2025 YIJC [H2] Prelim P2 (A) With EXAMINER REPORT

YISHUN INNOVA JUNIOR COLLEGE JC2 PRELIMINARY EXAM Higher 2 NAME INDEX NO CG BIOLOGY Paper 2 Structured Questions Candidates answer on the Question Paper. No Additional Materials are required. 9744/02 9 SEPTMBER 2025 2 hours READ THESE INSTRUCTIONS FIRST Write your name, index no. and CG on this cover page. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, 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 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 At the end of the examination, submit booklets A, B and C separately to the invigilator. This document consists of 23 printed pages and 1 blank page. For Examiner’s Use Section A 1 2 3 4 5 6 7 8 9 10 11 Total 100 ANSWERS

2 ©YIJC 9744/02/PE/2025 [Turn over] Answer all questions. 1 The uptake of glucose and the nutrient coupled transcellular ions traffic across epithelial cells in the small intestine has been an ongoing topic in physiological research. Fig. 1.1 shows how protein A and protein B work together for the uptake of glucose across the cell surface membrane. Fig. 1.1 (a) With reference to Fig. 1.1, (i) identify Protein A Sodium Potassium Pump (Na+–K+ pump) or protein pump / transport protein / carrier protein / channel protein [1] Examiner’s Comment: As Fig 1.1 does not provide sufficient information to conclude it is ATP pump. Transmembrane and integral protein are rejected as these answers are too vague in comparison to the information provided. (ii) describe how glucose enters the cell 1 (Idea of ATP Hydrolysis): Protein A facilitate the movement of K + and Na+ through the cell membrane where ATP is used /converted to ADP 2 (Idea of ion concentration gradient): A concentration gradient is created where there is more Na+ outside of the cell and more K+ within the cell 3 (Idea of simple diffusion of glucose): Na+ will enter the cell with glucose via protein B [3] (R) chemiosmosis, no ATP is generated. (R) concentration gradient of glucose (R) if student identify wrongly: i.e. W – potassium and X – sodium. Max:1mark Examiner’s Comment: Many students failed to refer to Fig1.1 to explain how glucose is brought into the cell, regurgitation of content resulted in the lost of marks. Protein A Protein B glucose 2 K+ 3 Na+ Outside of cell High Na+ concentration Low K+ concentration High glucose concentration Inside of cell High K+ concentration Low Na+ concentration Low glucose concentration glucose

3 ©YIJC 9744/02/PE/2025 [Turn over]

4 ©YIJC 9744/02/PE/2025 [Turn over] (b) Glucose may be used for the synthesis of secretory glycoproteins. Explain how such glycoproteins are made within the cell and secreted by the cell. 1 Glucose enters the cell via facilitate diffusion /active transport and made its way to the rER 2 Ribosomes attached to RER synthesised polypeptide chain by joining amino acids via peptide bond (idea of translation) 3 Polypeptide folds into 3D structure and undergoes initial glycosylation 4 Short chains of glucose/ oligosaccharides attached covalently to protein to form glycoprotein 5 The glycoproteins are packaged into rER / transport vesicles which pinch / bud off from the rER, transport to Golgi apparatus 6 The rER / transport vesicles transported to Golgi apparatus and fuse with the cis face of the Golgi apparatus, where the glycoproteins undergo further chemical modification and processing as they move from cis to trans face; 7 Glycoproteins are packaged into secretory vesicles which bud off from the trans face 8 transported to the cell surface membrane via microtubules; 9 Golgi vesicles then fuse with the cell surface membrane to release the glycoproteins via exocytosis; Any 5 of the 9 marking points [5] [Total: 9]

5 ©YIJC 9744/02/PE/2025 [Turn over] 2 Laccase, a copper-containing enzyme, catalyses the formation of lignin in plants. The copper atoms within the laccase enzyme are essential for its catalytic activity, accepting electrons from the substrate and transferring the electrons to monolignols. Fig. 2.1 is a diagram of the mode of action of laccase. Fig 2.1 (a) Describe and explain the mode of action of laccase when catalysing the formation of lignin. 1 Laccase has an active site with a specific 3D shape (lock and key hypothesis) that binds to monolignol substrates OR active site 3D shape will change to fit substrate (induce fit hypothesis ) 2 Substrates bind to the active site, forming an enzyme–substrate complex through temporary interactions 3 Laccase facilitates oxidation & reduction (redox) reactions by transferring electrons using copper 4 This results in the formation of bonds between monolignols OR initiation the polymerisation of monolignols into lignin chains; 5 Laccase remains unchanged at the end of the reaction and can catalyse further reactions; 6 This reaction lowers the activation energy required for lignin formation and increases the rate of reaction; [5] Any 5 Examiner’s Comment: Many students interpret the copper ion as a contact residue rather than a catalytic residue. Copper ion is responsible for the catalysis for bond formation rather than the change in conformation of active site for induced fit hypothesis.

6 ©YIJC 9744/02/PE/2025 [Turn over] In a recent study, researchers investigated how environmental pH affects the activity of laccase extracted from two plant species, Species A and Species B. The results are shown in Table 2. Table 2. Laccase activity at different pH levels (arbitrary units) pH 3 4 5 6 7 8 Species A 5 12 20 14 7 2 Species B 4 9 13 18 20 19 (b) Explain why the activity of laccase is lower at pH 3 and pH 8 for Species A. 1 At pH 3 and 8, the H⁺ or OH⁻ ions disrupt ionic and hydrogen bonds in the laccase enzyme; 2 This alters the 3D shape of active site/ enzyme is denatured reducing E+S complexes/ substrate binding and thus enzyme activity; [2] (c) Using Table 2, suggest and explain which species is likely to have a laccase adapted to alkaline conditions. 1 Species B is likely to have laccase adapted to alkaline conditions 2 Its enzyme activity remains high at 18 to 20 arbitrary units between pH 6 and 8, with an optimum at pH 7; 3 Species A shows a sharp drop in activity from 20 to 2 arbitrary units beyond its acidic optimum at pH 5 Marking point 1 + 2 or 3 If marking point 2/3 is present without data, 1 mark is given [3] Examiner’s Comment: Some student failed to identified an alkaline condition is above pH 7. There is also a lack of proper comparative data between A and B at the same pH. [Total: 10]

7 ©YIJC 9744/02/PE/2025 [Turn over] 3 Vinblastine is a chemotherapy drug used to treat cancers such as lymphoma and breast cancer. It works by binding to tubulin, preventing the formation of microtubules, which are essential for the formation of the mitotic spindle. Researchers investigated the effect of Vinblastine on the mitotic cell cycle of cancer cells. Each group was treated with a different concentration of Vinblastine. After 24 hours (approximately one cell cycle, the percentage of cells in stages of mitosis was calculated.