ASRJC 2023 JC2 H2 Biology Prelims Paper 2 (Ans)

ASRJC BIOLOGY DEPT 9744/2023/J2PRELIM/P2 [Turn over H2 ANDERSON SERANGOON JUNIOR COLLEGE HIGHER 2 ANSWERS 2023 JC2 PRELIMINARY EXAMINATIONS CANDIDATE NAME CLASS INDEX NUMBER BIOLOGY 9744/02 PAPER 2 SHORT STRUCTURED QUESTIONS Candidates answer on the Question Paper. No Additional Materials are required. 13 SEPTEMBER 2023 WEDNESDAY 2 HOURS READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graph Do not use paper clips, highlighters, glue or correction fluid. Answer all questions. 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. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 25 printed pages and 7 blank page For Examiner’s Use 1 2 3 4 5 6 7 8 9 10 11 Total /100

ASRJC BIOLOGY DEPT 9744/2023/J2PRELIM/P2 Answer all the questions. 1 Archaea are single-celled microorganisms which inhabit extreme environments, such as hot springs and volcanic vents at temperatures over 100C. Fig. 1.1 shows the molecular structure of a phospholipid found in the cell membrane of an archaea and that of a eukaryote. Fig. 1.1 (a) (i) With reference to Fig. 1.1, state two structural differences between the two phospholipids. [2] [Any one] 1. The fatty acid tails/ hydrocarbon chains of archaeal phospholipid are both saturated/ only contain C-C single bonds , but the fatty acid tails/ chains of eukaryote phospholipid consist of a saturated and an unsaturated chain/ contain a C=C double bond. 2. The fatty acid tails/ hydrocarbon chains of archaeal phospholipid are longer/ contained more C and H , but the fatty acid tails/ chains of eukaryote phospholipid are shorter/ contain fewer C and H. 3. Side chains are present in the fatty acid tails/ hydrocarbon chains of archaeal phospholipid, but these are absent in the fatty acid tails/ chains of eukaryote phospholipid.

3 ASRJC BIOLOGY DEPT 9744//2023/PRELIM/P2 [Turn over (ii) Explain why phospholipids are unable to provide structural support. [3]  Phospholipids is not a long, linear, macromolecule joined by strong covalent bonds which will then confer a high tensile strength.  Phospholipids are fluid as it is held by weak hydrophobic interactions between the hydrocarbon tails, and hence can move laterally.  (aka cellulose) Absence of numerous inter-chain hydrogen bonds between hydroxyl groups of adjacent parallel, linear chains, forming rigid cross-links between chains.  many cross-linked ce llulose chains are bundled together to form microfibrils which have high tensile strength. OR  Many tropocollagen molecules are covalently cross-linked with neighbouring tropocollagen molecules running parallel to them  to form a collagen fibril  fibrils then bundle to form collagen fibres. In an experiment to investigate the transport of ions into root cells, some pea plants were grown with their roots in a solution of ions. These ions were absent in the root cells at the beginning of the experiment. The concentration of five ions in the solution and in the cytoplasm of root cells were determined after one hour. The results are shown in Table 1.2. Table 1.2 ion concentration of ions / mmol dm-3 solution cytoplasm of root cells potassium (K+) 1.0 75.0 magnesium (Mg2+) 0.3 3.5 calcium (Ca2+) 1.0 1.0 phosphate (PO43-) 1.0 21.1 sulfate (SO42-) 0.3 19.7 (b) Explain which components of the cell surface membrane of root cells are responsible for the results shown in Table 1.2. [4] 1. There is the presence of various specific transport proteins on the cell surface membrane 2. Uptake of K+, Mg2+, PO43- and SO42- is via active transport 3. with the aid of carrier proteins, against their concentration gradients with energy from ATP hydrolysis (or quote data) 4. Uptake of Ca2+ is via facilitated diffusion

ASRJC BIOLOGY DEPT 9744/2023/J2PRELIM/P2 5. with the aid of channel proteins, down its concentration gradient (or quote data), until dynamic equilibrium is reached 6. Presence of hydrophobic core of phospholipid bilayer prevents simple diffusion of ions across/ hydrophobic core is impermeable to ions 7. hence maintaining differential distribution of ions across the cell surface membrane (or quote data) [Total: 9]

5 ASRJC BIOLOGY DEPT 9744//2023/PRELIM/P2 [Turn over 2 Fig. 2.1 is a photomicrograph of root tip cells at different stages in the cell cycle. A cell in interphase and telophase are labelled. Fig. 2.1 (a) (i) Complete Fig. 2.1 by naming the stage of mitosis shown in each of cells J and K. [1] (ii) State one feature of the cell in interphase, visible in Fig. 2.1, that shows this cell is not in early interphase. [1] 1. Large size / same size as cells in mito sis / same size as cells labelled in stages of mitosis; (b) Describe the events that occur in telophase. [2] 1. Spindle fibres (reject: kinetochore/non-kinetochore microtubules only, must be both) disintegrate (reject: disappears) 2. Chromosomes reach (opposite) poles and uncoil/decondense  form chromatin fibres 3. Nuclear envelope reforms around each set of chromosomes at each pole (reject: reform around two daughter cells) 4. Nucleolus reappears (reject: reforms) (c) Reduction division happens in meiosis. (i) Describe the events that cause reduction division. [2] 1. During metaphase I , homologous chromosomes align along metaphase plate in pairs / one homologue of each homologous pair facing each pole. 2. During anaphase I , spindle fibres (kinetocho re microtubules) shorten, pulling homologous chromosomes apart towards opposite poles of the cell. 3. During anaphase II , centromere divides , sister chromatids separate , each becomes a full-fledged daughter chromosome  spindle fibres / kinetochore microtubules shorten, pulling chromosomes apart towards opposite poles of the cell. (Reject phrasing: spindle fibres shorten, separating sister chromatids) metaphase (common wrong answer: anaphase) prophase (common wrong answer: telophase)

ASRJC BIOLOGY DEPT 9744/2023/J2PRELIM/P2 4. (Compulsory) Telophase I or II – nuclear envelope reforms leading to haploid number of chromosomes per nucleus (ii) Explain the need for reduction division during meiosis. [3] any three from: 1. (meiosis / reduction division) produces gametes which are haploid / are n / have half the normal number (of chromosomes); (reject: half amount of genetic material) 2. (two) gametes fuse / fertilisation occurs, to form a diploid zygote (reject: full set of chromosomes); 3. zygote will have maternal and pater