2026 NYJC H2 Biology Prelim Paper 2 MS

9477 / H2 Biology / 02 NANYANG JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME CLASS BIOLOGY 9477/02 Paper 2 Structured Questions 16 September 2026 Candidates answer on the Question Paper. No Additional Materials are required. 2 hours READ THESE INSTRUCTIONS FIRST Write your name and CT on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do no use staples, paper clips, highlighters, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. 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 no use appropriate units. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 7 2 10 3 5 4 9 5 10 6 9 7 10 8 8 9 10 10 6 11 6 Total 90 This document consists of 27 printed pages and 2 blank pages. [Turn over

9477 / H2 Biology / 02 2 Answer all the questions in this section. 1 Fig. 1.1 is a diagram of the cell surface membrane of a squamous epithelial cell lining an alveolus. Fig. 1.1 (a) A student measured the line A – B and calculated the actual width of the membrane in Fig. 1.1. (i) State the unit that the student should use for the actual width of the membrane. Nanometers / nm; [1]

9477 / H2 Biology / 02 3 (ii) With reference to Fig. 1.1, state how to identify the external surface of the cell surface membrane. A ora throughout (presence of) carbohydrate / sugar, chains / residues, on, (glyco)proteins / (glyco)lipids ; A there are no sugar chains on the inner surface A (presence of) glycoproteins / glycolipids (on external surface) A (presence of) cell surface antigens / receptors (on external surface) A (presence of) glycocalyx ® glucose chains, polysaccharide, LPS [1] (b) Name R and S in Fig. 1.1 and describe their roles in the membrane. component R: name cholesterol ; role maintains / regulates, fluidity of, membrane / phospholipid bilayer or at low temperatures, maintain / increase fluidity / prevents close packing A prevents hydrophobic ‘tails’ interacting at low temperatures or at high temperatures, stabilises the membrane / decreases fluidity / @prevent formation of large gaps ; 3 prevents passage (across membrane) of, hydrophilic / polar, substances ; component S name phospholipid (monolayer) ; R phospholipid bilayer role one from 5 forms a bilayer ; 6 (bilayer is a) barrier to, water soluble molecules / polar molecules / ions ; ora not a barrier to lipid soluble molecules / allows lipid soluble molecules to cross membranes A non-polar / AW 7 (bilayer is a) non- polar barrier between cytoplasm and (aqueous) surroundings; ® outside 8 hydrophilic ‘head’ forms hydrogen bonds with water ; [4] (c) The inner and outer membranes of the mitochondrion differ in the detail of their membrane components. The inner membrane is also much less permeable than the outer membrane. Suggest one way in which the structure of the inner membrane may differ from that of the outer membrane to produce a less permeable inner membrane. any one relevant suggestion: higher proportion of phospholipids with saturated fatty acids / ora ; fewer unsaturated fatty acids so, fewer ‘kinks’ in tails / closer packing ; higher proportion of cholesterol molecules ; fewer, channel / carrier / transport, proteins ; smaller diameter of channels in non-specific channel proteins ; fewer types of (specific), transport / carrier, proteins ; AVP ; e.g. fewer, aquaporins / channels for water ®absence of xxx

9477 / H2 Biology / 02 4 More phospholipids with saturated FA (proportion) ≠ more saturated FA tails ® less membrane proteins ® saturated carbons (refers to molecule, not the carbon itself) ® highly folded nature of inner membrane [1] [Total: 7] 2 (a) The lock‑and‑key hypothesis and the induced‑fit hypothesis are used to describe the interaction of enzymes and their substrates. Compare between the lock‑and‑key hypothesis and the induced‑fit hypothesis. one similarity ; e.g. substrate binds to active site enzyme-substrate complex forms product leaves active site interaction does not alter enzyme for re-use lower activation energy specific to one or very similar substrate(s) interaction can be affected by, action of inhibitors / pH / AW one difference ; e.g. induced-fit ora for lock-and-key active site not fully complementary to substrate change in, conformation / shape, of active site / AW context of binding substrate or after product release less specificity to substrate [2] (b) Phosphorylase enzymes can catalyse the synthesis of starch and the breakdown of starch in some plant tissues. A reaction catalysed by starch phosphorylase is shown in Fig. 2.1. Fig. 2.1 (i) Name the specific bond formed when the linear chain of starch is extended. α-(1,4) glycosidic bond [1]

9477 / H2 Biology / 02 5 (ii) Starch consists of two distinct polysaccharides, amylose and amylopectin. Explain why amylopectin allows for faster rate of glucose release by phosphorylase enzymes compared to amylose. Amylopectin contains α-(1,6) glycosidic bond creating multiple branches / terminal ends; ® amylopectin more branching than amylose Enzymes (starch phosphorylase) can only work on the non-reducing ends of starch chains, having more branches allows more enzyme molecules to bind and hydrolyze/cleave glucose units simultaneously; [2] A student carried out an experiment to study the synthesis of starch by phosphorylase found in potato tissue. The student was provided with a solution, E, extracted from potato tissue. The extract was filtered to remove all starch grains. Extract E contained biological molecules from the potato tissue, including phosphorylase but no starch. Iodine solution was used to confirm that starch was not present in extract E. Iodine solution turns from yellow to blue-black in the presence of starch. The student then added a small drop of a dilute starch solution and a solution of glucose 1‑phosphate to a test‑tube containing extract E. Samples of the reaction mixture in the test‑tube were removed every minute and a drop of iodine solution was added to each. The student used a colou rimeter to measure the absorbance of the solution in each sample. The results are shown in Fig. 2.2.

9477 / H2 Biology / 02 6 Fig. 2.2 (iii) Explain the results shown in Fig. 2.2. Idea of: any one phosphorylase (in extract) catalyses addition of glucose 1-phosphate to starch added to extract ; more starch is synthesised by phosphorylase as time progresses / AW ; idea of: any one: iodine solution added to mixture turns a, blue / black / blue- black, colour (in the presence of starch) ; the more starch present in the mixture, the, darker / more intense, the, colour / shade ; the darker the colour (of the positive result) the less light passes through, mixture / sample ; the darker the colour / AW, the higher the absorbance (reading of the colorimeter) ; [2] (iv) After 12 minutes, the student added a solution containing phosphate ions to the reaction mixture. The student continued taking samples every minute, adding iodine solution to each sample. The absorbance of the solution in each of these samples was measured. The results showed that the absorbance decreased over time. Suggest why the absorbance decreased. ref. to breakdown of starch catalysed by the enzyme / reaction in Fig. 3.1 goes in the opposite direction ;

9477 / H2 Biology / 02 7 AVP ; e.g. ref to equilibrium shift