YIJC 2023 H2 9744 P1 ANS

1 @YIJC 9744/01/PE/2023 [Turn over YISHUN INNOVA JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME CG INDEX NO BIOLOGY Paper 1 Multiple Choice Additional Materials: Multiple Choice Answer Sheet 9744/01 18 September 2023 1 hour READ THESE INSTRUCTIONS FIRST Do not use staples, paper clips, glue or correction fluid/tape. Write your name, index no. and CG on this cover page and OTAS 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. The use of an approved scientific calculator is expected, where appropriate. This document consists of 22 printed pages and 2 blank pages. ANSWERS

2 @YIJC 9744/01/PE/2023 [Turn over 1 Different processes occur in different organelles in a plant cell. 1 transcription of circular DNA 2 assembly of ribosomal subunits 3 breaking of covalent bonds by hydrolysis 4 polymerisation of monomers containing nitrogen Which processes occur in the following organelles? mitochondrion chloroplast nucleus A 1 1 3, 4 B 1, 3 1, 2 2, 4 C 2, 3, 4 2, 4 1, 2, 4 D 1, 3, 4 1, 3, 4 2, 3, 4 2 The diagrams show short sections of some common and modified polysaccharides. Which statement is correct? A 1 is found in cellulose as β-1,4-glycosidic bond is present. B 2 is found in amylopectin, which is a helical molecule found in starch. C 3 is found in cellulose, which is a linear unbranched chain of β-glucose monomers. D 4 is found in glycogen, which is made of chains of α-glucose linked by α-1,4-glycosidic bonds and α-1,6-glycosidic bonds at branch points.

3 @YIJC 9744/01/PE/2023 [Turn over 3 Molecule X is a lipid. Which row is correct for molecule X and a triglyceride? molecule X contains triglyceride contains A one unsaturated fatty acid saturated fatty acids B no ester bonds three ester bonds C one fatty acid three fatty acids D two fatty acids three fatty acids 4 Bread contains a mixture of polypeptides known as gluten. Two of the polypeptides found in gluten are glutenin and gliadin. Which statement describes the tertiary structure of a protein? A α-helical sections are found in glutenin and gliadin. B A large proportion of the amino acids in gliadin are glutamine. C Amino acids with hydrophobic R groups are found on the inside of glutenin. D Disulfide bonds form between glutenin and gliadin.

4 @YIJC 9744/01/PE/2023 [Turn over 5 Equal sized potato pieces were placed into a test -tube and covered with a sucrose solution. The test-tube was left for 30 minutes. All other variables were standardised. After 30 minutes, the potato pieces had not changed in size. What can be concluded from this result? A The concentration of sucrose is the same in the potato and in the solution and there is no more movement of water into or out of the potato. B The concentration of sucrose is the same in the potato and in the solution and there is no net movement of water into the potato. C The water potential is the same in the potato and in the sucrose solution and there is no more movement of water into or out of the potato. D The water potential is the same in the potato and in the sucrose solution and there is no net movement of water into or out of the potato. 6 Lipase is an enzyme that catalyses the hydrolysis of lipids. An experiment was carried out to investigate changes in pH when lipase is added to a food sample containing a high proportion of lipids. The results are shown in the graph. Which statements are possible explanations of the results of the experiment between 50 minutes and 60 minutes? 1 Enzyme concentration becomes the limiting factor. 2 Substrate concentration becomes the limiting factor. 3 All the enzyme active sites are occupied. 4 Denaturation of the enzyme by the products takes place. 5 There is end-product inhibition. A 1, 2 and 3 B 1, 4 and 5 C 2, 3 and 4 D 2, 4 and 5

5 @YIJC 9744/01/PE/2023 [Turn over 7 Which statements about adult stem cells are correct? 1 They are multipotent cells that have yet to express the genes and produce proteins characteristic of their differentiated state but do so when needed for repair of tissues and organs. 2 They are multipotent cells found in the adult and can give rise to all the cell types in the body. 3 They are undifferentiated cells that divide asymmetrically, giving rise to one daughter cell that remains a stem cell and one daughter cell that will differentiate to replace damaged and worn-out cells in the adult tissue or organ. 4 They are undifferentiated cells that reside under the surface of epithelial tissue, in position to take over the function of the tissue whe n the overlying cells become damaged or worn out. A 1 and 2 only B 3 and 4 only C 1, 3 and 4 only D 2, 3, and 4 only 8 Some of the events that occur when an influenza virus enters a host cell are listed. 1 The virus becomes enclosed in an endosome. 2 Haemagglutinin undergoes a change in conformational rearrangement that change its binding specificity. 3 A low pH is produced in the endosome. 4 Haemagglutinin in the viral envelope binds to sialic acid. 5 The viral envelope fuses with the endosomal membrane. Which sequence shows these events in the correct order? A 2  5  1  3  4 B 4  1  2  3  5 C 4  2  1  3  5 D 2  5  1  4  3

6 @YIJC 9744/01/PE/2023 [Turn over 9 Two new viruses E and Z, which infect eukaryotic cells, have been identified. In one experiment, the nucleic acid from each virus is isolated and analysed over a range of temperatures. The light absorbance of nucleic acids increases when denaturation occurs. The behaviour of the nucleic acid from each virus is shown in the graph. In a second experiment, it is found that treatment with reverse transcriptase inhibitors or with inhibitors of DNA synthesis blocks the ability of virus Z to infect cells. In contrast, reverse transcriptase inhibitors have no effect on the ability of virus E to infect cells but DNA synthesis inhibitors block infection by virus E. Which of the following conclusions can be drawn from the results of both experiments? A Virus E has double-stranded DNA as its genome while virus Z has single-stranded RNA. B Virus E has single-stranded RNA as its genome while virus Z has double-stranded DNA. C Virus E has double-stranded RNA as its genome while virus Z has single-stranded DNA. D Virus E has double-stranded DNA as its genome while virus Z has double-stranded RNA. 10 A student made the following statements about how genetic variation arises in prokaryotes. 1 Prokaryotes may undergo antigenic drift to result in genetically different populations. 2 Prokaryotes may accumulate mutations due to errors in DNA replication. 3 Crossing over and independent assortment of chromosomes could give rise to daughter cells which are different from parent cells. 4 Viruses may facilitate the transfer of bacteria