A student investigated the number of plants on a school field. This was done by examining small samples of the field using a square frame measuring $0.5\,\text{m} \times 0.5\,\text{m}$. Fig. 2.1 shows the plants in one of these $0.5\,\text{m} \times 0.5\,\text{m}$ samples. (a) The sample contained three different plant species. The student counted the number of plants of each of these species in this sample. One plant was not fully within the square frame. Suggest and…

(a) Include it + reason – more than half in the square / cannot have half an individual (or exclude it + reason – not fully in the square). · (b) 9 (or 10 if the boundary plant was included in (a)) · (c) 7200 · (d) Too time consuming to count them all / too difficult to keep an accurate tally / more likely to make a mistake. · (e) Plants not evenly distributed; larger proportion of field sampled / more representative sample. · (f) Bar chart on the printed grid: x-axis…

Biology 5090/31 — May/June 2024

Cambridge O-Level · Practical Test · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Experimental Contexts · Observations and Measurements · Microscopy and Biological Drawing · Analysis, Conclusions and Evaluation · Planning Experiments and Investigations

Q122MMediumMicroscopy and Biological DrawingExperimental ContextsObservations and Measurements

You are going to investigate the structure and nutrient content of a flower. You are provided with a flower and samples of substances A and B from a flower.

Carefully remove some petals from the flower so that you leave two petals that are next to each other.

(a)

(i)

Place the flower with two petals on the white tile. Ensure that the petals are resting on the tile so that you can see the internal parts of the flower clearly. Examine the flower carefully with a hand lens.

Make a large drawing of the two petals and the internal parts of the flower in the space below.

5M Similar

DifficultyMedium

Worked solution

Answer

A large drawing, at least 90 mm across, of the two petals and the internal parts of the flower: several stamens each with a filament and a clearly delimited anther, and a central style rising to a stigma. Drawn in clean, continuous lines with a sharp pencil, with no shading.

Final answer

Large labelled drawing of two petals and the internal parts of the flower

Detailed explanation

Walkthrough

This is the classic 5090 biological drawing task. The examiner is not testing your artistic skill — the five marks are awarded for following drawing conventions and for showing the right structures.

Line quality (1 mark). Every line must be clear, clean and continuous, drawn with a sharp pencil. There must be no shading, stippling or cross-hatching anywhere, and no ruled lines. If you make a mistake, do not scribble it out — a single clean line through the error is acceptable.
Size (1 mark). The drawing must be at least 90 mm wide in either direction. That is roughly the width of your hand — draw big. A small drawing loses this mark even if everything else is perfect.
Stamens (1 mark). Draw the male parts: each stamen has a stalk, the filament, topped by an anther. The anthers must be delimited — that is, each anther drawn as a distinct, clearly outlined structure at the tip of its filament, not a vague blob.
Style and stigma (1 mark). The style must be drawn with a double line (two parallel lines, because it is a tube-like structure) and it must end in a stigma at the top. Cell walls, midribs, stalks and tubes are always drawn with a double line.
Petals (1 mark). Exactly two petals, as instructed — the two you left next to each other. Draw only what was asked for and nothing else.

Key Takeaways

Biological drawings are marked on conventions: sharp pencil, continuous lines, no shading, minimum size, correct proportions.
Tubes and stalks (style, filament) are drawn with double lines.
Draw only the structures the question names — extra structures can lose marks.

Common Mistakes

Shading or stippling to show depth — this always loses the line-quality mark.
Drawing too small; the 90 mm minimum is a specific credited point.
Drawing the style as a single line instead of a double line.
Drawing vague, undelimited anthers.
Drawing more or fewer than two petals.

Things to Be Careful About

Use a sharp HB pencil, never pen.
Check the size before you start — measure 90 mm on the page so you know how much room the drawing needs.
The mark scheme credits "filaments + delimited anthers" and "style with double line + stigma" as paired points, so both halves of each pair must appear.

Techniques used

make a large labelled biological drawingapply drawing conventions for line quality and sizeobserve flower structure with a hand lens

(ii)

On your drawing, draw a line and label it P to show where pollen must land when pollination takes place.

1M Similar

(b)

10M

(i)

You are going to test substances A and B for their nutrient content using Benedict's solution, biuret reagent and iodine solution.

Draw a table in which to record the results of your tests in the space below.

4M Similar

DifficultyMedium-Easy

Worked solution

Answer

Test / reagent	Colour / observation for A	Colour / observation for B
Benedict's solution
Biuret reagent
Iodine solution

The table is drawn with ruled lines, all information written inside it, each header stated once only, and overarching headers naming the test and the colour observed.

Final answer

Ruled table with columns for substances A and B, rows for Benedict's, biuret and iodine, and overarching headers for test and colour/observation

Detailed explanation

Walkthrough

"Draw a table in which to record your results" is marked on table CONSTRUCTION, not on the results themselves — those come in (b)(iii).

The four marks are:

Ruled lines. Draw the table with a ruler; every row and column is ruled, and all information is written inside the table, not outside it.
Substance headers. A and B each appear once only, as column headers — never repeated in every cell.
Test headers. Benedict's, biuret and iodine each appear once only, as row headers.
Overarching headers. The top row must name the two variables: the test / reagent / solution on one side and the colour / observation / results on the other.

Key Takeaways

A well-constructed table has headers stated once, overarching headings naming each variable, and everything inside the ruled frame.
Plan the table before doing the tests so observations can be recorded straight in.

Common Mistakes

Repeating "A" and "B" in every cell instead of as column headers.
Writing observations outside the table.
Omitting the overarching header row naming the variables.
Freehand, unruled tables.

Things to Be Careful About

Leave the result cells empty at this stage — the observations are recorded in (b)(iii).

Techniques used

design a results table with ruled rows and columnsstate variables and units in table headers

(ii)

To carry out the tests, follow these instructions, making sure that you label the test-tubes so that you know which test is being done on which substance.

Raise your hand when you are ready to be supplied with hot water in your water-bath.

Measure and record the temperature of the water in the water-bath.

______

Pour about $1\,\text{cm}$ depth of substance A into each of three clean test-tubes.
Add about $1\,\text{cm}$ depth of Benedict's solution to one of these test-tubes and place it in your hot water-bath.
Add about $1\,\text{cm}$ depth of biuret reagent to another one of these test-tubes and place it in the test-tube rack.
Add a few drops of iodine solution to the third test-tube and place it in the test-tube rack.
After five minutes observe these test-tubes and observe the colour of the solution at the end of each test.
Repeat the above procedure with clean test-tubes and substance B. Raise your hand if you require more hot water. You do not need to record the temperature of the water again.

1M Similar

(iii)

Record your observations in the table you have drawn in (b)(i).

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Test / reagent	Colour for A	Colour for B
Benedict's solution	orange (green / yellow / orange / red all accepted)	blue
Biuret reagent	blue	purple
Iodine solution	brown	brown

Final answer

Benedict's: A orange, B blue; biuret: A blue, B purple; iodine: A brown, B brown

Detailed explanation

Walkthrough

Each reagent earns one mark for the correct FINAL colour observed:

Benedict's solution tests for reducing sugars. On heating in the water-bath it turns from blue through green, yellow and orange to brick-red, depending on how much sugar is present. Substance A gives a positive result (any of green/yellow/orange/red scores); substance B stays blue — no reducing sugar.
Biuret reagent tests for protein. A positive result turns from blue to purple (lilac/violet/mauve). Substance B is positive; substance A stays blue.
Iodine solution tests for starch. A positive result turns from brown to blue-black. Both A and B stay brown — no starch in either.

Record the final colour after five minutes, in the table you drew in (b)(i).

Key Takeaways

Benedict's: reducing sugars — blue to green/yellow/orange/brick-red on heating.
Biuret: protein — blue to purple.
Iodine: starch — brown to blue-black; staying brown means starch is absent.

Common Mistakes

Recording the starting colour instead of the final colour after five minutes.
Saying "blue-black" for iodine when the observation was brown (no starch present).
Confusing biuret (protein, purple) with Benedict's (reducing sugar, heated).

Things to Be Careful About

The mark is for the colour actually observed at the END of each test — wait the full five minutes before recording.

Techniques used

record the colour change in a food testcarry out Benedict's, biuret and iodine tests

(iv)

State the nutrients present in:

substance A ______

substance B ______

2M Similar

(c)

Fig. 1.1 is a photomicrograph of a pollen grain.

Fig. 1.1

(i)

Measure and record the diameter of the pollen grain at its widest point.

diameter = ______ $\text{mm}$

1M Similar

(ii)

Calculate the diameter of the actual pollen grain and record your answer to 2 decimal places.

diameter of the actual pollen grain = ______

3M Similar

DifficultyMedium-Easy

Worked solution

Working

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

Using the measured diameter from (c)(i), e.g. $26\text{ mm}$ :

\text{actual size} = \frac{26\text{ mm}}{400} = 0.065\text{ mm} = 65\,\mu\text{m}

Answer

diameter of the actual pollen grain = $0.07\text{ mm}$ (to 2 decimal places, using the measured value; e.g. $26\text{ mm} \div 400 = 0.065\text{ mm} \approx 0.07\text{ mm}$ ), with the unit mm (or $\mu\text{m}$ ).

Final answer

Measured diameter from (c)(i) divided by 400, given to 2 decimal places with the unit (e.g. 0.07 mm)

Detailed explanation

Walkthrough

The photomicrograph is magnified $\times 400$ , so the grain in the picture is 400 times larger than the real pollen grain. To find the real size, divide by the magnification:

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

The three marks are:

The method — your measured diameter from (c)(i) divided by 400.
The answer to 2 decimal places — the question demands this precision, so round only at the end. For example, $26 \div 400 = 0.065\text{ mm}$ , which rounds to $0.07\text{ mm}$ ; a measurement of $28\text{ mm}$ gives $0.07\text{ mm}$ exactly.
The unit — mm, or convert to $\mu\text{m}$ by multiplying by 1000 (e.g. $0.065\text{ mm} = 65\,\mu\text{m}$ ).

Key Takeaways

actual size = image size ÷ magnification; magnification = image size ÷ actual size.
$1\text{ mm} = 1000\,\mu\text{m}$ .
Round only once, at the end, to the precision asked for.

Common Mistakes

Multiplying by 400 instead of dividing — that gives the size of an impossible giant grain.
Forgetting the unit, which is a separate mark.
Giving more or fewer than 2 decimal places when the question states the precision.
Rounding the (c)(i) measurement before dividing.

Things to Be Careful About

The mark scheme awards ecf: if your (c)(i) measurement was inside the accepted range, dividing it correctly by 400 still scores, even if the final value differs from a classmate's.

Techniques used

calculate actual size from magnificationconvert mm to micrometresround to a stated number of decimal places

(iii)

Fig. 1.2 is a photomicrograph of a pollen grain from a different species of plant.

Fig. 1.2

Describe two visible differences in the structure of the pollen grains in Fig. 1.1 and Fig. 1.2.

Fig. 1.1 pollen grain	Fig. 1.2 pollen grain
1
2

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

	Fig. 1.1 pollen grain	Fig. 1.2 pollen grain
1	round / circular	oval
2	smooth surface, no spikes	spiky / rough surface, with many pointed projections

Final answer

Fig. 1.1 is round with a smooth surface; Fig. 1.2 is oval with a spiky surface

Detailed explanation

Walkthrough

Compare the two photomicrographs feature by feature. The scheme lists three possible differences and asks for two ("max 2"):

Shape. The Fig. 1.1 grain is round/circular; the Fig. 1.2 grain is oval (longer than it is wide).
Surface. Fig. 1.1 has a smooth surface with no spikes; Fig. 1.2 is covered in pointed projections (spines) all round its edge, giving a rough surface.
Internal division. Fig. 1.1 shows a furrow dividing the grain internally; Fig. 1.2 shows no such division.

Any two of these score. Each comparison must name BOTH grains — a statement about only one grain does not make a comparison.

Pollen grain surface texture is a real identification feature: smooth (psilate) grains versus spiny (spiny/echinate) grains belong to different species, and wind-pollinated plants tend to have smooth, light grains while insect-pollinated plants often have spiky grains that cling to insect bodies.

Key Takeaways

A comparison names both sides of each difference.
Visible features of pollen grains include shape, surface texture (smooth vs spiky) and internal divisions.

Common Mistakes

Describing only one grain ("Fig. 1.2 has spikes") without saying Fig. 1.1 has none.
Writing about size — the two photomicrographs may be printed at different magnifications, so apparent size is not a valid comparison.
Giving colour differences — the photomicrographs are greyscale.

Things to Be Careful About

Give exactly two differences; a third will not earn extra marks (max 2).

Techniques used

compare two specimens from photomicrographsdescribe visible structural differences

Q211MMediumExperimental ContextsObservations and MeasurementsAnalysis, Conclusions and Evaluation

A student investigated the number of plants on a school field.

This was done by examining small samples of the field using a square frame measuring $0.5\,\text{m} \times 0.5\,\text{m}$ .

Fig. 2.1 shows the plants in one of these $0.5\,\text{m} \times 0.5\,\text{m}$ samples.

Fig. 2.1

(a)

The sample contained three different plant species.

The student counted the number of plants of each of these species in this sample.

One plant was not fully within the square frame. Suggest and explain what you would do about this plant.

______

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Include it, because more than half of the plant lies within the square frame / you cannot count half an individual.

(Alternatively: exclude it, because it is not fully within the square frame.)

Final answer

Include it + reason – more than half in the square / cannot have half an individual (or exclude it + reason – not fully in the square).

Detailed explanation

Walkthrough

When using a quadrat, plants that straddle the frame edge are a classic problem: do they belong to this sample or not? The examiner accepts either decision, but only with a valid reason attached — the mark is awarded for the decision plus the reason as one combined point.

Include it is justified because more than half of the plant's area lies inside the square, or because you cannot score half a plant — every individual must be counted whole or not at all.
Exclude it is justified because the plant is not fully within the square, or again because you cannot have half an individual.

The important principle is that whichever rule you choose, you apply it consistently to every sample so that all samples are counted the same way.

Key Takeaways

Quadrat sampling needs a fixed, stated rule for boundary organisms.
The reason matters as much as the decision — 'include' and 'exclude' both score when paired with their justification.
Consistency across samples keeps the estimate unbiased.

Common Mistakes

Giving a bare decision ('count it') with no reason — the mark requires decision + reason joined together.
Saying 'count it because it looks like it's mostly inside' without stating the 'more than half' criterion or the 'cannot have half an individual' idea.
Inventing a third option such as 'estimate what fraction is inside' — the scheme offers only include or exclude.

Things to Be Careful About

The '+' in the mark scheme means both halves are needed for the single mark.
Either route scores equally; pick one and state it clearly rather than hedging between both.

Techniques used

apply a consistent rule for counting organisms on the quadrat boundaryjustify the rule with a reason

(b)

Count the number of plants of species E in Fig. 2.1, taking into account your answer to (a).

Record your answer in Table 2.1.

Table 2.1

plant species	number of plants in the sample	estimated number of plants in the whole field
C	7	5600
D	4	3200
E

1M Similar

DifficultyEasy

Worked solution

Answer

Number of plants of species E = 9

plant species	number of plants in the sample	estimated number of plants in the whole field
C	7	5600
D	4	3200
E	9

(The boundary plant of species E is excluded, following the answer to (a); a candidate who included it records 10.)

Final answer

9 (or 10 if the boundary plant was included in (a))

Detailed explanation

Walkthrough

This part simply applies your (a) decision to the figure. Counting the tulip-shaped flowers (species E) fully inside the square gives 9; the tenth tulip sits across the left-hand boundary line, so whether it counts depends on the rule you stated in (a).

The mark scheme credits 9 or 10, carried forward from your own (a) answer — there is no single 'correct' number, only a self-consistent one. The value goes into Table 2.1 in the species E row.

Key Takeaways

Counts from quadrat diagrams must follow the boundary rule stated earlier.
Examiners award error-carried-forward credit: a defensible 10 here is worth exactly as much as a defensible 9.

Common Mistakes

Counting all tulips including the boundary one after having said 'exclude' in (a) — inconsistency loses the mark.
Misidentifying species E: it is the tulip-shaped flower in the key, not the round-petalled daisy (C) or the many-petalled flower (D).

Things to Be Careful About

Check the key before counting — the three species look similar at a glance.
Record the number in the correct row of the table.

Techniques used

count individuals of one species from a diagramapply the boundary rule consistently

(c)

The whole field measured $10\,\text{m} \times 20\,\text{m}$ .

Use the sample in Fig. 2.1 to estimate the number of plants of species E in the whole field and record this value in Table 2.1.

Show your working.

2M Similar

DifficultyMedium

Worked solution

Working

Area of field $= 10\,\text{m} \times 20\,\text{m} = 200\,\text{m}^2$

Area of one sample $= 0.5\,\text{m} \times 0.5\,\text{m} = 0.25\,\text{m}^2$

\text{number of samples in the field} = \frac{200}{0.25} = 800

\text{estimated number of species E} = 9 \times 800 = 7200

Answer

7200 plants

plant species	number of plants in the sample	estimated number of plants in the whole field
C	7	5600
D	4	3200
E	9	7200

Final answer

7200

Detailed explanation

Walkthrough

To scale a quadrat count up to a whole-field estimate, you need to know how many quadrat-sized samples would fit in the field.

Field area: $10 \times 20 = 200\,\text{m}^2$ .
Sample area: $0.5 \times 0.5 = 0.25\,\text{m}^2$ .
Number of samples: $200 \div 0.25 = 800$ . This is the point the scheme rewards first ('ref. to 800').
If each sample holds 9 plants of species E, then 800 samples hold $9 \times 800 = 7200$ plants.

If you included the boundary plant in (b), the same method gives $10 \times 800 = 8000$ , which also scores.

Notice the printed table confirms the logic: species C has 7 per sample and an estimate of 5600, and $7 \times 800 = 5600$ — the same multiplier of 800 throughout.

Key Takeaways

Population estimate = mean count per quadrat × (total area ÷ quadrat area).
Always show the intermediate step (the number of samples) — it carries a mark of its own.
You can check your multiplier against rows already completed in the table.

Common Mistakes

Multiplying by 200 instead of 800 (forgetting to divide by the quadrat area).
Using $0.5\,\text{m}^2$ instead of $0.25\,\text{m}^2$ for the sample area.
Omitting the working — the question explicitly says 'Show your working', and the 800 step is separately credited.

Things to Be Careful About

Error carried forward applies: a wrong count in (b) used correctly here still earns the second mark.
Give the final value with no unit needed (it is a number of plants), and record it in the empty cell of Table 2.1.

Techniques used

calculate how many quadrats fit in the fieldscale up a sample count to a population estimate

(d)

Explain why the student counted the numbers of plants in samples of the field instead of counting the number of plants in the whole field.

______

1M Similar

(e)

Suggest two reasons why taking several samples would improve the accuracy of the estimate for the number of plants in the whole field.

______
______

2M Similar

(f)

Use the data in Table 2.1 to construct a bar chart to show the estimated number of plants of species C, D and E in the whole field.

grid

4M Similar

Q37MMediumPlanning Experiments and Investigations

(a)

Plants can be provided with fertiliser to help them grow.

Some students were provided with germinated seeds of the same species in some shallow dishes and a 10% fertiliser solution. One dish is shown in Fig. 3.1.

The students had access to any other common laboratory apparatus.

Fig. 3.1

Plan an investigation the students could carry out to find the effect of different fertiliser concentrations on plant growth.

6M Similar

DifficultyMedium

Worked solution

Answer

Measure the growth of the seedlings by measuring their height (or mass) at the start and again at the end of the experiment.
Prepare at least three different fertiliser concentrations (e.g. 2%, 4%, 6%, 8%), all less than or equal to 10%.
Add the same volume of each fertiliser solution to the respective dishes.
Leave the dishes for the same amount of time.
Ensure all dishes are kept under the same conditions (e.g. same light intensity, temperature, $\text{CO}_2$ levels, and volume of water).
Repeat the experiment with more dishes at each fertiliser concentration and calculate a mean for each concentration.
Compare the results (mean height or mass) against the fertiliser concentration.

Final answer

See working

Detailed explanation

Walkthrough

The question asks for a plan to investigate the effect of different fertiliser concentrations on plant growth. A complete plan must cover what will be measured, the range of the independent variable, the controlled variables, repeats, and how the results will be analysed.

Measurement: Decide how growth will be measured (e.g. height or mass) and state that it must be measured at the start and at the end to calculate the change.
Independent variable range: Use at least three different concentrations of fertiliser, ensuring none exceed the 10% solution provided.
Volume of treatment: Add the same volume of fertiliser solution to each dish to ensure a fair test.
Time: Leave the dishes for the same duration after applying the fertiliser.
Controlled variables: Keep other factors that could affect growth constant, such as light intensity, temperature, $\text{CO}_2$ levels, and the volume of water.
Repeats and mean: Use multiple dishes for each concentration and calculate a mean to make the results more reliable and identify anomalies.
Analysis: Plot or compare the mean growth against the fertiliser concentration to see the effect.

Key Takeaways

A 6-mark planning question requires a structured method. Always include: what you will measure and when, the range of the independent variable, at least three controlled variables, repeats with a mean, and how you will analyse the data.

Common Mistakes

Forgetting to measure growth at the start as well as the end (you need a baseline to calculate growth).
Suggesting concentrations greater than 10% (the mark scheme specifies all $\le 10\%$ ).
Giving vague controlled variables like "same conditions" instead of naming specific factors like light intensity or temperature.
Forgetting to calculate a mean from repeats.

Things to Be Careful About

The question provides a 10% fertiliser solution, so your planned concentrations must be $\le 10\%$ .
Ensure you state how you will measure growth (e.g. "measure height with a ruler") rather than just saying "measure growth".
Do not just list controlled variables; explain that they must be kept the same.
The final mark requires stating how you will use the data (e.g. "compare mean height to fertiliser concentration").

Techniques used

plan an investigationidentify controlled variablesdesign a method to measure growthcalculate a mean from repeats

(b)

Identify the independent variable in this investigation.

______

1M Similar