write the time in whole seconds next to the corresponding temperature. The expected trend is that time decreases as temperature increases, because diffusion of ascorbic acid is faster at higher temperatures.

Step-by-Step Reasoning

Independent-variable heading: 'temperature / $^\circ\text{C}$ '. The slash separates the quantity from its unit.
Dependent-variable heading: 'time / s'. Using seconds (not minutes) is the appropriate unit because the procedure instructs the candidate to stop the timer after 300 s, so the column is read in seconds.
Time recorded for each temperature: five rows, one per temperature tested.
Correct trend: the times should fall as temperature rises, with no obvious anomalies for the highest credibility. Because the actual numbers are student-dependent, the candidate simply needs the overall direction of change to be downward.
Whole seconds: values are recorded to the nearest second; no decimal places for time.

The representative values above (20 °C → 285 s; 30 °C → 175 s; 40 °C → 95 s; 50 °C → 55 s; 60 °C → 25 s) illustrate the expected shape of the relationship without prescribing the candidate's own readings.

Key Takeaways

Headings must include both quantity and unit.
Independent variable goes in the first column (left), dependent variable in the second column (right).
Data in a table should be in a form the reader can interpret without further work; whole seconds here is correct because the stopwatch reads in seconds.
A sensible trend in a diffusion-rate experiment with temperature as the independent variable is a decrease in time.

Common Mistakes

Omitting units in the column headings.
Using 'time/min' instead of 'time/s'.
Recording time to one decimal place or in minutes.
A table with times that do not fall with temperature — the mark scheme rejects this.

Things to Be Careful About

The candidate's own readings will not match the example values above; the mark is for the conventions and the trend, not for any specific number.

Techniques used

construct a results table with headings, units and recorded times

(iv)

State the dependent variable in this investigation.

1M Similar

(v)

Describe and explain the trend in your results.

2M Similar

DifficultyMedium

Worked solution

Answer

As temperature increases, the time taken for the agar cube to decolourise decreases (the rate of diffusion increases).
Ascorbic acid particles have more kinetic energy at higher temperatures, so they move faster and diffuse into the cube more quickly.

Final answer

As temperature increases the time to decolourise decreases because ascorbic acid particles have more kinetic energy and diffuse faster.

Detailed explanation

Background Concept

The kinetic theory of matter states that the temperature of a substance is a measure of the average kinetic energy of its particles. As temperature rises, particles move faster, collide more often, and the rate of any diffusion process (a passive process driven by random particle motion) increases. The diffusion of ascorbic acid through agar is a passive process and so follows this rule.

Understanding the Question

The candidate has collected times for five temperatures and must now describe the overall pattern in the data and explain why the pattern occurs. Two marks are available.

Approach

Separate the answer into a description (what the data show) and an explanation (why the data show what they do). Use the table that was just constructed.

Step-by-Step Reasoning

Description (mark 1): state the relationship explicitly. 'As temperature increases, the time taken for the agar cube to decolourise decreases.' This is a comparative statement that links the IV to the DV with the direction of change.
Explanation (mark 2): explain the cause. The ascorbic acid particles gain kinetic energy as temperature increases; this increases their speed and the frequency of collisions, and therefore the rate at which they diffuse into the agar. The agar cube is therefore decolourised more quickly.

Key Takeaways

A 'describe' point is a comparative statement of the trend.
An 'explain' point requires a reason rooted in the underlying biology or physics.
For temperature-dependent rate processes, the kinetic-energy link is almost always the right explanation.

Common Mistakes

Saying only 'diffusion is faster at higher temperatures' — this is the description, not the explanation. The reason is the increase in kinetic energy.
Stating that 'particles move more' without specifying kinetic energy or speed.
Bringing in unrelated ideas such as enzymes (the agar cube contains no active metabolism) or denaturation.

Things to Be Careful About

Avoid vague wording such as 'particles vibrate more' — ascorbic acid is in solution, so the particles are translating, not just vibrating. Saying 'particles have more kinetic energy and so diffuse faster' cleanly covers both ideas.

Techniques used

describe the trend with a clear comparative statementexplain the trend using kinetic energy and the rate of diffusion

(vi)

Explain why confidence in the results can be increased by repeating the procedure several times.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Repeating the procedure produces replicate times so that anomalous results can be identified (and excluded) and a statistical test (such as a t-test or correlation) can be carried out on the data.

Final answer

Repeats allow anomalous results to be identified/excluded and a statistical test to be performed.

Detailed explanation

Background Concept

A single measurement carries uncertainty — the human reaction time at the end-point, a momentary lapse in judging decolourisation, a slight difference in cube size. Repeats expose such anomalies: any one value that lies far from the others can be flagged. Once flagged, it can be excluded, or it can be left in but its effect on the mean and on a statistical test can be examined.

Understanding the Question

The candidate has just one set of times (one cube per temperature). They are asked why repeating the procedure several times would increase confidence in the results.

Approach

Identify what repeats actually allow. They give (a) a way to spot an anomalous result, (b) a way to exclude it, and (c) enough data to run a statistical test. The mark scheme accepts any of these (and explicitly says 'not mean' — calculating a mean is not the justification the examiner wants here).

Step-by-Step Reasoning

With several repeats per temperature, an extreme value can be seen as unusual and labelled anomalous.
Anomalous results can then be excluded from the analysis, leaving a more reliable data set.
More importantly, the data set is now large enough for a statistical test (e.g. a t-test comparing two temperatures, or a Spearman's rank correlation between temperature and time) to be performed, giving an objective measure of whether the trend is significant.

Key Takeaways

Repeats improve reliability, not validity.
The two principal benefits of repeats are identifying anomalies and enabling statistical testing.
'A mean' alone is not the answer the mark scheme is looking for.

Common Mistakes

Saying 'to calculate a mean' — explicitly rejected by the mark scheme.
Vague answers such as 'to be more accurate' or 'to avoid human error' — these are too imprecise.
Forgetting to link the repeats to the actual procedure (timing the decolourisation) and to what is gained from them.

Things to Be Careful About

The mark scheme permits 'identify anomalous results', 'exclude anomalous results' or 'allow a statistical test'. Pick one clear idea rather than a string of vague phrases.

Techniques used

justify the use of repeats in a practical investigation

(vii)

You used the procedure described in step 1 to step 8 to investigate the effect of temperature on the diffusion of ascorbic acid into agar cubes of the same size.

Describe how you would modify the procedure to investigate the effect of changing the surface area to volume ratio of agar cubes on the time taken to reach the end-point.

2M Similar

DifficultyMedium

Worked solution

Answer

Keep the temperature constant throughout (e.g. carry out all tests at $25\,^\circ\text{C}$ using the same water-bath setting) so that temperature is standardised.
Cut at least five agar cubes of different sizes (e.g. side lengths of 2, 4, 6, 8 and 10 mm) so that the surface area to volume ratio changes.

Final answer

Standardise temperature by using a single stated temperature and test at least five different sizes of agar cube.

Detailed explanation

Background Concept

The rate of diffusion into a cube depends on two physical factors: the surface area available for particles to cross into the cube, and the distance each particle must travel from the surface to the centre. A small cube has a much higher surface area to volume ratio than a large cube and so decolourises faster. To test the effect of surface area to volume ratio, that ratio must be varied while everything else (especially temperature) is held constant.

Understanding the Question

The candidate is asked how to modify the existing procedure so that the independent variable becomes the surface area to volume ratio of the agar cube, with time to reach the end-point still the dependent variable. Two marks are available.

Approach

Identify (1) what is to be varied (the cube size) and (2) what must be kept constant to make the test fair (the temperature). The mark scheme awards one mark for stating that temperature is standardised at a particular value, and one mark for using at least five different cube sizes.

Step-by-Step Reasoning

Mark 1 — standardising temperature: state that the water-bath is set to a single temperature (e.g. room temperature of $25\,^\circ\text{C}$ ) and that all cubes are tested at that one temperature. Without this, the change in time cannot be attributed to size alone.
Mark 2 — varying cube size: at least five cubes of different side lengths must be used. A range such as 2, 4, 6, 8, 10 mm gives five values and a clear spread of surface area:volume ratios. Each cube is then timed in the same way as before, and the results plotted or tabulated against surface area:volume ratio.

Key Takeaways

To change the IV, change only one thing at a time.
A modification question always asks for both the new IV (with at least five values) and the variable(s) that are now being controlled.
Surface area:volume ratio of a cube of side $a$ is $6/a$ , so smaller cubes have larger ratios.

Common Mistakes

Stating that the volume of ascorbic acid should be varied — the ascorbic acid is a control, not a variable.
Failing to give a specific temperature for standardisation.
Only suggesting two or three cube sizes — the mark scheme requires at least five.

Things to Be Careful About

The answer is about modifying the practical, not about a hypothesis or about a conclusion. Keep the answer tightly focused on what to change and what to keep constant.

Techniques used

swap the independent variable to surface area:volume ratiostandardise temperature while varying cube size

(b)

A scientist investigated the uptake of glucose into red blood cells. The red blood cells were put into a solution of radioactive glucose. The concentration of radioactive glucose in the red blood cells was measured over a period of 60 minutes.

The results are shown in Table 1.3.

Table 1.3

time / minutes	concentration of radioactive glucose / $\text{mmol dm}^{-3}$
0	0
10	48
20	71
30	83
40	94
60	102

(i)

Plot a graph of the data shown in Table 1.3 on the grid in Fig. 1.2. Fig. 1.2 is on page 7.

Use a sharp pencil.

Fig. 1.2

4M Similar

DifficultyMedium

Worked solution

Answer

A line graph with:

x-axis: time / minutes, scale 0–60 with major gridlines every 10 minutes (10.0 minutes to 2 cm), labelled at least every 2 cm.
y-axis: concentration of radioactive glucose / $\text{mmol dm}^{-3}$ , scale 0–120 with major gridlines every 20 (20 $\text{mmol dm}^{-3}$ to 2 cm), labelled at least every 2 cm.
All six points (0,0), (10,48), (20,71), (30,83), (40,94), (60,102) plotted as small crosses or dots in circles.
The six points joined with a thin smooth curve, plot-to-plot, passing through (or very close to) every point.

Final answer

Six correctly plotted points (0,0; 10,48; 20,71; 30,83; 40,94; 60,102) on correctly labelled and scaled axes, joined by a smooth thin curve.

Detailed explanation

Background Concept

A line graph is used when both variables are continuous and the dependent variable is a measured quantity that can take any value. The independent variable is on the x-axis and the dependent variable on the y-axis. Cambridge mark schemes for graph plotting reward four specific conventions: correct axis labels with units, a scale that uses at least half the grid and is not awkward, accurate plotting, and an appropriate line through the points.

Understanding the Question

The candidate is given six pairs of (time, concentration) values in Table 1.3 and asked to plot them on the grid provided in Fig. 1.2. Four marks are awarded.

Approach

Identify the IV and DV, choose scales, label axes, plot each point, and join them. The IV is time (0–60 minutes), the DV is concentration (0–102 $\text{mmol dm}^{-3}$ ). The two scales must be chosen to use at least half the available grid, and the units placed in the axis label.

Step-by-Step Reasoning

Axis labels (mark 1): x-axis 'time / minutes', y-axis 'concentration of radioactive glucose / $\text{mmol dm}^{-3}$ '. Both labels include the unit separated by a slash.
Scales (mark 2): a sensible x-scale is 10 minutes per 2 cm (so 60 minutes occupies 12 cm of a 15 cm grid), with labels every 2 cm (every 10 minutes). A sensible y-scale is 20 $\text{mmol dm}^{-3}$ per 2 cm (so 100 occupies 10 cm), with labels every 2 cm. Both scales are easy to read at the major gridlines.
Plotting (mark 3): mark each point as a small cross (×) or a dot in a circle (⊙) at the correct (x, y) position. All six points must be plotted.
Line (mark 4): join the points with a thin, unbroken line. Because the relationship is non-linear (it rises steeply at first and then flattens), a smooth curve is more appropriate than a straight line; the curve must pass through (or very close to) every plotted point.

Key Takeaways

Always put the independent variable on the x-axis.
Include units in axis labels, separated from the variable name by a slash.
A scale that uses at least half the grid and that is not awkward (multiples of 1, 2, 5, 10 of the chosen unit per 2 cm) is required.
A smooth curve is used when the relationship between variables is non-linear; the curve should pass through every point.

Common Mistakes

Forgetting the units in the axis label.
Choosing a scale that is awkward to use (e.g. 12 minutes per 2 cm) or one that uses less than half the grid.
Using a straight line of best fit when the data clearly curve.
Joining point to point with very thick or fuzzy lines.
Leaving a point off (e.g. forgetting (0,0) or the 60-minute point).

Things to Be Careful About

The mark scheme requires the line to be either 'a smooth curve' or 'joined plot to plot' — either is acceptable. The line must be thin and pass through (or very close to) all six points.

Techniques used

plot a line graph of concentration against timeselect suitable axis scales and label axes with units

(ii)

Use your graph in Fig. 1.2 to estimate the concentration of radioactive glucose in the red blood cells at 50 minutes.

Show on your graph how you estimated this value.

concentration of radioactive glucose = ______ $\text{mmol dm}^{-3}$

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Draw a vertical line from 50 on the x-axis up to the curve, then a horizontal line across to the y-axis. Read the value at the intersection with the y-axis.

\text{concentration of radioactive glucose at 50 minutes} \approx 98\ \text{mmol dm}^{-3}

Final answer

≈ 98 mmol dm⁻³

Detailed explanation

Background Concept

A smooth curve through the plotted data can be used to estimate intermediate values that were not measured directly. The process is called interpolation: it is reliable only between two measured points, not outside the range. Cambridge examiners award two marks here: one for showing on the graph where the value is read, and one for the actual numerical read-off.

Understanding the Question

The candidate has a completed line graph of concentration against time. They must read off the concentration at 50 minutes, which lies between the measured 40-minute point (94 $\text{mmol dm}^{-3}$ ) and the 60-minute point (102 $\text{mmol dm}^{-3}$ ).

Approach

Use a sharp pencil and a ruler. From 50 on the x-axis, draw a vertical line up to the curve, then from that intersection draw a horizontal line across to the y-axis. The value where the horizontal line meets the y-axis is the answer. The mark scheme requires both lines to be drawn on the graph for mark 1, and a sensible read-off for mark 2.

Step-by-Step Reasoning

50 minutes is exactly midway between 40 and 60 minutes.
At 40 minutes the concentration is 94; at 60 minutes it is 102. A straight-line interpolation would give 98, but the curve is flattening slightly so the true read is marginally above 98.
A reasonable answer from a correctly drawn curve is in the range 97–99 $\text{mmol dm}^{-3}$ , with about 98 being the most likely value.

Key Takeaways

Interpolation is the use of a curve to find a value between two measured points.
Always show the construction lines on the graph for the marker to follow.
The read-off is sensitive to where the curve is drawn; small differences in graph drawing give small differences in the read-off, and the mark scheme accepts this with the phrase 'according to candidate's graph'.

Common Mistakes

Trying to read directly above the '50' label on the x-axis without drawing a construction line — the first mark is lost.
Reading off a value below 95 or above 101 (clearly inconsistent with the data).
Using the units $\text{mmol/l}$ or omitting units.

Things to Be Careful About

The candidate's own curve may differ slightly from a model curve, and the mark scheme explicitly allows the read-off to match the candidate's own line. Always include the unit $\text{mmol dm}^{-3}$ in the answer.

Techniques used

read an interpolated value from a smooth curve at a given time

(iii)

Explain why the concentration of radioactive glucose in the red blood cells increases over time.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The radioactive glucose moves from the external solution, where its concentration is high, into the red blood cells, where its concentration is low (i.e. down the concentration gradient).
This movement is by diffusion (a passive process requiring no ATP).

Final answer

Glucose moves down a concentration gradient (from high outside to low inside) by diffusion.

Detailed explanation

Background Concept

Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration, driven by the random kinetic motion of the particles. It is a passive process that requires no energy input from ATP. Glucose is a small, uncharged molecule and crosses the red blood cell membrane by simple diffusion through the lipid bilayer (with the assistance of GLUT transporters in real biology, but at A-level the simple model of diffusion down the concentration gradient is the expected answer).

Understanding the Question

Table 1.3 shows that the concentration of radioactive glucose inside the red blood cells rises from 0 to 102 $\text{mmol dm}^{-3}$ over 60 minutes, with the rise becoming slower as time goes on (the curve flattens). The candidate is asked to explain why the concentration rises at all.

Approach

Identify (1) the direction of net movement and (2) the process responsible. The direction is from high (outside) to low (inside); the process is diffusion. The mark scheme awards one mark for the concentration-gradient idea and one for naming diffusion.

Step-by-Step Reasoning

At time 0, the inside of the cell has no radioactive glucose while the outside solution has a high concentration. There is therefore a concentration gradient from outside to inside.
Glucose particles move down this gradient, into the cells, by simple diffusion (random kinetic motion).
As the inside concentration rises, the gradient becomes shallower; this is why the curve levels off — the rate of entry decreases as the gradient is reduced. Eventually (well beyond 60 minutes) the gradient would approach zero and the inside concentration would approach the outside concentration.
The two marking points are therefore 'movement from high to low concentration' and 'diffusion'.

Key Takeaways

'Movement down a concentration gradient' and 'diffusion' are the two ideas that together explain passive entry.
A graph that rises quickly at first and then levels off is characteristic of a process that depends on a concentration gradient — the rate decreases as the gradient decreases.
A common error is to invoke active transport or ATP; the data (glucose equilibrating rather than being concentrated inside) rule these out.

Common Mistakes

Saying 'particles move from low to high concentration' — this is wrong; glucose is not being concentrated inside the cell.
Naming 'osmosis' — osmosis is the diffusion of water, not of a solute like glucose.
Naming 'active transport' — this would require ATP and would allow glucose to be accumulated against a gradient, which the data do not support.
Saying 'particles move because of kinetic energy' without naming the process as diffusion.

Things to Be Careful About

The mark scheme requires both ideas. State the direction of movement first and then name the process, in that order, for clarity.

Techniques used

explain the increase in intracellular glucose concentration using diffusion

Q2MediumUse of the Light MicroscopeAnalysis, Conclusions and Evaluation

P1 is a slide of a stained transverse section through a plant organ.

(a)

10M

(i)

Draw a large plan diagram of the region on P1 indicated by the shaded area in Fig. 2.1.

Use a sharp pencil.

Use one ruled label line and label to identify the xylem.

Fig. 2.1

5M Similar

DifficultyMedium

Worked solution

Answer

A large plan diagram drawn with a sharp pencil that:

fills most of the available space;
shows only the tissue layers of the stem sector (epidermis, cortex, vascular ring with xylem and phloem, pith) — no individual cells drawn;
has the epidermis drawn as two lines drawn close together;
shows the layers in the correct relative proportions (epidermis a thin layer at the outside, cortex reasonably wide, vascular tissue a narrower ring with xylem to the inside and phloem to the outside, pith a wide central region);
contains one ruled label line ending on the xylem with the label xylem.

No shading anywhere in the drawing.

Final answer

Plan diagram of the stem sector (epidermis, cortex, vascular bundle with xylem labelled, pith) drawn with correct proportions and no cells.

Detailed explanation

Background Concept

A plan diagram is a low-magnification outline drawing of a specimen that records the shapes and relative positions of the different tissues but contains no cellular detail. It is the standard way to record the gross organisation of a stained section at low power. A cell drawing (the next sub-question) is different: it shows the shapes of a small number of individual cells at high power.

The stem shown on slide P1 is a typical young dicotyledonous stem in transverse section. From the outside in, the tissue layers are: epidermis (a single layer of cells, drawn as two lines because the cell wall has an outer and an inner face), cortex (a band of parenchyma), vascular tissue arranged as a ring of bundles, with phloem on the outside and xylem on the inside of each bundle, and a central pith of large parenchyma cells. Xylem vessels typically appear as the larger, more conspicuous empty lumens in the inner part of each vascular bundle.

Understanding the Question

You are asked to draw a plan diagram of the shaded sector shown in Fig. 2.1 — a wedge of about a quarter of the whole circular section. The shading simply identifies which part of the section you must draw; your diagram itself must not be shaded. Only the region indicated should appear, not the entire cross-section. One ruled label line and label must identify the xylem.

Approach

Plan-diagram conventions to obey:

Use a sharp HB pencil, single thin lines, no shading.
Use most of the available space — make the drawing large.
Draw the outline only of tissues; never draw individual cells.
Show layers in the correct relative proportions (do not, for example, draw a thick epidermis or a narrow pith if the slide shows the opposite).
Draw the epidermis as two close parallel lines (it is one cell thick but each cell wall has an outer and inner face).
Add one ruled label line ending exactly on the tissue labelled; the label word sits at the end of the line.

Step-by-Step Reasoning

Frame: lightly sketch a rectangle the size you intend to use, so the wedge fits inside.
Outline of the sector: draw the outer arc of the wedge (the curved epidermis boundary) and the two straight radial cut edges.
Epidermis: draw a second arc parallel to and just inside the first, leaving a thin gap — this is the inner epidermal wall. Two close parallel lines.
Cortex: draw a third arc inside, leaving a noticeably wider gap to represent the cortex thickness as seen on the slide. Draw only the outer and inner boundaries of the cortex.
Vascular ring: draw the inner and outer boundaries of the ring of vascular tissue, leaving a narrower band than the cortex.
Pith: enclose the central area inside the vascular ring up to the two straight cut edges.
Check proportions: relative to the whole wedge, the pith should be the widest zone, the cortex moderately wide, the vascular ring narrower, and the epidermis a thin line.
Labelling: outside the diagram, write the word xylem. From the end of the word draw a single ruled horizontal line ending exactly on the inner part of the vascular ring, where the xylem vessels lie. Do not draw arrows.

Marks are awarded for: (1) using most of the space and no shading, (2) drawing the correct region with no cells, (3) epidermis as two close lines, (4) correct proportions of tissue layers, (5) one ruled label line ending on the xylem.

Key Takeaways

A plan diagram is tissues, not cells.
The epidermis is two lines because one cell layer has two walls.
Layers must be drawn in the correct relative thicknesses seen on the slide.
Labels need a single ruled line that ends precisely on the structure named.

Common Mistakes

Drawing individual cells inside the tissue zones (this turns a plan into a high-power cell drawing and loses marks).
Shading the diagram (plan diagrams are never shaded).
Drawing the epidermis as a single thick line instead of two close lines.
Labelling by arrow or with a line that ends in empty space rather than on the xylem.
Drawing only one cut edge or the whole circle instead of the shaded sector.

Things to Be Careful About

A pencil line that ends on or near the xylem region (inner half of the vascular ring) is what earns the labelling mark — not on the phloem, cortex or pith.
"Correct proportions" means the eye-test match to the slide: do not let the cortex be drawn thicker than the pith, etc.
Keep all lines thin, single, continuous — no sketchy or fuzzy outlines.

Techniques used

produce a low-power plan diagram of a stem sectorrepresent tissue layers with correct relative proportionsuse a ruled label line to identify a tissue (xylem)

(ii)

Observe the cells in the cortex of the organ on P1.

The cortex is the tissue beneath the outer layer of cells (epidermis) of the organ on P1.

Select a group of four adjacent cells from within this tissue, making sure that each of the four selected cells is touching at least two of the other cells.

Make a large drawing of this group of four cells.
Use one ruled label line and label to identify the cell wall of one of the cells that you have drawn.

5M Similar

DifficultyMedium

Worked solution

Answer

A large high-power drawing of four adjacent cortical cells with the following features:

drawn with a sharp pencil, using continuous, thin and sharp lines;
each of the four cells touches at least two of the others;
every cell is enclosed by two lines (representing the two cell walls on either side of the middle lamella) and where two cells meet there are three lines (each cell contributes two walls, one of which is shared);
the cells have detailed shapes that reflect what is actually visible under the microscope (roughly polygonal, with the slight curvature of real parenchyma cell walls);
one ruled label line ending on the wall of one of the cells, with the label cell wall.

No shading, no nucleus, no contents drawn unless clearly visible.

Final answer

Drawing of four adjacent cortical cells (each touching at least two others) drawn with double lines around each cell and three lines where cells meet, with cell wall labelled.

Detailed explanation

Background Concept

At high magnification, plant cells are bounded by cellulose cell walls. Where two cells are adjacent, each contributes its own cell wall, so the boundary seen under the microscope consists of the wall of cell A, a middle lamella, and the wall of cell B. This is why adjacent plant cells appear separated by a double line — there is a wall on each side of the shared interface.

Cortex cells of a young dicot stem are roughly isodiametric parenchyma: polygonal, slightly rounded at the corners, with thin primary walls. They have large central vacuoles and, in a stained preparation, often visible nuclei and some cytoplasmic detail. Only what is actually visible should be drawn.

Understanding the Question

You must look down the microscope at the cortex of P1 (the tissue layer immediately inside the epidermis) and find a group of four cells in which each cell touches at least two of the others. Then draw that group, large, and label the cell wall of one of the cells using one ruled label line.

The "each touches at least two" rule is the examiner's check that you have selected a real cluster, not four cells scattered around the field.

Approach

Cell-drawing conventions to apply:

Sharp HB pencil, continuous, thin lines (no sketchy hatching).
Two lines around every cell — each cell is enclosed by two strokes, one for the wall on one side and one for the wall on the other side of the middle lamella.
Three lines where cells meet — the two cells on either side of a shared boundary each contribute their wall, so the boundary is three strokes (cell A's wall, middle lamella line, cell B's wall).
Draw the shape you actually see — irregular polygons with slightly curved sides — not textbook-perfect hexagons.
One ruled label line to one cell wall, with the label cell wall.
No shading, no organelles unless you can see them clearly.

Step-by-Step Reasoning

Under the microscope, scan the cortex for a clean cluster of four cells where each cell is in contact with at least two others (a 2 × 2 group, or a rosette-like arrangement, both work).
Mentally map the relative sizes and shapes of the four cells.
Lightly frame the area you will draw.
Draw each cell with two boundary lines running all the way around. Where two cells share a boundary, draw three close parallel strokes along that interface.
Keep the proportions of the four cells accurate — they will not all be the same size or shape.
Outside the cluster, write cell wall. Draw one horizontal ruled line ending on one cell's wall.
Double-check: continuous lines ✓, four cells ✓, each touches ≥ 2 others ✓, double-line walls ✓, label ✓.

The five marks are awarded for: (1) continuous thin sharp lines, (2) every cell touches at least two others, (3) two lines around each cell AND three lines where cells touch, (4) detailed (real) shapes, (5) one ruled label line ending on the cell wall.

Key Takeaways

Adjacent plant cells share a wall, hence the two-lines-per-cell / three-lines-at-a-junction rule.
A cell drawing shows only what is visible through the eyepiece — shapes, proportions, anything you can resolve clearly.
Labels always sit at the end of a single ruled line.

Common Mistakes

Drawing a single thick line around each cell (loses the two-wall mark).
Drawing textbook regular hexagons rather than the irregular shapes actually seen.
Including organelles (nucleus, chloroplasts) that are not clearly visible.
Adding shading or colour — drawings on CIE papers are line diagrams only.
Labelling cytoplasm, nucleus or vacuole instead of (or as well as) the cell wall — the question specifies the cell wall.

Things to Be Careful About

The four cells must be adjacent in the actual tissue, not four cells drawn from different parts of the field.
"Each touches at least two" is satisfied by any cluster where no cell is isolated — a small 2 × 2 group works perfectly.
The label line must end on the wall itself, not on the inside of the cell or in the space between two cells.

Techniques used

produce a high-power drawing of four adjacent cellsapply the cell-drawing convention of double lines at cell boundariesuse a ruled label line to identify the cell wall

(b)

Fig. 2.2 is a photomicrograph of a stained transverse section of the same organ shown on P1 from a different species of plant. This species of plant has thorns. One of the thorns has been labelled on Fig. 2.2.

Fig. 2.2

Identify three observable differences, other than colour, size and presence or absence of thorns, between the section on P1 and the section shown in Fig. 2.2.

Record these three observable differences in an appropriate table.

4M Similar

DifficultyMedium

Worked solution

Answer

The three observable differences (excluding colour, size and thorns) between P1 and Fig. 2.2:

feature	P1	Fig. 2.2
location of vascular tissue	(vascular bundles) towards the centre of the section	peripheral (vascular tissue near the outside, inside the cortex)
overall shape of section	circular / round	triangular
trichomes (root hairs / surface hairs)	present on the epidermis	absent
width of cortex	wide	narrow

Any three of the rows above are credited. The table must have either three columns (feature / P1 / Fig. 2.2) or two columns with the feature clearly stated and compared row-by-row.

Final answer

Three observable differences (other than colour, size and thorns), e.g. (1) vascular tissue central in P1 vs peripheral in Fig. 2.2; (2) P1 circular vs Fig. 2.2 triangular; (3) trichomes present on P1 but absent from Fig. 2.2.

Detailed explanation

Background Concept

When comparing two biological specimens under the microscope, observations must be restricted to what can actually be seen — features such as shape, position, presence or absence of structures, and relative sizes. Comparisons should not include colour (stain intensity varies), absolute size (depends on magnification) or any feature the question has explicitly excluded. The comparison is then recorded in a table so each feature is set against both specimens side-by-side.

The two specimens are transverse sections of stems from two different plant species. Fig. 2.2 belongs to a thorn-bearing plant: it is the section used in parts (c)(i) and (c)(ii). P1 is a different stem section (used for the drawings in part (a)). Both can be inspected — P1 on the slide, Fig. 2.2 as a printed photomicrograph.

Understanding the Question

You are asked to find three observable differences between P1 and Fig. 2.2, excluding:

colour (e.g. stain colour);
size (absolute size depends on magnification);
presence or absence of thorns (Fig. 2.2 has thorns, P1 does not — this is excluded).

The answer must be presented in a table.

Approach

Scan each specimen systematically — overall shape, epidermis, cortex, vascular arrangement, pith, surface features.
Build a feature list: write down every difference you can see, then cross out the three banned categories.
Pick the three strongest observable differences — those that are unambiguous from the images alone.
Present them in a table — either a 3-column table (feature / P1 / Fig. 2.2) or a 2-column table where each row clearly states the feature being compared.

Step-by-Step Reasoning

Looking at the photomicrograph of Fig. 2.2:

The outline of the section is triangular (it has three thorn-bearing points).
The vascular tissue forms a ring that sits near the outside, just inside a narrow cortex, with a large central pith.
The epidermis has no hairs / trichomes projecting from it (apart from the thorns, which are excluded).

Looking at P1 (a typical young dicot stem in transverse section):

The outline is circular.
The vascular bundles are arranged in a ring towards the centre of the section, with a wide cortex outside them and a central pith.
The epidermis bears trichomes (small hair-like outgrowths), clearly visible as fine projections from the surface.
The cortex is a wide band of parenchyma between the epidermis and the vascular ring.

Choosing any three from:

feature	P1	Fig. 2.2
overall shape	circular	triangular
location of vascular tissue	towards the centre	peripheral
trichomes	present	absent
width of cortex	wide	narrow

Key Takeaways

Comparisons must use observable features only — no assumptions about physiology, ecology, or taxonomy that are not visible.
A comparison table needs a clear feature column and one column per specimen.
Always re-read the question to check which features are banned (here: colour, size, thorns).

Common Mistakes

Writing "P1 has thorns, Fig. 2.2 has thorns" (or vice versa) — thorns are excluded.
Comparing stain intensity or overall section colour — colour is excluded.
Quoting absolute sizes ("P1 is 5 mm across") — size is excluded; use shape and proportions instead.
Putting differences in a list instead of a properly formatted table (loses the layout mark).
Including more than three differences without prioritising the strongest — the mark scheme credits any three.

Things to Be Careful About

The table layout mark is awarded for either a 3-column table (feature, P1, Fig. 2.2) or a 2-column table in which each row clearly states the feature being compared. A plain list does not earn this mark.
"Observable" means the feature must be visible in the images; do not invent features you cannot see.
Where a feature is absent, state "absent" — do not leave the cell blank.

Techniques used

compare two transverse sections of plant stems by observable featuresrecord a comparison in a correctly formatted tabledistinguish features other than colour, size and presence/absence of thorns

(c)

Fig. 2.3 is the same photomicrograph as that shown in Fig. 2.2. Labels have been added to show where to measure the height and width of thorn Q.

Fig. 2.3

(i)

The section through thorn Q is the shape of a triangle. Line R–S is the height of the triangle and line T–U is the width of the triangle.

Measure the lengths of line R–S and line T–U on Fig. 2.3.

length of line R–S: ______

length of line T–U: ______

Calculate the actual height and actual width of the section through thorn Q using your measurements for the lengths of line R–S (height) and line T–U (width).

Show your working.

actual height = ______

actual width = ______

3M Similar

DifficultyMedium-Easy

Worked solution

Working

Measure the lengths of R–S (height) and T–U (width) on Fig. 2.3 in millimetres. Representative measurements are given below; the candidate's own measurements will depend on the printed size of the figure.

\text{measured length of R–S (image)} = 56\ \text{mm}

\text{measured length of T–U (image)} = 28\ \text{mm}

The magnification is ×14. Because

\text{magnification} = \frac{\text{image size}}{\text{actual size}}

then

\text{actual size} = \frac{\text{image size}}{\text{magnification}}

\text{actual height of thorn Q} = \frac{56\ \text{mm}}{14} = 4.0\ \text{mm}

\text{actual width of thorn Q} = \frac{28\ \text{mm}}{14} = 2.0\ \text{mm}

Answer

Measured R–S = 56 mm; measured T–U = 28 mm.
Actual height = 4.0 mm.
Actual width = 2.0 mm.

(Exact values depend on the candidate's measurement of the printed figure.)

Final answer

Actual height ≈ 4.0 mm, actual width ≈ 2.0 mm (image measurements divided by 14; representative values shown).

Detailed explanation

Background Concept

A photomicrograph records an image of a specimen at a particular magnification. If you measure the image (the printed figure) and know the magnification, the true size of the specimen is recovered by:

\text{actual size} = \frac{\text{image size}}{\text{magnification}}

This is the same relationship that gives magnification = image size / actual size, rearranged. Units must be handled consistently: if you measure in mm, your actual size is in mm; if you measure in µm, the actual size is in µm.

Fig. 2.3 carries the line magnification ×14, so each millimetre on the printed figure represents 1/14 mm of the real specimen.

Understanding the Question

You are given Fig. 2.3, the same photomicrograph as Fig. 2.2, but with line R–S drawn across the height of thorn Q and line T–U across the width of its base. You must:

Measure both lines on the figure (in mm).
Divide each measurement by 14 (the magnification) to obtain the actual height and width of the section through thorn Q.

Show your working and quote units.

Approach

Use a ruler with mm divisions.
Lay the ruler along each line carefully, reading to the nearest mm.
Apply actual = image / magnification for each.
Quote both measurements with units, and quote both actual sizes with units.

Step-by-Step Reasoning

Measure R–S: place the zero of the ruler at R, read the position of S. In the printed figure, R–S is the longer line (height of the thorn). A representative reading is 56 mm.
Measure T–U: place the zero at T, read the position of U. T–U is the shorter line (base of the thorn). A representative reading is 28 mm.
Convert to actual size: the magnification is ×14, so each mm on the figure corresponds to $\frac{1}{14}$ mm in reality.
$\text{actual height} = \frac{56\ \text{mm}}{14} = 4.0\ \text{mm}$ $\text{actual width} = \frac{28\ \text{mm}}{14} = 2.0\ \text{mm}$
Quote units: both actual sizes are in mm, because the measurements were in mm and 14 is dimensionless.

Marks are for (1) measuring both lines correctly, (2) quoting units, (3) showing each measurement divided by 14.

Key Takeaways

The triangle relationship magnification = image / actual is the workhorse for converting measurements on photomicrographs to real specimen sizes.
Units must be carried through: the unit of the actual size equals the unit of the measured image length (magnification is dimensionless).
Always show the division — the third mark is explicitly for showing the length divided by 14.

Common Mistakes

Multiplying instead of dividing by the magnification.
Converting mm to cm part-way through and losing the unit (e.g. writing "0.40" instead of "4.0 mm").
Quoting only one of the two actual sizes.
Failing to show the working — the third mark is for the visible division by 14.

Things to Be Careful About

The candidate's measured values will differ slightly depending on the printed size of Fig. 2.3. The mark scheme accepts a range that scales correctly by 14.
Keep the measured value and the actual value on separate lines, with units, so the examiner can see both.
Do not round the measured image length to one significant figure — keep at least two s.f. so the actual size has a sensible number of s.f.

Techniques used

measure a length on a printed photomicrograph in mmconvert a measured image length to actual size using the stated magnificationpresent working with units and the correct relationship

(ii)

Calculate the actual area of the section through thorn Q using your answers to (c)(i).

\text{area of a triangle} = 0.5 \times \text{height} \times \text{width}

Show your working and give your answer to two significant figures.

actual area of section through thorn Q = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using the actual dimensions from (c)(i):

\text{actual height} = 4.0\ \text{mm}, \quad \text{actual width} = 2.0\ \text{mm}

The area of the triangular cross-section is:

\text{area} = 0.5 \times \text{actual width} \times \text{actual height}

\text{area} = 0.5 \times 2.0\ \text{mm} \times 4.0\ \text{mm} = 4.0\ \text{mm}^2

(2 significant figures, since both inputs are quoted to 2 s.f.)

Answer

Actual area of the section through thorn Q = 4.0 mm² (to 2 s.f.).

(The numerical answer scales with the candidate's own measurements in (c)(i); the working and units are the marks.)

Final answer

4.0 mm² (to 2 significant figures)

Detailed explanation

Background Concept

The question describes the section through thorn Q as a triangle, with line R–S as the height and line T–U as the width of the base. The area of a triangle is half the product of base and perpendicular height:

\text{area} = \frac{1}{2} \times \text{base} \times \text{height}

Because the two lengths are in mm, their product is in mm × mm = mm². Always state the unit explicitly.

Understanding the Question

Using the actual dimensions you calculated in (c)(i), find the area of the triangular section through thorn Q. Show your working and give the answer to two significant figures with the correct unit (e.g. mm²).

Approach

Take the actual height and actual width from (c)(i).
Substitute into area = 0.5 × width × height.
Quote the result to 2 s.f. with units.

Step-by-Step Reasoning

Using representative values (height = 4.0 mm, width = 2.0 mm):

\text{area} = 0.5 \times 2.0 \times 4.0\ \text{mm}^2 = 4.0\ \text{mm}^2

If the candidate's own measured values were, say, height = 4.1 mm and width = 2.3 mm (from image measurements of 57 mm and 32 mm), then:

\text{area} = 0.5 \times 2.3 \times 4.1 = 4.715\ \text{mm}^2 \approx 4.7\ \text{mm}^2

The candidate's answer scales with their own (c)(i) values; error carried forward is allowed by the mark scheme, so a correctly executed calculation from earlier incorrect measurements still earns the marks.

Significant figures: with two inputs each given to 2 s.f., the result should be given to 2 s.f.

Marks: (1) showing 0.5 × actual width × actual height; (2) correct numerical answer to 2 s.f. with appropriate unit (mm²).

Key Takeaways

Triangle area = ½ × base × height.
Always carry the unit into the final answer: mm × mm = mm².
2 significant figures — do not give three or four, and do not give one (e.g. "4 mm²" loses a s.f.).

Common Mistakes

Multiplying instead of taking half — the 0.5 factor is essential.
Forgetting the unit, or writing "mm" instead of "mm²".
Quoting too many significant figures (e.g. "4.715 mm²" should be "4.7 mm²" to 2 s.f.).
Re-measuring the figure instead of carrying forward the (c)(i) values.

Things to Be Careful About

The mark scheme awards an ecf (error carried forward): a candidate whose own measurements in (c)(i) were slightly off still gains credit here, provided the working uses those values correctly.
Make sure the substitution is visible: write 0.5 × … × … so the examiner can credit the method mark even if the final number is miscalculated.

Techniques used

calculate the area of a triangle from two perpendicular dimensionscarry forward the candidate's own measurements from the previous partround a final answer to two significant figures with the correct unit