2025 TMJC H2Bio PE P2 (A)

CANDIDATE NAME CIVICS GROUP H2 BIOLOGY 9744/02 Paper 2 Structured Questions 17 September 2025 2 hours Candidates answer on the Question Paper. No additional materials are required. ______________________________________________________________________________________ This document consists of 29 printed pages and 1 blank page. TAMPINES MERIDIAN JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION SUGGESTED ANSWERS

2 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Biology Answer all questions. 1. Fig. 1.1 shows a diagram of a glycosphingolipid, a type of lipid found in the cell surface membranes of most eukaryotic organisms. Fig. 1.1 (a) Compare the structures of the glycosphingolipid and phospholipids. [2] No. feature glycosphingolipid phospholipid 1 no. of fatty acid tails / hydrocarbon chains two 2 head group sugar molecule / monosaccharide phosphate head 3 bond joining sugar / glycerol to fatty acid ether bond / glycosidic bond ester bond 4 elements present C, H, O, P C, H, O, N 5 presence of glycerol absent present (b) Suggest a role of the glycosphingolipid in the cell surface membrane. [1] • Cell-to-cell recognition / cell-to-cell adhesion / regulate membrane fluidity Accept: (precursor) to form second messenger (c) Glucocerebrosidase is a glycosidase enzyme that breaks down glycosphingolipids by removing the saccharide portion of the molecule. Another example of a glycosidase enzyme is maltase that breaks down maltose into glucose molecules. Draw an arrow in Fig. 1.1 to show which bond glucocerebrosidase hydrolyses in the glycosphingolipid. [1]

3 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Biology (d) Fig. 1.2 shows the results of an experiment conducted to investigate the activity of glucocerebrosidase at various pH levels. Fig. 1.2 Explain why there is no enzyme activity below pH 3. [3] 1. Below pH 3, charged and/or polar R -groups of amino acids in the enzyme are neutralised. 2. Ionic and hydrogen bonds (ecf from point 1) between amino acids are disrupted. 3. There is a change in 3D conformation of the enzyme, hence the active site is no longer complementary to the substrate 4. No enzyme-substrate complex can be formed to catalyse the hydrolysis of the bond. Accept: less

4 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Biology (e) Another experiment was conducted to investigate the effect of substrate concentration on the activity of glucocerebrosidase, with and without a chemical called isofagomine. The results are shown in Fig. 1.3. Fig. 1.3 (i) Describe and explain the results of the experiment at substrate concentrations above 10 µmol dm-3 for the reaction without isofagomine. [2] [Describe] 1. Above 10 µmol dm-3, rate of reaction remains constant/plateaus at 40 a.u. [Explain] 2. All active sites of enzyme molecules are saturated with substrates / Enzyme concentration is the limiting factor. (ii) With reference to Fig. 1.3, explain the effect of isofagomine on the rate of enzymatic activity. [HI-2] [4] 1. Isofagomine is a competitive inhibitor. 2. Isofagomine is structurally similar to glycosphingolipids. 3. Isofagomine competes with glycosphingolipids to bind to the active site of the glucocerebrosidase enzyme, forming an enzyme -inhibitor complex / is complementary in shape to the active site of glucocerebrosidase. 4. The glycosphingolipids cannot bind to glucocerebrosidase , fewer enzyme- substrate complexes are formed per unit time, hence rate of reaction decreases. 5. With isofagomine added, the rate of reaction reaches Vmax at 10 µmol dm -3 substrate concentration. [Total: 13]

5 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Biology 2. Fig. 2.1 shows the movement of three different substances across a cell surface membrane. Fig. 2.1 (a) Explain why transmembrane proteins are needed to transport glucose and potassium ions across the cell surface membrane. [3] 1. Glucose is polar and potassium ions are charged. 2. Polar and charged molecules cannot pass through the hydrophobic fatty acid core of the cell surface membrane. 3. Transmembrane proteins provide a hydrophilic channel/pathway/passage for glucose and potassium ions to pass through the cell surface membrane. (b) Describe one difference between the mechanism of transport of glucose and potassium ions across the cell surface membrane. [1] 1. Idea of conformational change in glucose carrier but not potassium ion channel. 2. Only one molecule of glucose can be transported across the cell surface membrane at a time while multiple potassium ions can be transported across the cell surface membrane together. (c) Fig. 2.1 shows insulin being released out of the cell via exocytosis. Describe how insulin synthesised by ribosomes is transported to the cell surface membrane. [3] 1. Insulin molecules synthesised in the rough endoplasmic reticulum bud off from the rER in ER vesicles. 2. ER vesicles move to and fuse with the cis face of the Golgi apparatus 3. Insulin is sorted and packaged into secretory vesicles and buds off at the trans face of the Golgi body. 4. Ref to movement of vesicles along microtubules within the cell. [Total: 7]

6 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Biology 3. When there is insufficient oxygen in the cell, ATP is synthesised via anaerobic respiration. Fig. 3.1 shows one of the reactions occurring during anaerobic respiration in humans. This reaction is catalysed by lactate dehydrogenase. Fig. 3.1 (a) Identify (i) molecule A [1] pyruvate / pyruvic acid (ii) molecules B and C [1] B NADH C NAD+ (b) State the location in the cell where this reaction occurs. [1] cytosol / cytoplasm (c) Explain the importance of this reaction in the production of ATP. [2] 1. To regenerate NAD+ for glycolysis. 2. Net 2 molecules of ATP are produced via substrate level phosphorylation during glycolysis.

7 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Biology (d) Fig. 3.2 shows the structure of lactate dehydrogenase 3 (LDH -3), which is found in the lungs. Fig. 3.2 With reference to Fig. 3.1 and Fig. 3.2, describe the structural differences between LDH-3 and haemoglobin. [2] No. feature LDH-3 haemoglobin 1. secondary structure α-helices and β-pleated sheets only α-helices and β- pleated sheets 2. cofactor bound to the enzyme NADH heme group 3. type of binding site active site oxygen binding site (e) Lactate dehydrogenase is an allosteric enzyme. Explain how allosteric activators increase the rate of enzymatic activity in lactate dehydrogenase. [2] 1. Allosteric activators bind to an allosteric site / site away from active site on one subunit of lactate dehydrogenase. 2. Idea of conformational change in (all) subunits into active form. [Total: 9]

8 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Biology 4. Meristematic tissue is found in the growing region of plants, such as root tips. Fig. 4.1 shows a section through the meristematic region of a root tip of onion, Allium cepa. Fig. 4.1

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