2026 DHS H2 Biology Prelim Paper 2 answers and examiner's comments

© DHS 2026 9477/02 [Turn over Name: Centre/Index Number: Class: DUNMAN HIGH SCHOOL Preliminary Examination Year 6 H2 BIOLOGY 9477/02 Paper 2 Structured Questions (PART 1 of 2) 26 August 2026 2 hours Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST: Write your name, centre number, index number and class at the top of this page. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. This exam consists of two parts: Part 1 and Part 2. You must complete both Part 1 and Part 2. This is Part 1. Answer all questions in the spaces provided on the Question Paper. The use of an approved scientific calculator is expected, where appropriate. You may lose marks if you do not show your working or if you do not use appropriate units. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 8 2 10 3 13 4 11 This document consists of 28 printed pages and 6 blank pages.

2 © DHS 2026 9477/02 Answer all questions. 1 Fig. 1.1 shows two different cells. Scale bars are shown to illustrate the relative sizes of the cells. Fig. 1.1 (a) Using your knowledge of cell structure, complete Table 1 which compares the locations of translation and size of ribosomes in the two cells. Table 1.1 Features of Comparison Bacteria Cell Animal Cell Location(s) of ribosomes Cytoplasm Rough endoplasmic reticulum and cytoplasm / mitochondrion Size of ribosome 70S 80S / 70S [2] Marker’s comment: The question already gives a clue “Locations(s)”, so try to give more answers. Also, size of ribosome is measured in Svedberg unit.

3 © DHS 2026 9477/02 [Turn over Both cells in Fig. 1 .1 contain membranes. The animal cell has considerably more membrane than the bacterial cell. (b) Use Fig. 1.1 and your knowledge of the structure of these cells to suggest two reasons for this difference. Animal cell has a larger surface area than bacteria cells.; Animal cells contain membrane-bound organelles, but bacterial cell does not.; [2] Both cells in Fig. 1.1 can store carbohydrates as glycogen. However, sweet potato cells store carbohydrates as sucrose instead. (c) (i) Explain why glycogen is generally a better storage molecule than sucrose. Glycogen is composed of thousands of α glucose while sucrose is a disaccharide. Hence, one molecule of glycogen stores a larger amount of energy compared to sucrose.; In glycogen, most of the -OH groups are occupied in intra -chain hydrogen bonding, but in sucrose, the - OH groups are free to interact with water. Hence, glycogen is more insoluble in water and osmotically inactive as compared to sucrose.; [2] Marker’s comment: Many students just give a rote response. You had memorised advantage of storing glycogen. BUT this question needs you to compare with sucrose. So, if you use the point about branched and compact, it is irrelevant when compared to sucrose since it is already a small molecule and can be compact even more. (ii) Suggest one reason why sweet potato cells store carbohydrates as sucrose. Sucrose is soluble; So can be easily transport in the phloem; or sucrose is a disaccharide; so it can be easily metabolised and provide energy; or sucrose is less reactive than redusing sugars like glucose; so it does not interfere with cell metabolism; or Sucrose has a lower osmotic effect per unit of energy stored; so storing carbohydrate in this form helps avoid excessive water influx and cell damage; [2] Total: [8]

4 © DHS 2026 9477/02 2 Seed germination is the process by which a dormant seed resumes growth and develops into a seedling. The key event that occurs during seed germination include hydrolysis of starch by amylase for respiration and biosynthesis. An investigation on the effect of temperature on germination was carried out using seeds from three species of North American woodland plants. Fig. 2.1 shows the percentage germination of these seeds at the mean spring temperature for these woodlands compared with an increase of 5°C. Fig. 2.1 (a) With reference to Fig. 2.1, explain the effect of an increase in temperature on germination of these seeds. The increase in spring temperature increased germination for all three species + quote any 1 data.; This is due to the increased in kinetic energy of starch and amylase ; Resulting in the increase in frequency of effective collisions between enzymes and substrates, and thus the rate of hydrolysis of starch.; Therefore, there is a greater supply of glucose to increase the rate of respiration.; Marker’s comment: Well done. Many students got full marks. The data is printed for you in this question and it is great that many of you use the data. SO, it is important to quote data for data-response question like this one. Use ruler and find correct data on your own. [4] 30 78 12 37 5 75

5 © DHS 2026 9477/02 [Turn over This temperature increase also affects the cell surface membrane. Another experiment was carried out to investigate the position of phospholipids in the membrane of a seedling cell as shown in Fig. 2.3. All phospholipids in the membrane were labelled with a fluorescent marker dye. When the marker dye absorbed light, it stopped fluorescing. The membrane in the area exposed to light then appeared as a dark spot. The level of fluorescence of the membrane over a period of time is shown in Fig. 2.2. Fig. 2.2 (b) Use one of the letters (A – D) shown on Fig. 2.2 to suggest when the cell membrane was exposed to the light and explain your answer. B; The fluorescence level suddenly drops to 0 au. This indicates the moment when the membrane was exposed to light, because the m arker dye stopped fluorescing, producing a dark spot.; [2]

6 © DHS 2026 9477/02 Fig. 2.3 shows the top and side view of an area of the membrane before, during and after exposure to a light source. Fig. 2.3 (c) With reference to Fig. 2.2, Fig. 2.3 and your knowledge of membrane structure, explain the recovery of the fluorescence. [Knowledge] Cell membrane is fluid / has a fluid mosaic structure.; [Knowledge] This allows phospholipids to move laterally within the bilayer.; [Fig. 2.3] The phospholipids with fluorescent marker dye move into dark spot (where phospholipids have stopped fluorescing).; [Fig. 2.2] This results in the increase in fluorescence from 0 au to 0.88 au and reaches a plateau at point D, when phospholipids are evenly redistributed across the membrane.; Marker’s comment: You have 3 references to make: Fig. 2.2, Fig. 2.3 and your knowledge of membrane structure (bilayer, fluid mosaic model). Your marks will come from this references, so it is important to write specifically about these 3 references. [4] Total: [10]

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8 © DHS 2026 9477/02 3 A researcher investigates a mutation in a gene using molecular techniques. Fig. 3.1 shows part of a DNA molecule. Strand 1: 5' — A T G C C A — 3' | | | | | | Strand 2: 3' — T A C G G T — 5' Fig. 3.1 (a) (i) Using Fig. 3.1, describe the structure of DNA and explain how this structure enables accurate replication. Double-stranded, complementary base pairing , A–T, C–G ; Each parental strand