Biology 5090/32 — October/November 2024
Cambridge O-Level · Practical Test · worked solutions for every part, with the mark scheme
Topics Observations and Measurements · Experimental Contexts · Analysis, Conclusions and Evaluation · Use of Techniques, Apparatus and Materials · Planning Experiments and Investigations · Microscopy and Biological Drawing
Carbohydrates are found in different plant organs in the form of starch and sugars (such as glucose and maltose).
You are going to test two different organs from two plants for the presence of starch and sugar.
You are provided with a piece of potato and a piece of apple.
State which reagent you will use for the test for starch and describe how you will use it.
Details of possible results are not required.
______
Answer
Apply iodine solution to the cut / crushed surface of the sample.
Iodine solution, applied to the cut or crushed surface of the sample.
Walkthrough
The starch test uses iodine solution. The mark scheme gives one mark for the reagent (iodine solution) and one mark for how it is used — you must apply it to the cut surface, or cut or crush the sample first. This is because starch is stored inside cells, so the reagent must reach the starch granules inside the tissue; a whole intact piece would give a weak or misleading result.
Key Takeaways
- Iodine solution is the reagent for starch; it turns blue-black if starch is present.
- The sample must be cut or crushed so the iodine can reach the starch inside the cells.
Common Mistakes
- Naming Benedict's solution (that tests for reducing sugars, not starch).
- Describing the colour change — the question explicitly says details of possible results are not required.
- Saying just 'add iodine' without saying it is applied to a cut or crushed surface — that second mark is lost.
Things to Be Careful About
- The question says 'Details of possible results are not required', so do not waste time describing blue-black colour here — that comes in (a)(ii).
Carry out a test for starch on a piece of the apple and a piece of the potato. Record your observations and what you can conclude from them in Table 1.1.
Table 1.1
| plant organ | observation | conclusion |
|---|---|---|
| apple | ||
| potato |
Answer
| plant organ | observation | conclusion |
|---|---|---|
| apple | yellow-brown / only slightly black (compared with the potato) | little or no starch (much less starch than potato) |
| potato | blue-black | starch present / large amount of starch |
Apple: yellow-brown, little or no starch; potato: blue-black, starch present.
Walkthrough
Iodine solution turns from yellow-brown to blue-black when it meets starch. The potato stores starch in abundance, so its cut surface goes strongly blue-black. The apple stores its carbohydrate mainly as sugars, so the iodine stays yellow-brown or only darkens slightly. The mark scheme insists that a comparison is implied in both the observation and the conclusion to reach full marks — 'less black than potato', 'less starch than potato'. The conclusion must be consistent with the observation you record.
Key Takeaways
- Iodine: yellow-brown → blue-black in the presence of starch.
- Potato = starch store; apple = sugar store.
- Conclusions in a results table must follow logically from the recorded observation.
Common Mistakes
- Writing 'black' for the potato without any comparison — the scheme caps the conclusion at 1 mark if no comparison is made.
- Giving a conclusion that does not match the observation (e.g. 'no starch' when you wrote 'slightly black').
- Describing the apple as 'no colour change' when a slight darkening is usually seen — 'yellow-brown' or 'less black than potato' is safer.
Things to Be Careful About
- The scheme note: 'comparison must be implied in observation and conclusion to access full marks' and 'max 1 for conclusion if no comparison in the conclusion'. Build the comparison into both columns.
You are going to use Benedict's solution to test for sugar in the potato and apple.
Follow these instructions:
- Cut a cube of potato.
- Cut the cube into small pieces and place these in a large test-tube.
- Label the test-tube.
- Add of distilled water and use a stirring rod to gently crush the pieces and mix them with the water.
- Clean your tile, cutting device and stirring rod.
Repeat this procedure with the apple using a clean test-tube.
- Add of Benedict's solution to both test-tubes.
Then raise your hand and the supervisor will add hot water to your water-bath.
Take care as the water will be hot.
Answer
Record the temperature of the water-bath in (e.g. — the value read from the thermometer).
The temperature read from the thermometer, recorded with the unit °C.
Walkthrough
This is a candidate-dependent reading: you read the thermometer in the water-bath and write the value with its unit, . The mark scheme awards the mark for a temperature recorded with the unit. Read to the precision the thermometer allows, usually the nearest whole degree.
Key Takeaways
- Always attach the unit to a recorded measurement — a bare number loses the mark.
Common Mistakes
- Omitting the unit .
- Recording a temperature without actually reading the thermometer (the supervisor checks the real bath temperature).
Things to Be Careful About
- The mark scheme says 'temperature recorded ' — the unit is part of the answer.
Place both test-tubes in the water-bath and leave them for ten minutes.
Record your observations and your conclusions in Table 1.2 after the test-tubes have been in the water-bath for ten minutes.
Table 1.2
| plant organ | observation | conclusion |
|---|---|---|
| apple | ||
| potato |
Answer
| plant organ | observation | conclusion |
|---|---|---|
| apple | solution turns green / yellow / orange / brick-red | sugar (reducing sugar) present |
| potato | solution stays blue | no sugar present |
Apple: green/yellow/orange/red, sugar present; potato: blue, no sugar.
Walkthrough
Benedict's solution is blue. On heating with a reducing sugar it changes colour through green, yellow and orange to brick-red, depending on how much sugar is present. The apple is full of sugars (glucose, fructose), so its tube changes colour. The potato stores starch, not reducing sugar, so its tube stays blue. One mark is for the observations, one for the conclusions, and each conclusion must be consistent with the observation recorded.
Key Takeaways
- Benedict's solution + heat: blue → green/yellow/orange/brick-red if reducing sugar is present.
- Staying blue means no reducing sugar.
- Observations and conclusions must match.
Common Mistakes
- Writing 'red' for the apple when the actual colour seen was only green or yellow — record what you saw.
- Concluding 'no sugar' for the potato when the observation recorded was a colour change, or vice versa.
Things to Be Careful About
- The scheme says 'Check supervisor's results' — your recorded observations must match what actually happened in the bath.
Suggest why it was important to:
- use a cube for each plant organ
______
- cut each cube up into small pieces
______
- crush the small pieces and mix them with water.
______
Answer
- Same-sized cube: so the results for the two organs are comparable (same amount of tissue tested).
- Cut into small pieces: to increase the surface area / break open the cells and release the sugars.
- Crush and mix with water: to release the sugars into the water, forming a sugar solution that the Benedict's solution can react with.
Comparable results; increased surface area / cells broken open to release sugars; sugars dissolve into the water to form a sugar solution.
Walkthrough
Each bullet is one mark.
- The 1 cm cube makes it a fair test: both organs contribute the same volume of tissue, so the colour intensities can be compared.
- Cutting into small pieces increases the surface area exposed to the water and Benedict's solution, and breaks cells open so the sugars inside are released.
- Crushing and mixing with water dissolves the released sugars, forming a sugar solution — Benedict's solution reacts with sugars dissolved in water, not with sugar locked inside intact tissue.
Key Takeaways
- Fair testing: keep the quantity of sample the same when comparing.
- Grinding/cutting increases surface area and ruptures cells, releasing the contents.
- Benedict's test works on sugars in solution.
Common Mistakes
- Writing vague answers like 'to make it work better' — name the quantity affected (surface area, sugars released).
- Confusing the cube (fair comparison) with the crushing (release of sugars) — each bullet has its own reason.
Things to Be Careful About
- The scheme phrases are 'results comparable', 'increase surface area / release sugars / break open cells', and 'sugars into solution / form a sugar solution' — mirror these ideas.
Explain why the tile, cutting device and stirring rod were cleaned after using them on the potato.
______
Answer
To prevent transfer of potato (starch / sugar) to the apple sample, i.e. to prevent cross-contamination.
To prevent cross-contamination of the apple sample with potato starch or sugar.
Walkthrough
Any potato residue left on the tile, cutting device or stirring rod could be carried into the apple tube. That would add starch or sugar from the potato to the apple sample and give a false result — the apple might appear to contain starch or more sugar than it really does. Cleaning prevents this cross-contamination.
Key Takeaways
- Clean apparatus between samples to avoid cross-contamination and false results.
Common Mistakes
- Writing 'to be hygienic' or 'for safety' — the issue is contamination of the sample, not cleanliness for its own sake.
- Naming only 'dirt' instead of the specific transferred material (potato starch or sugar).
Things to Be Careful About
- The scheme accepts 'prevent transfer of potato / starch / sugar or prevent cross contamination' — use the word 'cross-contamination' for a secure mark.
Describe how you would test the apple for protein, and state what you would observe if protein was present and not present.
test = ______
protein present = ______
protein not present = ______
Answer
test = cut / crush the apple, then add biuret reagent
protein present = solution turns mauve / purple / lilac
protein not present = solution remains blue
Crush apple and add biuret reagent; present = mauve/purple/lilac; absent = stays blue.
Walkthrough
The protein test uses biuret reagent. Like the Benedict's test, the sample must first be cut or crushed so the protein is released into the solution — biuret reagent reacts with proteins in solution. Biuret reagent is blue; if protein is present it turns mauve (purple, lilac or violet). If there is no protein, it stays blue. Each of the four elements — preparing the sample, naming the reagent, the positive colour and the negative colour — carries one mark.
Key Takeaways
- Biuret test: blue → mauve/purple/lilac/violet if protein is present; stays blue if not.
- Solid samples must be crushed so the nutrient dissolves before testing.
Common Mistakes
- Writing 'blue to purple' for 'not present' — no protein means the reagent stays blue.
- Naming Benedict's solution or iodine instead of biuret reagent.
- Omitting the crushing step, losing the first mark.
Things to Be Careful About
- Give the positive colour in the scheme's own words: mauve / purple / lilac / violet — 'pink' is not credited.
Photosynthesis in green plants can be summarised by the equation:
A student used the apparatus shown in Fig. 2.1 to investigate the rate of photosynthesis in an aquatic plant.
When the student switched the lamp on, bubbles of oxygen formed and entered the syringe. The student recorded the volume of oxygen produced in ten minutes. The syringe was then refilled with sodium hydrogencarbonate solution and the oxygen collected for another ten minutes.
The student then decided to investigate the effect of different colours of light on the rate of photosynthesis. Leaving the same plant in the glass jar, a red transparent filter was wrapped around the glass jar so that the plant received only red light. The volume of oxygen produced in ten minutes was measured. This measurement was then repeated.
The red transparent filter was replaced by a green transparent filter and then by a blue transparent filter. For each filter, the volume of oxygen produced in ten minutes was also measured twice.
The measurements are shown in Table 2.1
Table 2.1
| filter | volume of gas collected in ten minutes / | ||
|---|---|---|---|
| measurement 1 | measurement 2 | mean | |
| no filter | 1.1 | 1.3 | 1.2 |
| red | 0.7 | 0.9 | 0.8 |
| green | 0.3 | 0.3 | 0.3 |
| blue | 0.6 |
The syringe at the end of ten minutes for measurement 2 with the blue filter is shown in Fig. 2.2.
Record the volume of oxygen produced in measurement 2 with the blue filter in Table 2.1.
Answer
0.4
0.4
Walkthrough
The syringe scale in Fig. 2.2 is graduated in with major marks at 1 and 2. The gas-liquid meniscus sits just under halfway between 0 and 1, at the 0.4 mark. Read the bottom of the meniscus and record it in the measurement 2 column for the blue filter. The table already carries the unit () in its header, so only the number goes in the cell.
Key Takeaways
- Read a liquid or gas meniscus at its lowest point, to the precision of the scale (here 0.1 ).
- Units belong in the table header, not repeated in every cell.
Common Mistakes
- Reading the meniscus as 0.5 or 0.45 by misjudging the scale divisions.
- Writing 'cm³' in the cell — the unit is already in the column heading.
Things to Be Careful About
- Check the direction of the scale: here 0 is at the top (the blocked end) and numbers increase downwards, so the meniscus at 0.4 means 0.4 of gas has been collected.
- Record to the same precision as the other readings in the column (one decimal place).
Calculate the mean volume of oxygen produced with the blue filter and record it in Table 2.1.
Working
Answer
0.5
0.5
Walkthrough
A mean is found by adding the repeated readings and dividing by the number of readings. Measurement 1 for the blue filter is 0.6 and your reading in (a)(i) is 0.4 , so the mean is . Record it in the mean column to one decimal place, matching the other means in the table. Note the mark scheme allows error carried forward from (a)(i), so a correct mean from a slightly wrong reading still scores.
Key Takeaways
- mean = sum of readings ÷ number of readings.
- Repeating a measurement and taking a mean reduces the effect of an anomalous single reading.
Common Mistakes
- Dividing by the wrong number of readings, or adding without dividing.
- Giving the answer to more decimal places than the data (e.g. 0.50) — match the table's precision.
Things to Be Careful About
- The mean must be consistent with your (a)(i) value; the ecf note means the calculation itself is what is being marked.
Using the mean value, calculate the rate of photosynthesis per minute when no filter was used.
______
Working
Answer
0.12 per minute
0.12 cm³ per minute
Walkthrough
A rate is a quantity per unit time. The mean volume with no filter is 1.2 collected in 10 minutes, so the rate is per minute. The blank in the question asks for a number with a unit, so both must be given.
Key Takeaways
- rate = volume ÷ time.
- Always attach the unit to a rate: here per minute.
Common Mistakes
- Using 1.2 without dividing, or dividing by the wrong time.
- Omitting the unit — the answer line expects a rate with units.
Things to Be Careful About
- Use the mean (1.2), not one of the individual measurements.
- Write the unit as per minute; 'cm³/min' is also accepted.
Construct a bar chart on the grid to show the mean volume of gas collected with no filter and with the different coloured filters.
Answer
Bar chart of mean volume of gas collected (cm³) against filter colour, with four correctly plotted, ruled, equally spaced bars of equal width
Walkthrough
A bar chart is the right presentation because the independent variable (filter colour) is discontinuous — the categories are no filter, red, green and blue, not numbers on a continuous scale.
Step by step:
- Put the filter colour on the x-axis and the mean volume of gas collected on the y-axis. Label both axes fully, including the unit on the y-axis — an unlabelled axis loses a mark.
- Choose a linear scale that starts at 0 at the origin and stretches so the bars use at least half the grid in both directions. The largest mean is 1.2 , so a scale of 2 cm per 0.2 works well.
- Plot the four means: 1.2 (no filter), 0.8 (red), 0.3 (green), 0.5 (blue).
- Draw each bar with ruled vertical sides, all of equal width, with equal gaps between bars and the bars not touching. Label each bar centrally with its filter colour (or label the x-axis categories centrally under each bar).
Key Takeaways
- Bar charts are for discontinuous (categorical) data; line graphs are for continuous data.
- Every mark on a chart-construction question maps to a convention: labels, scale, plots, bar style.
Common Mistakes
- Omitting the unit on the y-axis label.
- Drawing a line graph instead of a bar chart.
- Bars of different widths, or bars touching each other.
- A scale that does not start at 0, or that uses only a small corner of the grid.
Things to Be Careful About
- The mark scheme splits the marks as: axes fully labelled + bars labelled centrally; linear scale with value at origin and at least half the grid used in both directions; mean values plotted correctly; bars ruled, equal width, equal spacing, not touching.
- Plot the MEANS (1.2, 0.8, 0.3, 0.5), not the individual measurements.
- Use a sharp pencil and ruled lines throughout.
Use the data given and the bar chart to state one conclusion that can be made from the results of this investigation.
______
Answer
Green light caused the lowest rate of photosynthesis (smallest volume of oxygen collected).
Green light gave the lowest rate of photosynthesis / different colours of light gave different rates of photosynthesis
Walkthrough
A conclusion must be a statement about what the investigation shows, supported by the data. The mean volumes fall in the order no filter (1.2) > red (0.8) > blue (0.5) > green (0.3), so any of these conclusions scores: green light gave the least photosynthesis; the different colours caused different rates; or no filter (white light) gave the most photosynthesis. The reason green light gives the lowest rate is that green light is reflected by the chlorophyll in the leaves rather than absorbed, so little light energy is available for photosynthesis.
Key Takeaways
- A conclusion links the independent variable to the dependent variable using the data.
- Chlorophyll absorbs red and blue light strongly and reflects green light, which is why leaves look green.
Common Mistakes
- Restating a single number ('green gave 0.3 cm³') without saying what it means about the rate of photosynthesis.
- Saying 'green light is best for photosynthesis' — the data show the opposite.
Things to Be Careful About
- Only ONE conclusion is required; give one clear statement.
- Refer to the rate of photosynthesis (or the volume of oxygen as a measure of it), not just to the volumes.
Plan an investigation that you could do to determine the effect of light intensity on the rate of photosynthesis, using the apparatus shown in Fig. 2.1.
Answer
- Place the lamp at several different distances from the glass jar (e.g. 10 cm, 20 cm, 30 cm, 40 cm) to give at least three different light intensities.
- Keep the same plant in the same sodium hydrogencarbonate solution of the same concentration throughout.
- Keep the temperature constant (the lamp may heat the water — use a water bath or check with a thermometer).
- At each distance, allow the plant time to adjust to the new light intensity before measuring.
- Refill the syringe with fresh sodium hydrogencarbonate solution at each light intensity.
- Measure the volume of gas collected in the syringe over the same length of time (10 minutes) at each distance.
- Repeat each measurement and calculate a mean volume, then calculate the rate of photosynthesis (volume per minute) at each light intensity and compare.
See working — a method varying lamp distance to change light intensity, controlling plant, temperature and sodium hydrogencarbonate concentration, measuring gas volume over a fixed time with repeats
Walkthrough
The planning question asks how to investigate the effect of light intensity on the rate of photosynthesis using the Fig. 2.1 apparatus. The mark scheme lists eight creditable points for six marks, so aim to cover every category:
- Independent variable — light intensity. The easiest way to change it with this apparatus is to move the lamp to different distances from the jar; changing the bulb power or using a dimmer switch would also work. Use at least three different intensities so a pattern can be seen.
- Dependent variable — rate of photosynthesis, measured as the volume of oxygen collected in the syringe in a fixed time (or the time to collect a fixed volume).
- Controlled variables — keep everything else the same: the same plant, the same concentration of sodium hydrogencarbonate solution (which supplies carbon dioxide so it is not limiting), and the same temperature, because heat from the lamp could otherwise become the limiting factor.
- Fair testing details: refill the syringe before each reading so only the gas produced at that intensity is measured, and let the plant equilibrate at each new intensity before timing — bubbles released immediately after a change reflect the old conditions.
- Reliability: repeat each measurement and take a mean.
Key Takeaways
- Every plan must state what is changed, what is measured, and what is kept the same — and how each is done.
- Light intensity decreases with distance from the lamp, so distance is a valid proxy for intensity.
- Sodium hydrogencarbonate solution maintains the carbon dioxide concentration so that carbon dioxide does not limit the rate.
Common Mistakes
- Saying 'change the light intensity' without saying HOW (distance, bulb power, dimmer).
- Forgetting to keep the temperature constant — lamp heat is the obvious uncontrolled variable here.
- Using a different plant or different solution concentration between readings.
- Not specifying a fixed time period for collecting gas, so rates cannot be compared.
- Omitting repeats and a mean.
Things to Be Careful About
- The mark scheme wants at least THREE different light intensities — two is not enough.
- 'Allow time for the plant to adjust' is a separate mark point; do not merge it into the measuring step.
- Write the plan as a numbered method a candidate could follow, not as prose.
Fig. 3.1 shows photographs of a male and a female of an insect species. Both the male and female insects are green in colour.
Answer
The male has larger / thicker / swollen hind legs (femora) than the female.
Male has larger / thicker / swollen hind legs than the female.
Walkthrough
Look carefully at the two photographs side by side. The most striking difference is in the hind legs: the male's back legs have noticeably swollen, bulbous thighs (femora), while the female's are slender. The mark scheme also accepts the reverse comparison — a wider or longer abdomen in the female. Whichever difference you choose, you must make it COMPARATIVE: say which insect has the feature relative to the other. Writing 'the male has swollen legs' without comparing to the female does not fully answer 'difference between'.
Key Takeaways
- When asked for a difference between two specimens, always name both sides of the comparison.
- The swollen hind femora of the male Oedemera nobilis is its classic identifying feature.
Common Mistakes
- Giving a non-comparative statement ('the male has big legs') — the scheme insists answers must be comparative.
- Describing colour: both insects are green, so that cannot be a difference.
Things to Be Careful About
- Only one difference is required; do not write a paragraph.
- Use precise terms such as 'hind legs' or 'femora' rather than just 'legs'.
In the space below, make a large drawing of the male insect as it appears in Fig. 3.1.
Answer
A large drawing of the male insect, drawn in sharp pencil:
- clear, clean, continuous outline lines; no shading, stippling or cross-hatching;
- drawn in the same orientation as the photograph, at least 75 mm long from tip of head to end of abdomen;
- all six legs shown, each leg drawn with double lines at least to the first joint, with the hind femora (thighs) drawn swollen;
- two antennae attached to the head, and two clearly separate wing cases that are shorter than the abdomen.
Large pencil drawing of the male insect, ≥75 mm long, continuous unshaded lines, six double-lined legs with swollen hind femora, two antennae and two separate wing cases shorter than the abdomen.
Walkthrough
This is a standard 5090 biological drawing task, and most of the marks come from following drawing CONVENTIONS, not artistic skill.
Step 1 — Set up. Use a sharp HB pencil on plain (unruled) paper. Place the drawing so the finished insect will be at least 75 mm long from the tip of the head to the end of the abdomen — roughly three-quarters of the width of an A4 page.
Step 2 — Outline. Draw the body outline in one continuous movement where possible: head, thorax, elongated abdomen tapering to the posterior tip. Keep every line clean and continuous — never sketch over a line repeatedly, never use shading, stippling (dots) or cross-hatching to show tone, and never use a ruler.
Step 3 — Orientation and proportion. Draw the insect exactly as it appears in Fig. 3.1 — same orientation, same proportions. The abdomen must be the longest part, and the wing cases must be visibly SHORTER than the abdomen so the abdominal tip projects beyond them.
Step 4 — Appendages. Add all SIX legs, three per side, each drawn with a double line at least as far as the first joint. Critically, draw the hind pair with swollen, thickened femora — this is the male feature identified in part (a)(i). Add the two long antennae attached to the head.
Step 5 — Wing cases. Draw the two wing cases (elytra) as clearly separate structures running down the back, ending before the abdominal tip.
Key Takeaways
- Biological drawings are marked on conventions: continuous single lines, no shading, minimum size, correct proportions and orientation.
- Legs and tubes are drawn with double lines.
- Draw only what is specified — here, no labels were required.
Common Mistakes
- Shading or stippling to show the dark areas of the photograph — this loses the convention mark outright.
- Drawing too small (under 75 mm).
- Omitting one of the six legs, or drawing thin hind legs instead of the swollen male femora.
- Drawing the wing cases longer than the abdomen — the key in part (b) depends on them being shorter.
- Using ruled lines for the body outline.
Things to Be Careful About
- The scheme awards four separate points: line quality + orientation; minimum length; six legs + double lines + swollen femora; two antennae + separate wing cases shorter than the abdomen. Check each against your drawing before moving on.
- Measure your drawing with a ruler — 'large' means a specific 75 mm here.
C and D indicate the length of the female insect. D indicates the end of the abdomen.
Draw a line to join C and D on the photograph.
Measure the length of the line and record it.
______
Calculate the actual length of the insect and record it to the nearest whole number.
______
Working
Draw a straight line joining C (anterior end of head) to D (end of abdomen) on the female photograph.
Measured length of line = (accepted range 29–31 mm)
Answer
Actual length of the insect = (to the nearest whole number)
Line measured 29–31 mm; actual length ≈ 8–9 mm, given to the nearest whole number (e.g. 30 ÷ 3.5 = 8.57 → 9 mm)
Walkthrough
Step 1 — Draw the line. Use a ruler to join C, at the front of the head, to D, at the tip of the abdomen, with a single STRAIGHT line. A curved line following the body would not score the 'straight line' point.
Step 2 — Measure. Measure the line in mm with a ruler. Any measurement between 29 and 31 mm is accepted, because print sizes vary slightly.
Step 3 — Convert to actual size. The photograph is magnified ×3.5, meaning image length = 3.5 × actual length. So:
Divide your measured length by 3.5. For example, 30 ÷ 3.5 = 8.57 mm.
Step 4 — Round. The question asks for the answer to the nearest whole number, so 8.57 becomes 9 mm. Note the real insect is only about 8–9 mm long — consistent with the key in part (b), where Oedemera nobilis is 7–11 mm.
Key Takeaways
- Actual size = image size ÷ magnification; magnification = image size ÷ actual size.
- Always check the precision demanded — here, nearest whole number.
- Straight-line measurements between marked points, not curved ones.
Common Mistakes
- Multiplying by 3.5 instead of dividing — that gives ~105 mm, an absurd insect length; sanity-check against the key (7–11 mm).
- Forgetting the unit mm.
- Measuring along a curve instead of a straight line between C and D.
- Rounding too early or giving decimals when whole numbers are demanded.
Things to Be Careful About
- The accepted measured range is 29–31 mm; if your print measures outside this, re-check that you measured C to D exactly.
- ecf applies: a slightly wrong measurement still earns the calculation marks if divided correctly by 3.5.
- Write the division explicitly (measured length ÷ 3.5) — the method mark is separate from the answer mark.
Use the key below to identify and record the name of the insect.
1 wings longer than abdomen ............. Oedemera femoralis
wings shorter than abdomen ............ go to 2
2 colour brown ..................................... Oedemera barbara
colour green or grey ......................... go to 3
3 length 5–6.5 mm .............................. Oedemera lurida
length 7–11 mm ................................ Oedemera nobilis
name of insect = ______
Answer
Couplet 1: wings shorter than abdomen → go to 2.
Couplet 2: colour green → go to 3.
Couplet 3: length about 9 mm, i.e. within 7–11 mm → Oedemera nobilis.
name of insect = Oedemera nobilis
Oedemera nobilis
Walkthrough
Work through the dichotomous key one couplet at a time, using evidence from Fig. 3.1 and your own results.
Couplet 1: Are the wings longer or shorter than the abdomen? From the photograph (and your drawing in (a)(ii)), the wing cases end before the abdominal tip — wings SHORTER than abdomen → go to 2.
Couplet 2: Colour? Both insects are stated to be green (not brown) → go to 3.
Couplet 3: Length? Your calculation in (a)(iii) gave about 8–9 mm, which falls in the 7–11 mm range, not 5–6.5 mm → Oedemera nobilis.
This shows how the parts of a practical question link together: the identification depends on the length you calculated earlier.
Key Takeaways
- A dichotomous key offers two choices at each step; always follow the route supported by your observations.
- Species names are written in italics with a capital genus and lower-case species epithet.
Common Mistakes
- Choosing Oedemera lurida by ignoring the calculated length.
- Writing the name without italics or with a capital species name (Oedemera Nobilis) — binomial convention requires lower-case.
- Stopping at 'Oedemera' without the species name.
Things to Be Careful About
- Write the full binomial: genus capitalised, species lower-case, both italicised.
- Justify each step from the figure or your own measurement — the key only works if the observations feeding into it are right.