Cambridge IGCSE Biology 0610: Leaf structure (by subtopic) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 6.2, Leaf structure, of the Cambridge IGCSE Biology 0610 syllabus for examination in 2026, 2027 and 2028. They test outcomes 6.2.1 (leaf shape), 6.2.2 (identifying leaf tissues) and 6.2.3 (how each tissue adapts the leaf for photosynthesis). All of 6.2 is Core, so every candidate needs it. One part, marked (Extended), uses the Supplement skill of converting millimetres to micrometres from section 2.2.

The topic-level plant nutrition practice questions cover photosynthesis itself. For the teaching, use the plant nutrition study guide and the plant nutrition revision notes. Course pages: Cambridge IGCSE Biology hub and Biology checklist. Whole-course check: Core diagnostic or Extended diagnostic.

Questions

1. Most leaves are thin and have a large surface area. Explain how each feature adapts a leaf for photosynthesis. [4]

2. A student describes six structures, P to U, in a photograph of a section through a dicotyledonous leaf. Identify each structure.

(a) P: a thin, shiny, non-cellular layer on the upper surface. [1] (b) Q: tall, closely packed cells just below the upper surface, full of chloroplasts. [1] (c) R: rounded, irregular cells with large gaps between them, below Q. [1] (d) S: a pair of curved cells in the lower surface, with a small pore between them. [1] (e) T: thick-walled, empty-looking cells on the upper side of a vein. [1] (f) U: smaller cells on the lower side of the same vein. [1]

3. Which list gives the leaf layers in order from the upper surface? [1]

• A upper epidermis, cuticle, palisade, spongy, lower epidermis
• B cuticle, upper epidermis, palisade, spongy, lower epidermis
• C cuticle, upper epidermis, spongy, palisade, lower epidermis
• D cuticle, palisade, upper epidermis, spongy, lower epidermis

4.

(a) The cuticle is waxy and transparent. Explain how each of these properties helps the leaf. [2] (b) The cells of the upper epidermis have no chloroplasts. Explain how this helps photosynthesis in the leaf. [1]

5. Describe the route carbon dioxide takes from the air into a chloroplast in a palisade cell, and explain why it moves in this direction. [4]

6. This question is about a vascular bundle in a leaf.

(a) Name the tissue that brings water into the leaf. [1] (b) Explain why the leaf needs a supply of water for photosynthesis. [1] (c) Name the carbohydrate that is carried away from the leaf in the phloem. [1] (d) State which tissue, xylem or phloem, is nearer the upper surface. [1]

7. Give three differences between palisade mesophyll cells and spongy mesophyll cells. [3]

8. A drawing of one palisade cell has a magnification of ×600. The cell is 42 mm long in the drawing.

(a) Calculate the actual length of the cell. Give your answer in mm. [2] (b) (Extended) Convert your answer to (a) into micrometres (μm). [1] (c) The drawn width is 12 mm. Calculate the actual width in mm. [1]

9. A student makes clear nail-varnish peels of the upper and lower surfaces of a leaf. Under a microscope she counts the stomata in five fields of view on each surface. Each field of view has an area of 0.20 mm².

Field of view	1	2	3	4	5
Stomata, lower surface	41	38	44	40	37
Stomata, upper surface	7	9	6	8	10

(a) Calculate the mean number of stomata per field of view for each surface. [2] (b) Calculate the number of stomata per mm² on the lower surface. [2] (c) Calculate how many times more stomata the lower surface has than the upper. [1] (d) Name the cells that open and close each stoma. [1] (e) Suggest one advantage of having most stomata on the lower surface. [1]

10. A plant is destarched in the dark for 48 hours. Three of its leaves are then treated:

• leaf 1: petroleum jelly spread over the lower surface
• leaf 2: petroleum jelly spread over the upper surface
• leaf 3: no petroleum jelly

Gases cannot pass through petroleum jelly. After six hours in bright light, each leaf is tested with iodine solution. Leaf 1 stays orange-brown. Leaves 2 and 3 turn blue-black.

(a) Explain the result for leaf 1. [3] (b) Suggest why leaf 2 still made starch. [1] (c) State the purpose of leaf 3. [1]

11. Leaf X is traced onto a grid of 1 cm squares. The outline completely covers 52 squares, covers more than half of another 18 squares and covers less than half of another 11 squares. Leaf Y has an area of 35 cm². Leaf X is 0.2 mm thick and leaf Y is 0.4 mm thick.

(a) Estimate the area of leaf X. Count squares more than half covered. [2] (b) Calculate the volume of each leaf in cm³. Use volume = area × thickness. [2] (c) Use your answers to explain which leaf is better adapted for photosynthesis. [3]

12. Explain how the internal structures of a dicotyledonous leaf adapt it for photosynthesis. Refer to light, carbon dioxide, water and products. [7]

Answers

1. Large surface area: more light can be absorbed [1], and there is more surface for carbon dioxide to enter the leaf [1]. Thin: carbon dioxide has only a short distance to diffuse to the mesophyll cells [1], and light can reach all the cells with chloroplasts [1]. Examiner insight: “Explain” needs each feature linked to photosynthesis; “thin so gases get in easily” is too vague without a short diffusion distance.

2. (a) Cuticle [1] (b) Palisade mesophyll [1] (c) Spongy mesophyll [1] (d) Guard cells [1] (e) Xylem [1] (f) Phloem [1] Examiner insight: A bare “mesophyll” earns neither (b) nor (c), and “epidermis” for (d) is too general: use the exact syllabus name.

3. B [1] Examiner insight: Check both pairs before choosing: the cuticle is on top of the epidermis, and palisade is above spongy.

4. (a) Waxy: it is waterproof, so it reduces water loss from the leaf [1]. Transparent: it lets light through to the cells below [1]. (b) Light is not absorbed there, so more light reaches the palisade mesophyll [1]. Examiner insight: Two properties means two separate marks; an answer about wax alone cannot earn the transparency mark.

5. Carbon dioxide diffuses in through an open stoma in the lower epidermis [1]. It moves through the air spaces in the spongy mesophyll [1]. It dissolves in the moisture on a palisade cell and diffuses into the chloroplast [1]. It moves this way down a concentration gradient, because photosynthesis uses up carbon dioxide in the chloroplasts, so its concentration inside is lower than in the air [1]. Examiner insight: “Absorbed through the leaf” gains nothing: name the structures in order and say diffusion is down a concentration gradient, not “towards where it is needed”.

6. (a) Xylem [1] (b) Water is a raw material of photosynthesis: it combines with carbon dioxide to make glucose [1]. (c) Sucrose [1] (d) Xylem [1] Examiner insight: In (c), “glucose” or “sugar” is not credited, because the syllabus specifies sucrose for phloem transport.

7. One mark per difference:

• palisade cells are tall and column-shaped; spongy cells are rounded or irregular [1]
• palisade cells are closely packed; spongy cells are loosely packed with air spaces between them [1]
• palisade cells contain more chloroplasts than spongy cells [1]

Also credited: palisade is nearer the upper surface. Examiner insight: Each difference needs both cell types or a comparative; “palisade cells have lots of chloroplasts” alone is not a difference.

8. (a) actual size = image size ÷ magnification [1] = 42 ÷ 600 = 0.07 mm [1] (b) 1 mm = 1000 μm, so 0.07 × 1000 = 70 μm [1] (c) 12 ÷ 600 = 0.02 mm [1] Examiner insight: Multiplying instead of dividing gives 25 200 mm, an impossible cell length; an answer that size loses both marks in (a).

9. (a) Lower surface: (41 + 38 + 44 + 40 + 37) ÷ 5 = 200 ÷ 5 = 40 [1] Upper surface: (7 + 9 + 6 + 8 + 10) ÷ 5 = 40 ÷ 5 = 8 [1] (b) stomata per mm² = mean per field ÷ area of field [1] = 40 ÷ 0.20 = 200 stomata per mm² [1] (c) 40 ÷ 8 = 5 times [1] (d) Guard cells [1] (e) The lower surface is shaded and cooler, so less water is lost by evaporation through the stomata [1]. Examiner insight: Error carried forward is usually allowed in (b), so a wrong mean divided correctly by 0.20 still earns the method mark if the division is shown.

10. (a) The petroleum jelly blocks the stomata [1]. Most stomata are on the lower surface, so carbon dioxide cannot enter the leaf [1]. Without carbon dioxide the leaf cannot photosynthesise, so no starch is made (the plant had been destarched in the dark) [1]. (b) There are few stomata on the upper surface, so carbon dioxide could still enter through the uncovered lower surface [1]. (c) A control: an untreated leaf under the same conditions [1]. Examiner insight: Each step (stomata blocked, no carbon dioxide, no photosynthesis) is a separate mark, so skipping the middle step costs a mark.

11. (a) 52 + 18, ignoring the 11 squares under half covered [1] = 70 cm² [1]

(b) 0.2 mm = 0.02 cm and 0.4 mm = 0.04 cm. Leaf X: 70 × 0.02 = 1.4 cm³ [1] Leaf Y: 35 × 0.04 = 1.4 cm³ [1] (c) Leaf X: same volume of tissue, but twice the surface area of Y, so it can absorb more light [1]. X is half as thick, so carbon dioxide has a shorter distance to diffuse to the mesophyll cells [1], and light reaches more of the cells [1]. Examiner insight: “Use your answers” means quoting the calculated comparison; a general statement about thin leaves without the data earns fewer marks.

12. One mark per linked point:

• Cuticle and upper epidermis are transparent, so light reaches the mesophyll [1].
• Palisade mesophyll is near the upper surface, with closely packed cells full of chloroplasts, so it absorbs most light [1].
• Chloroplasts contain chlorophyll, which transfers light energy for making carbohydrates [1].
• Stomata, opened by guard cells, let carbon dioxide in and oxygen out [1].
• Air spaces in the spongy mesophyll let carbon dioxide diffuse to the mesophyll cells [1].
• Xylem brings water, a raw material, to the leaf [1].
• Phloem carries sucrose away from the leaf [1].

Other correct structure-plus-function points are also credited. Examiner insight: Marks go to separate structure-plus-function points, so naming structures without saying how each helps photosynthesis scores little.

Where marks are usually lost

• Writing “mesophyll” without saying palisade or spongy.
• Putting the cuticle under the upper epidermis, or the spongy layer above the palisade layer.
• Writing “stomata” when the question asks for the cells that open and close them (guard cells).
• Saying carbon dioxide is “absorbed by the leaf” instead of naming the stoma, the air spaces and diffusion.
• Calling the phloem product “glucose” or “food” instead of sucrose.
• Placing phloem above xylem in a leaf vascular bundle.
• Multiplying by the magnification when finding an actual size, or forgetting units.
• Listing leaf features without the “so that…” link to light, carbon dioxide, water or products.

Next steps

• Plant nutrition revision notes
• Plant nutrition study guide
• Plant nutrition practice questions
• Cambridge IGCSE Biology hub
• Biology checklist
• 0610 Core diagnostic and 0610 Extended diagnostic
• Book a free trial class

Official syllabus

Cambridge IGCSE Biology 0610 syllabus for examination in 2026, 2027 and 2028, published by Cambridge Assessment International Education (Cambridge University Press & Assessment). Topic 6, Plant nutrition, section 6.2, Leaf structure (Core outcomes 1 to 3). The unit conversion in question 8(b) is Supplement outcome 2.2.3.