2026 NJC H2 Bio Prelim P2 (MS)

1 ©NJC 2026 9477/02/SH2 Preliminary Examination NATIONAL JUNIOR COLLEGE, SINGAPORE Senior High 2 Preliminary Examination Higher 2 CANDIDATE NAME BIOLOGY CLASS 2bi2_____ REGISTRATION NUMBER Biology Paper 2 Structured Questions Candidates answer on the Question Paper. No Additional Materials are required. 9477/02 21 August 2026 2 hours READ THESE INSTRUCTIONS FIRST Write your name, Biology class and registration number in the spaces at the top of this page. Write in dark blue or black pen. You may use an HB for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. 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 workings or if you do not use appropriate units. The number of marks is given in the brackets [ ] at the end of each question or part of question. For Examiner’s Use Section A 1 9 2 9 3 11 4 11 5 8 6 11 7 13 8 10 9 8 Total 90

2 ©NJC 2026 9477/02/SH2 Preliminary Examination This document consists of 31 printed pages and 1 blank pages. Answer all the questions. 1 Fig. 1.1 shows some organelles and storage molecules in a mouse liver cell. Fig. 1.1 (a) Name the storage polysaccharide A. A glycogen [1] Examiner’s comments: Several students did not read the first statement in the question that the image is a liver cell, so “starch” was a common incorrect answer. (b) (i) Name organelle B. B mitochondrion (accept: mitochondria) [1] Examiner’s comments: Some students thought that cristae in mitochondria will always appear as very clear folding of the inner membrane. (ii) B and C are the same organelle. Suggest why B and C appear to have different shapes. Different angles / views / planes / orientation; B is transverse section and C is longitudinal section; (no ECF) [1] (c) D is a ribosome. Identify two features visible in Fig. 1.1 that distinguishes D from A.

3 ©NJC 2026 9477/02/SH2 Preliminary Examination 1. D is attached (reject embedded) to a membrane / rER, while A exists freely scattered in the cytoplasm / is not attached to membrane; 2. D exists in a linear array / line (accept singly), while A are clustered / clumped together; if combine pt1&2, only award 1m (e.g. D is attached to membrane but A clumped together in cytoplasm) 3. D are all the same size, while A varies in size; accept A larger in size than D Examiner’s comments: A few students provided differences that are not visible or only described features of D without comparing to A. [2] (d) Scientists investigated the effect of ethanol on the changes in cell organelles in mice by using electron micrographs to estimate the volume of organelles as a percentage of the total cell volume. Table 1.1 shows the volumes of smooth endoplasmic reticulum (sER) and mitochondria as a percentage of the total cell volume in liver cells of control mice (unexposed to ethanol) and mice chronically exposed to ethanol. Table 1.1 organelle volume of organelles as a percentage of the total cell volume control mice mice chronically exposed to ethanol smooth endoplasmic reticulum (sER) 5.5 8.4 mitochondria 21.3 26.5 With reference to Table 1.1, describe and explain the changes in volume of sER and mitochondria in the mice chronically exposed to ethanol compared to the control mice. [Describe] sER increased by around 1.53 times or percentage increase of 53%; accept increase by 2.9% [Explain] increase sER is needed for detoxification/breakdown of ethanol ; (accept: to make ethanol less harmful) [Describe] mitochondria increased by around 1.24 times or percentage increase of 24%; accept increase by 5.2% [Explain] increase mitochondria increases the production of ATP by aerobic respiration for synthesis of more sER / detoxification enzymes; [4] Examiner’s comments: Many students did not manipulate the data to show changes but quoted the numbers directly from the table. A few students were confused with alcohol fermentation during anaerobic respiration in plants/yeast. Even if alcohol fermentation occurred, that is in the cytoplasm not mitochondria. [Total: 9]

4 ©NJC 2026 9477/02/SH2 Preliminary Examination 2 Fig. 2.1 is a simplified diagram showing the structure of callose. Not all hydrogen atoms are shown. Fig. 2.1 (a) Complete Table 2.1 to compare the structure of callose with cellulose. Table 2.1 feature callose cellulose monosaccharide used to synthesise the polysaccharide β-glucose β-glucose bond connecting monosaccharides β-1,3-glycosidic bond β-1,4-glycosidic bond orientation of monosaccharides in the molecule glucose molecules are oriented in the same direction / not rotated alternate glucose molecules are rotated 180° [3]

5 ©NJC 2026 9477/02/SH2 Preliminary Examination (b) Plasmodesmata are tiny pores in the cell walls of plants that connect neighboring cells. Fig. 2.2 shows the regulation of plasmodesmata pore size by callose. Fig. 2.2 (i) With reference to Fig. 2.2, describe how the synthesis of callose regulate plasmodesmata pore size. Signalling (max 2): Signal molecule/ligand bind to specific receptor, ref to signal transduction leading to activation of transcription factor; results in increase expression/transcription of genes involved in callose synthesis; Callose deposition (at least 1): Callose is deposited at the neck region of plasmodesmata / corners of the cell wall near the plasmodesmata; narrowing / closing the plasmodesmata pore/channel; As a result, the space between cell surface membrane and ER available for molecules to pass through is reduced, so movement of molecules between the two cells is restricted; Examiner’s comments: The question focus is on how callose made changes pore size, not the signalling pathway (though points were [3]

6 ©NJC 2026 9477/02/SH2 Preliminary Examination accepted). Also, question only focus on the synthesis of callose, not the degradation. Most students did well for this question as there are many marking points allowed, however, not many understood the pathway shown in Fig. 2.2. The signal molecule/receptor does not pass through the membrane and enter the cell. The dotted line is to represent signal transduction and not entry. Similarly, the genes do not more out to the cell wall. The genes involved in callose synthesis code for enzymes needed to catalyse the formation of callose. The genes do not code for callose since it is not a protein. (ii) Suggest one advantage to the plant in regulating plasmodesmata pore size through callose synthesis and degradation. The synthesis and degradation is reversible, allowing the plant to respond temporarily to a specific signal by closing the pore, and restore the molecular flux after the signal has passed; Prevents a permanent seal of the plasmodesma, which would irreversibly isolate the two cells and preventing future communication/transport; Examiner’s comments: Many students did not read the second half of the question and simply answered the purpose of regulating plasmodesmata pore size for transport, but did not consider the use of callose for regulation. Generic advantages such as reuse resources/glucose were rejected as it is not contextualised to regulation of pore size. Speed was not accepted also, as the process is not fast since it involves gene expression before callose can be synthesised. [1] (iii) Suggest why cellulose, despite also being a structural polysaccharide, would not be suitable for the regulation of plasmodesmata pore size shown in Fig. 2.2. cellulose chains are extensively hydrogen bonded into microfibrils (-OH unavailable for H-bond with water); ref to cellulose is highly stable and resistant to enzym