2026 NJC H2 Bio Prelim P3 (QP)

©NJC 2026 9477/03/SH2 Preliminary Examination [Turn over NATIONAL JUNIOR COLLEGE, SINGAPORE Senior High 2 Preliminary Examination Higher 2 CANDIDATE NAME BIOLOGY CLASS 2bi2_____ REGISTRATION NUMBER Biology Paper 3 Long Structured and Free-response Questions Candidates answer on the Question Paper. No Additional Materials are required 9477/03 25 August 2026 2 hours READ THESE INSTRUCTIONS FIRST Write your name, Biology class and registration number 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 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 the brackets [ ] at the end of each question or part of question. For Examiner’s Use Section A 1 35 2 12 3 8 Section B 4 or 5 20 Total 75 This document consists of 18 printed pages and 2 blank pages.

2 ©NJC 2026 9477/03/SH2 Preliminary Examination Section A Answer all the questions in this section. 1 Glucose-6-phosphate dehydrogenase (G6PD) deficiency is the most common genetic disorder worldwide. (a) G6PD is a enzyme present in all cells . It catalyses the oxidation of glucose-6-phosphate (G6P) and reduces NADP+ to NADPH simultaneously. Fig. 1.1 shows the structure of G6PD which is composed of two identical polypeptide chains, each with an active site for G6P and NADP+, as well as a binding site for a cofactor NADP+. Fig. 1.1 (i) State and explain the highest level of protein structure of G6PD. [2]

3 ©NJC 2026 9477/03/SH2 Preliminary Examination [Turn over (ii) With reference to Fig 1.1, explain why enzymes like G6PD must be made from a hundred amino acids or more in order to fulfil their functions. [4] Fig. 1.2 shows the biochemical pathway involving G6PD. The letters P and Q represent the enzymes that catalyse a reaction in the pathway. Fig. 1.2 (iii) In red blood cells, G6PD is the only source of NADPH, which is required for generating reduced glutathione, an important antioxidant. Using Fig. 1.2 and the information given, describe how G6PD deficiency can lead to haemolysis (rupturing of red blood cells). [3]

4 ©NJC 2026 9477/03/SH2 Preliminary Examination (b) Researchers have identified over 230 variants of the G6PD gene. (i) Table 1.1 shows the classification of G6PD variants with varying G6PD activity relative to normal and the associated symptoms. Table 1.1 G6PD variant class G6PD activity relative to normal activity / % symptoms A <20 chronic haemolytic anaemia B <45 acute haemolytic anaemia triggered by certain medicines, fava beans or infection C >60 no haemolysis Justify the claim that the G6PD deficiency phenotype is the result of genotype and the environment acting together. [2]

5 ©NJC 2026 9477/03/SH2 Preliminary Examination [Turn over (ii) Researchers investigated the effect of substrate concentration on the activity of two G6PD variants, J and K. Variant J is the normal form of the enzyme, while variant K is resulted from a gene mutation. The results are shown in Fig. 1.3. Fig. 1.3 With reference to Fig. 1.3, describe the effect of the mutation on the activity of G6PD. [3]

6 ©NJC 2026 9477/03/SH2 Preliminary Examination (c) Fig. 1.4 is a pedigree showing the inheritance of G6PD deficiency in a family. Fig. 1.4 (i) Explain how Fig. 1.4 provides evidence that the inheritance of G6PD deficiency is X-linked recessive. [3] (ii) In Kilifi County of Kenya, the G6PD A- allele is the only cause of G6PD deficiency. The frequency of G6PD deficiency in males is 12%, and the rest of the males in the population have the G6PD A+ allele which codes for normal G6PD enzyme. Assuming the Hardy-Weinberg principle applies to this population and that allele frequencies in males and females are the same, complete the Table 1.2 to show the frequency of A+ allele and heterozygotes in females. Use the equations: p + q = 1 and p2 + 2pq + q2 = 1 Table 1.2 algebraic representation frequency value frequency of A- allele q 0.12 frequency of A+ allele p frequency of heterozygotes in females [3]

7 ©NJC 2026 9477/03/SH2 Preliminary Examination [Turn over (d) Malaria is caused by parasites from the genus Plasmodium. (i) There are approximately 229 million cases of malaria worldwide per year. The number of malaria cases found in Africa account for 94% of these cases, but these are not caused by Plasmodium vivax. Outside Africa, P. vivax causes 61% of the cases of malaria. Calculate the number of cases worldwide caused by P. vivax each year. Show your working and give your answer to three significant figures. cases of malaria caused by P. vivax = [3] (ii) When an infected mosquito injects Plasmodium into the human bloodstream. Plasmodium invade and multiply in the liver and red blood cells, causing them to rupture. Individuals with G6PD deficiency or carriers of G6PD deficiency show a reduced risk of severe malaria. Explain how evolution could have resulted in G6PD deficiency being more common in malaria-endemic regions compared to other populations. [4]

8 ©NJC 2026 9477/03/SH2 Preliminary Examination (e) (i) Primaquine is a prescription antimalarial drug used to eliminate dormant malaria parasites in the liver to prevent relapse. However, primaquine is a strong oxidising agent which can cause acute haemolytic anaemia in individuals with G6PD deficiency or female carriers with reduced G6PD enzymes. Hence, testing for G6PD activity is necessary before treatment. The most common test for G6PD activity is Fluorescent Spot Test (FST) which involves the following steps. • A patient’s blood sample is obtained and red blood cells are lysed. • A small volume of the blood is added to a reaction mixture containing glucose-6-phosphate (G6P) and NADP+. • The mixture was incubated at 37°C for 10 minutes, then a drop of the mixture is spotted onto a filter paper and allowed to dry. • Once dried, the sample is viewed under ultraviolet light lamp. NADPH fluoresces under ultraviolet light whereas NADP+ does not. The results are interpreted as follows. • Normal: bright green fluorescence (high NADPH production). • G6PD deficient: faint or completely absent fluorescence. • Partial deficiency: intermediate fluorescence (often seen in heterozygous females). State two strengths of using FST to test for G6PD activity. [2] (ii) A malaria patient experienced a recent haemolytic episode, causing a large number of red blood cells to lyse. This prompts the bone marrow to quickly produce more red blood cells to compensate. Newly produced red blood cells have less exposure to oxidative damage. Predict and explain the effect of this haemolytic episode on the result of the FST for a female patient who is a carrier for G6PD deficiency. [2]

9 ©NJC 2026 9477/03/SH2 Preliminary Examination [Turn over (iii) Suggest, in outline, a procedure that could determine whether or not a G6PD deficient allele is present. [4] [Total: 35]

10 ©NJC 2026 9477/03/SH2 Preliminary Examination 2 Influenza virus and human immunodeficiency virus (HIV) are important human pathogens that have resulted in pandemics. (a) (i) Compare the structure and organisation of genomes of influenza virus and HIV. [2] (ii) Compare how