2026 NJC H2 Bio Prelim P1 (QP)

1 ©NJC 2026 9477/01/SH2 Preliminary Examination NATIONAL JUNIOR COLLEGE, SINGAPORE Senior High 2 Preliminary Examination Higher 2 CANDIDATE NAME BIOLOGY CLASS 2bi2_____ REGISTRATION NUMBER Biology Paper 1 Multiple Choice Additional Materials: Multiple Choice Answer Sheet 9477/01 16 September 2026 1 hour READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name, Biology class and registration number on all the work you hand in. There are thirty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. Read the instructions on the Answer Sheet very carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. The use of an approved scientific calculator is expected, where appropriate. Your five-digit OAS index number is made up of the second and last four digits of your registration number. For example: Registration no. OAS Index no. 2205123 25123 This document consists of 19 printed pages and 1 blank page.

2 ©NJC 2026 9477/01/SH2 Preliminary Examination 1 The electron micrograph shows some cell structures. Some cell structures are listed. 1 nucleolus 2 chloroplast 3 microtubules 4 ribosomes 5 mitochondria Which cell structures can be seen in the electron micrograph? A 1, 2, 3 and 4 B 1, 4 and 5 C 2, 3 and 5 D 2, 4 and 5 Explanation: The nucleus contains a dense, granular nucleolus. The cytoplasm has many mitochondria, seen as double-membrane profiles with cristae, and ribosomes as small dark granules. There are no chloroplasts: no thylakoid membranes, grana or starch grains are visible. Microtubules are not visible at this magnification.

3 ©NJC 2026 9477/01/SH2 Preliminary Examination 2 Cell structures can be isolated by centrifuging the cells in a sucrose density gradient. This breaks up the cells and separates the cell structures into fractions. The diagram shows the fractions in a sucrose density gradient after centrifugation. cellular components density chloroplasts and mitochondria 1.18 endoplasmic reticulum 1.16 Golgi bodies 1.06 lysosomes 1.14 ribosomes 1.60 Which fractions contain cell structures that carry out protein synthesis? A 1, 2 and 3 B 2, 3 and 4 C 2, 3 and 5 D 3, 4 and 5 Explanation: Each component settles where the sucrose density equals its own density. Reading the density scale: Golgi (1.06) is in fraction 1, lysosomes (1.14) in 2, endoplasmic reticulum (1.16) in 3, chloroplasts and mitochondria (1.18) in 4, ribosomes (1.60) in 5. Protein synthesis happens on free ribosomes (5) and on rough ER (3). It also happens in mitochondria and chloroplasts (4), which have their own 70S ribosomes and make some of their own proteins.

4 ©NJC 2026 9477/01/SH2 Preliminary Examination 3 The diagrams show three types of covalent bonds. Which bonds will be found in glycolipids and glycoproteins? bond 1 bond 2 bond 3 A glycolipids only glycolipids and glycoproteins glycoproteins only B glycolipids and glycoproteins glycolipids only glycoproteins only C glycoproteins only glycolipids only glycolipids and glycoproteins D glycoproteins only glycolipids and glycoproteins glycolipids only Explanation: Bond 1 is an ester bond. It joins fatty acids to glycerol in the lipid part of a glycolipid, and glycoproteins have no ester bond in their structure. Bond 2 is a glycosidic bond. It joins sugar residues to each other and joins the carbohydrate to the lipid or the protein, so it is in both molecules. Bond 3 is a peptide bond, which links amino acids in the polypeptide of a glycoprotein only.

5 ©NJC 2026 9477/01/SH2 Preliminary Examination 4 The table shows the results of an experiment on the rate of uptake of sugars by rat intestine. sugars rate of absorption under aerobic conditions under anaerobic conditions hexose sugars glucose 100 30 galactose 106 32 pentose sugars xylose 32 32 arabinose 30 30 Which conclusion may be drawn from the results? A All four sugars can be absorbed by active transport. B Only hexose sugars can be metabolised by the rat. C The intestine is freely permeable to hexose sugars. D The rate of passive diffusion of all four sugars into the gut is similar. Explanation: Glucose and galactose uptake falls sharply without oxygen (100 to 30, and 106 to 32). Some of their uptake therefore depends on ATP, which means active transport. Under anaerobic conditions the hexoses are absorbed at about 30 to 32. This matches the pentoses, xylose and arabinose, whose uptake (30 to 32) is unaffected by oxygen and therefore passive. The remaining uptake of all four sugars is by passive diffusion at a similar rate.

6 ©NJC 2026 9477/01/SH2 Preliminary Examination 5 Some students investigated the effect of pH on the rate of starch digestion by amylase. • They added 2 cm3 of a 1% amylase solution and 1 cm3 of pH 4 buffer to a test-tube. • They added 2 cm3 of a starch solution, containing 0.02 g of starch, to the test-tube. • They measured the time taken for the starch to be digested (to disappear). The students then repeated these steps with pH 5, pH 6, pH 7 and pH 8 buffers. The table shows the results. pH time taken for starch to be digested/ minutes 4 5 6 7 8 4.0 1.5 1.0 2.5 6.0 Which graph is correct for these data? Shortest time = highest rate, so pH 6 should be the optimum pH. Time taken and rate are inversely related, hence the graph should have opposite shape compared to the table of results. Since time taken to digest starch at pH 6 is 1 min, it means that it will reach the maximum rate, producing the total mass of starch in 1 min. Hence, the peak at pH 6 = 0.02 g = 20 mg.

7 ©NJC 2026 9477/01/SH2 Preliminary Examination 6 Scientists cultured two populations of human stem cells for many generations. • Population P consisted of adult stem cells. • Population Q consisted of embryonic stem cells. Which prediction about telomere length is most likely? population P population Q A telomeres are completely lost telomeres lengthen continuously B telomeres lengthen continuously telomere length is unchanged C telomere length is largely maintained telomeres shorten gradually D telomeres shorten gradually telomere length is largely maintained Explanation: Telomeres shorten with each round of replication because DNA polymerase cannot fully replicate the 3′ end of the lagging strand template (the end-replication problem). Embryonic stem cells have high telomerase activity, which adds telomeric repeats and keeps length largely constant, so they can divide indefinitely in culture. Adult stem cells have lower telomerase activity, so their telomeres shorten gradually and their division potential is limited. 7 The nucleic acids present in a cell of the bacterium Escherichia coli were analysed. Some of the results are shown in the table. type of nucleic acid number of different variants number of molecules per cell percentage of dry mass of cell 1 600 2500 2 2 60 160 000 3 3 2 20 000 21 Which row identifies each type of nucleic acid, 1, 2 and 3? 1 2 3 A DNA mRNA tRNA B mRNA tRNA rRNA C rRNA DNA mRNA D tRNA rRNA DNA Explanation: Type 1 has the most variants (600) but few copies of each, and a small share of dry mass. This fits mRNA: many different genes, each with few, short-lived transcripts. Type 2 has about 60 variants and many small molecules (160 000 per cell) but only 3% of dry mass. This fits tRNA, which comes in a limited number of types, each with a different anticodon. Type 3 has just 2 variants but m