2024 NYJC Prelim 9744 Biology P2 (A)(2)

9744 / H2 Biology / 02 NANYANG JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME CLASS BIOLOGY 9744/02 Paper 2 Structured Questions 10 September 2024 Candidates answer on the Question Paper. No Additional Materials are required. 2 hours READ THESE INSTRUCTIONS FIRST Write your name and CT 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 no use staples, paper clips, highlighters, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Answer all questions 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 no 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. For Examiner’s Use 1 8 2 11 3 9 4 13 5 8 6 8 7 12 8 10 9 10 10 5 11 6 Total 100 This document consists of 30 printed pages and 0 blank pages. [Turn over

9744 / H2 Biology / 02 2 Answer all the questions in this section. 1 Candida albicans is a yeast-like fungus that lives in human lungs. It is the causati ve agent of one of the opportunistic infections that may develop during AIDS. C. albicans is eukaryotic. Fig. 1.1 shows its structure. Fig. 1.1 (a)(i) Name H to L. H nucleolus ; J Golgi (body / apparatus) ; K cell wall ; R murein / peptidoglycan ignore cellulose or chitin L vacuolar membrane / vacuole ; A tonoplast R cell sap; Any two 1m; [2] (ii) State two ways in which the structure of a prokaryotic cell differs from that shown in Fig. 1.1. no double membrane-bound organelles ; no, nucleus / nuclear membrane / nuclear envelope / nucleolus ; A DNA lies free in the cytoplasm no mitochondrion ; no (large) vacuole ; no, ER / RER / SER ; no Golgi (body / apparatus) ; smaller / 70S / 18nm, ribosomes ; cell wall made of, murein / peptidoglycan / different compounds (from eukaryote) ; circular DNA / plasmid(s) / no linear DNA ; no histones / not complexed with proteins ; A naked DNA / no chromosomes AVP ; e.g. mesosomes ; pili / no 9+2 microtubule pattern

9744 / H2 Biology / 02 3 [2] C. albicans uses a transport protein, TMP1, to absorb sugar molecules from the inside of the mouth. TMP1 is encoded by a gene within the nucleus and is produced when sugars are present in the surroundings. (b) Explain how the structures within the cell shown in Fig. 1.1, are involved with the production of functioning TMP1. nucleus, transcription / described as DNA to complementary RNA code / AW ; nuclear pore, mRNA to, cytoplasm / ribosome / RER ; RER / ribosome, assembly of amino acids / translation / polypeptide or protein synthesis ; RER, transports protein to Golgi (apparatus / body) / modifies protein ; Golgi adds, carbohydrates / sugars, to proteins ; A glycosylation A post translational modification / other e.g.s [max 1]: Golgi, packages protein / makes vesicle(s) ; (Golgi) vesicle fuses with cell (surface) membrane ; mitochondrion, provides / produces / synthesises, ATP in correct context ; [4] [Total: 8]

9744 / H2 Biology / 02 4 2 Table 2.1 contains statements about four molecules. (a) Complete the table by indicating with a tick ( ✓ ) or a cross ( ✘ ) whether the statements apply to haemoglobin, DNA, phospholipids or antibodies. You should put a tick or a cross in each box of the table. Table 2.1 statement haemoglobin DNA phospholipids antibodies contains phosphate ×   × able to replicate ×  × × hydrogen bonds stabilise the molecule   ×  Contains nitrogen     [4] 1m each row; (b) Haemoglobin is a globular protein that shows quaternary structure. It is composed of two types of polypeptide, known as α and β globin. (i) Explain how a globular protein differs from a fibrous protein, such as collagen. Assume answers are about globular proteins Soluble vs insoluble; Hydrophilic amino acids on the exterior and hydrophobic amino acids in the interior vs hydrophobic amino acids on the exterior; Spherical vs chain-like; Compact; Ref tertiary structure; [2]

9744 / H2 Biology / 02 5 Fig. 2.1 shows part of the base sequence of the mRNA that codes for t he first ten amino acids of β globin. Table 2.1 shows some of the codons and the amino acids for which they code. Fig. 2.1 Table 2.2 (ii) Use the information in Table 2.1 to complete the sequence of amino acids at the beginning of β globin using the first three letters of each amino acid. Some of them have been done for you. val his leu thr pro glu glu lys ser ala [2] 2 marks if all correct, 1 mark if one wrong, no marks if two or more wrong (iii) β globin has a tertiary structure that consists of eight helices arranged to give a precise three-dimensional shape. Describe how the precise three-dimensional shape of a polypeptide is maintained. (8 helices) further folds into (sp 3D structure); Hydrogen bond between polar groups; ionic bond between amino and carboxylic acid groups/ acidic/ basic / positive/negative group; Hydrophobic interactions between non-polar side chains; (Idea of) no Disulphide/ covalent bonds present; [max 1] if H2I betw R groups; [3] [Total: 11]

9744 / H2 Biology / 02 6 3 Pepsin is an enzyme that hydrolyses proteins (protease). Some students used pepsin from the stomach of a mammal. The activity of the pepsin was investigated by placing a small quantity of the enzyme with a known concentration of the protein albumen. Fig. 3.1 shows the progress of the enzyme-catalysed reaction that was carried out at 20 °C. Fig. 3.1 (a) Calculate the initial rate of the reaction. any two from: anything within range 0.6 to 0.8 ; µmol dm–3 min–1 / µmol per dm3 per min ; A µmol per dm3 / min or µmol dm–3 / min initial rate of reaction = ......................................................... [2]

9744 / H2 Biology / 02 7 (b) The procedure was repeated to find the effects on the activity of the pepsin using a N-acetyl-statine, a competitive inhibitor at the same temperature, 20 °C. (i) Predict the results that will be obtained using the competitive inhibitor. ignore any explanation any two from: 1 initial rate will be, lower / slower ; 2 takes longer to reach the, plateau / end concentration / 10 μmol dm–3 ; A longer to complete the reaction / maximum concentration 3 idea that the final / end, concentration of product will be, the same / 10 μmol dm– 3 ; I refs to the shape of the graph [2] (ii) Explain how N-acetyl-statine inhibits the enzyme pepsin. N-acetyl-statine has a same / similar shape as protein / albumen; complementary in shape to the active site of pepsin; binds/ attaches / fits into active site of enzyme pepsin; protein cannot bind to active site/ block substrate; @no/few E-S complexes formed amino acid production decreases / stops; [3] (c) Enzyme chymotrypsin is another protease synthesized by mammals. However chymotrypsin and pepsin are structurally different with different amino acid sequences. Explain how two enzymes with different amino acid sequences can catalyse the same reaction. Same sequence of amino acids in active site means same R-groups which can form the same bonds between substrate and active site; Also gives rise to active sites which are of the same shape and are complementary in terms of shape, size, charge and orientation to that of substrate / bond; [2] [Total: 9]

9744 / H2 Biology / 02 8 4 Telomere length has been associated with cell division and cell cycle arrest. Fig 4.1 shows the telomere length over time in various cell types. If telomeres are shortened t o a ‘critical le