TJC 2023 H2 BIO P2 ANS

1 [TURN OVER READ THESE INSTRUCTIONS FIRST Write your Center number, index number and name in the spaces 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. 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 any working or if you do not use appropriate units. TEMASEK JUNIOR COLLEGE 2023 JC2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME ANSWERS CENTRE NUMBER S INDEX NUMBER BIOLOGY 9744/02 Paper 2 Structured Questions (Part I) 25 AUGUST 2023 2 hours Candidates answer on the Question Paper. No Additional Materials are required. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 / 12 2 / 11 3 / 11 4 / 10 5 / 8 Answer all questions. This document consists of 16 printed pages.

2 Answer all questions. 1 Fig. 1.1 shows an electron micrograph of a plant cell. Fig. 1.1 (a) Identify the organelles labelled A, B, and C in Fig. 1.1. A: Nucleus B: Chloroplast C: Vacuole [3] (b) Use a line to label the cell wall, D. [1] (c) The magnification of the photomicrograph is 560x. Calculate the actual length of organelle C in μm, along the line P-Q. Show your working. Length of PQ = 74 mm [1/2] (A: 73, 74, 75 mm) Actual length of organelle C = length of image magnification [1/2] = 𝟕𝟕𝟕𝟕×𝟏𝟏 𝟎𝟎𝟎𝟎𝟎𝟎 𝝁𝝁𝝁𝝁 𝟓𝟓𝟓𝟓 𝟎𝟎 = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟏𝟏 𝝁𝝁𝝁𝝁 (A: 132.14 for 2 d.p.) actual length of organelle C: 132.1 [1] μm [2] [1/2] – formula [1/2] – state measurement of length PQ [1] – correct final answer (number) on the line provided D

3 [TURN OVER (d) Compare the structural features of organelle A and organelle B. [2] Similarities [any 1]: 1. Both of them are bound by double membrane. 2. Both of them contain DNA (and RNA). Differences [any 1]: 3. Nucleus has linear DNA while chloroplast has circular DNA. 4. Nucleus does not thylakoids while chloroplast has thylakoid membranes inside. 5. Nucleus does not contain ribosomes but chloroplast contain 70S ribosomes. 6. Presence of chlorophyll pigments/photosystems in chloroplasts vs absent in nucleus 7. Nucleus is larger / denser than chloroplast. AVP: A outer membrane is continuous with RER, but not B B has electron carriers / ATP synthase but not A B has starch grains but not A (e) A factor that can limit the rate of photosynthesis is the rate of regeneration of RuBP. Sedoheptulose-1,7-bisphosphatase (SBPase) is an enzyme in the Calvin cycle that controls the rate of regeneration of RuBP. SBPase is coded for by the gene SBPase , which is present in most plants. In an experiment, 2 wheat plants were studied. • one was genetically modified to make more SBPase by introducing the SBPase gene from another grass species. • one was not modified (wild type). Fig. 1.2 shows the mean mass of plant for the wild type plants and genetically modified plants. Fig. 1.2 Suggest and explain why genetically modified plants have a different mean mass than wild type plants. [4] mean mass of plant / g genetically modified wild type

4 1. Must mention: genetically modified plants (GMP) has higher biomass of 28g than wild type with 20.5g [1] [1/2 each] 2. GMP plants have an extra copy of SBPase gene / new type of SBPase gene 3. (overall) increased, expression / transcription, of SBPase gene (so more SBPase) 4. increased rate of / more, regeneration of RuBP 5. (so) increased / more, carbon fixation / Calvin cycles / light independent reaction / TP / GP 6. m ore , glucose synthesized, for cellular respiration [1/2] 7. m ore starch is stored [1/2] 8. more amino acids [1/2] / proteins for growth [1/2] [Total: 12] 2 Fig. 2.1 shows an electron micrograph of a mitochondrion. The labelled arrows X and Y both represent a structural feature of this organelle. Fig. 2.1

5 [TURN OVER The table below shows the protein composition of various areas in the mitochondrion in Fig. 2.1. Table 2.1 labels protein composition / % X 6 Y 21 region between X and Y 6 inside mitochondria 67 total 100 (a) Using the information in Table 2.1 above, (i) state the name of the structures labelled X and Y; X outer mitochondrial membrane Y inner mitochondrial membrane [2] (ii) account for the abundance of protein inside the mitochondrion. [2] 1. Protein composition inside W is high at 67% [1] 2. as there are abundant enzymes needed for Krebs cycle and link reaction [1] R: If candidate mentioned all processes e.g., glycolysis, link reaction, Krebs cycle, oxidative phosphorylation. Newborns have a large amount of brown fat tissue, which contains abundant mitochondria. Brown fat cells express the protein, thermogenin, which is embedded in the inner mitochondrial membrane. Protons flow through the channel in thermogenin instead of ATP synthase. As a result, the proton gradient is less steep, and energy is released in the form of heat. This keeps the babies warm. The mitochondrial matrix has a pH of about 7.8. The intermembrane space of mitochondria in different cells exhibits different pH values, as shown in Table 2.2. Table 2.2 cells from which mitochondria are isolated pH in intermembrane space resting muscle 7.0 muscle during exercise 6.8 brown fat 7.4 (b) (i) Explain the difference in pH values in the intermembrane space and the matrix of the mitochondria in the resting muscle cells. [3] 1. QF: The intermembrane space has a lower pH value of 7.0 compared to pH 7.8 in the matrix [1]

6 2. Reduced NAD and FAD release high energy electrons [1/2] 3. electrons are passed down the electron transport chain (electron carriers of decreasing energy levels), energy is released [1/2 ] 4. Energy is used to pump proton from the matrix into the intermembrane space, [1/2] 5. building a proton pool / maintain a proton gradient [1/2] (ii) Explain how low oxygen concentration will result in the newborns suffering from a drop in body temperatures. [2] any 4 – ½ each 1. Oxygen acts as the final electron and proton acceptor in the electron transport chain 2. When the concentration of oxygen is low, less reduced NAD and FAD molecules are able to release electrons into the ETC / less transfer of electrons down the electron carriers 3. H ence, less pumping of H+ from matrix to IMS 4. The proton concentration will reduced / reduction of proton gradient OWTTE 5. Less protons flow through thermogenin and ATP synthase, and less heat is produced (iii) The respiratory processes in the mitochondria require oxygen. Explain how oxygen is transported into the mitochondria. [2] 1. Oxygen is a small [1/2] 2. and non-polar molecule [1/2] 3. Hence, it can diffuse across the hydrophobic core of the phospholipid bilayer of the mitochondrial membrane [1] [Total: 11] 3 Glycogen phosphorylase is an enzyme involved in glycogenolysis. Fig 3.1 shows a model of the enzyme glycogen phosphorylase. Glycogen phosphorylase catalyses the hydrolysis of glycosidic bonds to break up glycogen into glucose subunits. The active site of glycogen phosphorylase is located in a slight depression on one side of the molecule. The three amino acids that form the active site are Serine, Histidine and Aspartic Acid. These three amino acids are some distance apart on the polypeptide chain but close together in the active site.

7 [TURN OVER Fig. 3.1 (a) Explain how the substrate may be attached to the enzyme. [2] 1. (Tempor