2026 HCI 9477 H2 Bio Prelim P3 QP

HWA CHONG INSTITUTION JC2 Preliminary Examinations Higher 2 CANDIDATE NAME CT GROUP 25S7 __ CENTRE NUMBER S INDEX NUMBER BIOLOGY Paper 3 Long Structured and Free-response Questions Candidates answer on the Question Paper. No Additional Materials are required. 9477/03 15 September 2026 2 hours READ THESE INSTRUCTIONS FIRST There are four question booklets ( I - IV ) to this paper. Write your name , CT group , Centre number and index number in the spaces provided at the top of this cover page and on the lines provided at the top of the cover page of Booklets II, III and IV . 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. Section A Answer all questions in the spaces provided on the Question Paper. Section B Answer any one question 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. You are reminded of the need for good English and clear presentation in your answers. This document consists of 23 printed pages. © Hwa Chong Institution 2026 9477 H2 Biology / JC2 Preliminary Examinations / Paper 3 For Examiners' Use Section A 1 / 34 2 / 10 3 / 11 Section B 4 or 5 / 20 Final Mark / 75

2 Section A Answer all the questions in this section. QUESTION 1 The human papilloma virus (HPV) is sexually transmitted and can cause the development of cervical intraepithelial neoplasia (CIN). Patients with CIN are observed to have abnormal growth of cells that line the cervix. This can lead to the development of cervical cancer. Fig. 1.1 shows some key genes present in the HPV genome. E6 and E7 are structural genes while E1 and E2 are regulatory genes. The effect of the E2 protein on the transcription of viral genes is also shown. Fig. 1.1 (a) Suggest why the E2 gene is a regulatory gene . [2] © Hwa Chong Institution 2026 9477 H2 Biology / JC2 Preliminary Examinations / Paper 3

3 (b) Upon infection, HPV enters cells of the cervical mucosal epithelium, where its genome is integrated into the host chromosome. During the integration process, the E2 gene is cleaved and joined to a segment of the host chromosome, as shown in Fig. 1.2. Fig. 1.2 When expressed, viral E6 and E7 proteins interfere with host tumour suppressor proteins p53 and RB, as shown in Fig. 1.3. E2F is a host transcription factor while HDAC is a chromatin remodelling enzyme. © Hwa Chong Institution 2026 9477 H2 Biology / JC2 Preliminary Examinations / Paper 3

4 Fig. 1.3 © Hwa Chong Institution 2026 9477 H2 Biology / JC2 Preliminary Examinations / Paper 3

5 (b) (i) HPV genome remains in the cytoplasm until the initial stages of mitosis, when it is integrated into the host chromosome. Suggest why the integration of HPV genome into the host chromosome can only occur during the initial stages of mitosis . [2] (b) (ii) Outline the fate of the p53 protein after viral E6 protein binds to it . [2] (b) (iii) State the level of regulation of gene expression exerted by viral E7 protein on the RB protein and S-phase genes . RB protein S-phase genes [2] © Hwa Chong Institution 2026 9477 H2 Biology / JC2 Preliminary Examinations / Paper 3

6 (b) (iv) With reference to Fig. 1.1, 1.2 and 1.3, explain how infection with HPV could result in the development of CIN . [5] © Hwa Chong Institution 2026 9477 H2 Biology / JC2 Preliminary Examinations / Paper 3

7 Methotrexate is a common chemotherapy drug used for the treatment of cancers. It interacts with the enzyme dihydrofolate reductase, which catalyses the synthesis of nucleotides from its substrate, folic acid. Fig. 1.4 shows the structures of methotrexate and folic acid. methotrexate folic acid Fig. 1.4 (c) Describe the mode of action of methotrexate as a chemotherapy drug . [3] © Hwa Chong Institution 2026 9477 H2 Biology / JC2 Preliminary Examinations / Paper 3

8 (d) Research on the differences in respiratory metabolism between normal cells and cancer cells may provide an alternative molecular target for cancer treatment. Fig. 1.5 is a simplified diagram comparing the respiratory metabolism in normal differentiated cells and cancer cells. In the presence of oxygen, normal differentiated cells first metabolise glucose to pyruvate via glycolysis and then completely oxidise pyruvate in the mitochondria to CO 2 during the process of oxidative phosphorylation. Up to 36 molecules of ATP are synthesised for each molecule of glucose. When oxygen is limiting, normal differentiated cells redirect the pyruvate to generation of lactate (anaerobic glycolysis). This generation of lactate allows anaerobic glycolysis to continue but results in minimal ATP production. In this case, 2 molecules of ATP are synthesised for each molecule of glucose. Cancer cells tend to convert most glucose to lactate regardless of whether oxygen is present (aerobic glycolysis). Studies have shown that about 85% of the glucose available is directed to generation of lactate, while 5% of the glucose available is directed to oxidative phosphorylation. Fig. 1.5 © Hwa Chong Institution 2026 9477 H2 Biology / JC2 Preliminary Examinations / Paper 3

9 (d) (i) Describe the importance of converting pyruvate to lactate for the maintenance of glycolysis in the absence of oxygen. [2] (d) (ii) Using the informatio