NYJC 2023 H2 Biology 9744 P1 QP

NANYANG JUNIOR COLLEGE PRELIMINARY EXAMINATIONS Higher 2 CANDIDATE NAME CLASS BIOLOGY 9744/01 Paper 1 Multiple Choice 22 September 2023 1 hour Additional Materials: Multiple Choice Answer Sheet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, highlighters, glue or correction fluid. Write your name and CT on the Answer Sheet in the spaces provided unless this has been done for you. 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. Calculators may be used. This document consists of 25 printed pages. [Turn over

9744 / H2 Biology / 01 2 1 A eukaryotic 80S ribosome consist of a 60S large and a 40S small subunit. Which sequence of events concerning ribosomes is correct? A rRNA and proteins are synthesi zed and subuni ts are formed within the nucleolus . They become membrane bound as they are exported through the nuclear envelope to the cytoplasm and rough endoplasmic reticulum (rER). B rRNA and proteins assemble to form the 60S subunit in the nucleus while those of the 40S subunit are assembled independently in the cytoplasm. C rRNA is synthesized in the nucleolus and proteins are synthesized by the rough endoplasmic reticulum (rER). Subunit formation occurs within the cytoplasm for free ribosomes and on the surface of the rER for attached ribosomes. D rRNA synthesised within the nucleolus is complexed with prot eins that have been impor ted from the cytoplasm. The subunits formed are exported separately to the cytoplasm v ia the nuclear pores.

9744 / H2 Biology / 01 3 2 Human and mouse cells were fused to make hybrid cells. Anti-human and anti-mouse antibodies, carrying different coloured fluorescent dyes, were added. The antibodies bind to the protein of the cell surface membrane. The fused cells were incubated for 40 minutes. The locations of the human and mouse membrane proteins were identified at intervals using the fluorescent dyes. The diagram represents the results of the experiment by showing the positions of the human and mouse proteins on the surface of the cells. What does this experiment show? A Movement of the phospholipids pushes the membrane proteins apart. B Some membrane proteins move through the phospholipids to different places. C The phospholipids of the human and mouse cells surface membranes do not mix. D The proteins of human cell surface membranes can move further than those of mouse cells.

9744 / H2 Biology / 01 4 3 The peptidoglycan layer in bacterial cell walls is a crystal lattice structure formed from linear chains of two alternating amino sugars, N-acetylglucosamine (NAG) and N-acetylmuramic acid (NAM). Each NAM is attached to a short amino acid chain. The figure below shows the molecular structure of NAM and NAG. Which of the following statements correctly compares peptidoglycan and cellulose? similarity difference A both contain β-1,4 glycosidic bonds two different types of monomers are present in cellulose; one type of monomer is present in peptidoglycan. B monomers are rotated 180° to each other cross-links between cellulose chains are hydrogen bonds; cross-links between carbohydrate chains in peptidoglycan are peptide bonds. C Both provide high tensile strength that resist osmotic lysis Cellulose contains only C, H, O and N atoms; Peptidoglycan contains C, H, O, N and P atoms. D both comprise chains lying parallel to one another β-glucose is found in cellulose; α-glucose is found in peptidoglycan.

9744 / H2 Biology / 01 5 4 A student made observations of the structures and functions of the molecules W, X, Y and Z. Which row correctly describes the structure and function of the specified molecule? molecule structure function A W helical and unbranched allows for the development of turgidity B X helical and branched allows for quick release of respiratory substrates C Y long unsaturated chains compartmentalizes enzymes D Z branched and saturated maintains osmotic potential in the cells

9744 / H2 Biology / 01 6 5 The graph shows the results of an investigation using invertase, an enzyme that breaks down sucrose into glucose and fructose. 1g of sucrose was dissolved in 100 cm3 of water and 2 cm3 of a 1% invertase solution was added. Which conclusion can be drawn from this information? A Between 0 and 60 min, the concentration of the substrate remains constant. B After 60 min, the concentration of enzymes becomes the limiting factor. C At 140 min, some of the enzyme molecules are denatured. D Between 60 and 140 min, the concentration of the substrate is the limiting factor.

9744 / H2 Biology / 01 7 6 Catechol is oxidized to benzoquinone, as shown in the equation, resulting in darkening of peeled fruits. Catechol oxidase is an enzyme which is inhibited by parahydroxybenzoic acid (PHBA). The structure of PHBA is shown below. Catechol oxidase is also inhibited by phenylthiourea (PTU) which binds to a copper atom in the enzyme. The copper atom is essential for the oxidative activity. Which of the following statements are not correct? I PHBA acts as a competitive inhibitor because its structure is similar to benzoquinone. II PHBA acts as a competitive inhibitor, and in the presence of PHBA, the Michaelis constant, K M of the reaction can be restored by increasing the concentration of catechol. III PTU acts as a non-competitive inhibitor because it does not change the shape of the active site of catechol oxidase. IV PTU acts as a non-competitive inhibitor, preventing the formation of enzyme-substrate complex between catechol oxidase, catechol and O 2. A I and II only B III and IV only C I, II and III only D I, II, III and IV

9744 / H2 Biology / 01 8 7 The diagram below shows the relative amounts of DNA in a Meselson and Stahl experiment that demonstrates semi -conservative DNA replication. 14N and 15N represents the type of nitrogen isotope present in the DNA strands. Time / min proportion of DNA / % 14N14N 14N15N 0 0 100 20 50 50 40 75 25 60 87.5 12.5 80 93.75 6.25 Which of the following statements are false about the process that occurred during the experiment? I Each parental strand has the same number of nucleotides and base sequence as its daughter strand. II DNA nucleotides and RNA nucleotides were used in the synthesis of each daughter strand. III The number of 14N15N DNA molecules observed decreased by 50% with each successive generation. IV 14N was introduced only at the start of the experiment. V Original strands of DNA at the start are no longer observed if the experiment is extended to 120 minutes. A I, II B I, III, V C II, IV, V D I, II, III, V

9744 / H2 Biology / 01 9 8 Fig. 8.1 and Fig. 8.2 are electron micrographs that show RNA synthesis. One of the diagrams depicts the process in a prokaryotic cell, while the other in a eukaryotic cell. The dark circular structures in Fig. 8.2 represent ribosomes. Fig. 8.1 Fig. 8.2 Which of the following statement(s) is/are false? I The arrow in Fi g. 8.1 is pointing to a type of RNA while F ig. 8.2 is pointing to the chromosomal DNA. II Fig. 8.1 shows transcription and translation occurring simultaneously in a membranous compartment within the cell.