2021 EJC Promo JC1 P1 Bio (A)

©EJC 2021 9744/01/J1H2PROMO/2021 [Turn over EUNOIA JUNIOR COLLEGE JC1 Promotional Examinations 2021 General Certificate of Education Advanced Level Higher 2 H2 Biology Paper 1 Multiple Choice 9744/01 05 October 2021 1 hour Additional Materials: Multiple Choice Answer Sheet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use paper clips, glue or correction tape/fluid. Write your name, civics group and registration number on the Answer Sheet in the spaces provided. There are thirty Multiple Choice Questions in 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 18 printed pages. ANSWERS

2 ©EJC 2021 9744/01/J1H2PROMO/2021 1 Raffinose is a trisaccharide which can be degraded by enzymes. The results of two different enzymatic incubations are shown here: enzyme used products sucrase melibiose and fructose galactosidase galactose and sucrose Which statements are consistent with the results shown above? 1 Raffinose is composed of two different monosaccharides. (False – from the table, besides fructose and galactose, we can derive that the other monosaccharide is glucose, as sucrose = fructose + glucose) 2 Melibiose is a dissaccharide. (True – from digestion by sucrose; melibiose is made up of glucose and galactose) 3 Acid hydrolysis of raffinose would yield glucose. (True – the high heat and acid would hydrolyse the glycosidic bond joining glucose and fructose in sucrose , yielding glucose as stated) 4 The products of raffinose digestion by sucrase and galactosidase will each yield an orange - red precipitate when heated with Benedict's reagent. (True – since both fructose and glucose will be released and these reducing sugars will give an orange-red ppt with Benedict’s test). A 1 and 3 only B 2 and 4 only C 2, 3 and 4 only D 1, 2, 3 and 4 2 In triglyceride molecules, where are carbon-carbon double bonds located? A between fatty acids and glycerol (the ester bond has a carbon-oxygen (C=O) double bond) B within fatty acids and within glycerol (no carbon-carbon double bonds within glycerol) C within fatty acids only (only the fatty acid chains have the carbon-carbon (C=C) double bonds) D within glycerol only (no carbon-carbon double bonds)

3 ©EJC 2021 9744/01/J1H2PROMO/2021 [Turn over 3 Which correctly matches the functional and structural features of cellulose, collagen, glycogen and triglyceride? key: ✓= true ✗= false function structure fibrous molecule held together by hydrogen bonds branched chains A cellulose triglyceride support energy source   ✓    B collagen cellulose strengthening support     (only inter- chain)   (x) C collagen glycogen strengthening storage      (x)  D glycogen triglyceride storage energy source    (x)  (x)   4 Which row about the structure of proteins is correct? primary structure tertiary structure quaternary structure A is the number (x) of amino acids present in a protein is the result of cross bonding between all (x) the amino acids in the primary structure is the polypeptides that link together to form a protein (maybe - unclear) B is the order of amino acids present in a protein encoded by DNA (maybe) is the shape formed by folding of a polypeptide and held together by hydrogen bonds (not just hydrogen bonds) contains two types of polypeptide that interact forming the shape of a protein (vague about ‘two types of polypeptide’) C is the result of translation of an mRNA molecule by a ribosome into a chain of amino acids (does not describe the structure!) is the result of ionic and hydrogen bonds, disulfide bridges and hydrophobic interactions between amino acids (correct but missing reference to side chains of amino acids) is formed by four polypeptides and an additional reactive group attached to the protein (not necessary must be four polypeptides and reactive group vague) D is the sequence of amino acids in a protein coded by an mRNA molecule is formed as a result of interaction of the side chains of amino acids in the primary structure is formed by the linking together of more than one polypeptide to form a protein

4 ©EJC 2021 9744/01/J1H2PROMO/2021 5 During the development of HIV, the polyprotein is hydrolysed by a HIV protease enzyme , producing several smaller peptides. This viral enzyme is the target of new anti-AIDS drugs. Which feature is essential for the success of these drugs? (Specifically targeting HIV protease) A A complex structure that inhibits many types of viral and non-viral enzymes. (Not good as the drug is not specific and may target the host’s enzymes too!) B A molecule containing a heavy metal atom that is a non -competitive inhibitor of enzymes. (Speculative that heavy metal atom is a non-competitive inhibitor of all enzymes! Even if true, still not specific and may target host’s enzymes!) C A protein that can act as a competitive inhibitor of protease enzymes. (Again not specific and could target host’s enzymes!) D A specific structure that inhibits only HIV protease. 6 The graphs show how the concentration of different components of an enzyme-catalysed reaction (e.g. substrate, active sites) changes with time. (Take note!) Which graph represents enzymes with empty active sites? B A – Concentration of empty active sites should not increase with time continuously. B – Concentration of empty active sites decreases initially as substrates bind to the empty active sites (and are converted to products). However, as more substrates are converted to products, concentration of substrate decreases over time, so more active sites become empty again. Hence, concentration of empty active sites increases. C – (see explanation for option B) D – Concentration of empty active sites should not increase during the initial course of the reaction.

5 ©EJC 2021 9744/01/J1H2PROMO/2021 [Turn over 7 The diagram shows a stage micrometer scale viewed through an eyepiece containing a graticule. Each small division of the stage micrometer scale is 0.1 mm. (100 μm) The stage micrometer scale is replaced by a slide of a plant cell . (Note position of eyepiece graticule as compared to previous diagram, its position has not changed, so the scale above it in the previous diagram must have been stage micrometer.) What is the actual length of the nucleus in the plant cell? (10 graticule units or 25 μm – option B) A 8 µm B 25 µm C 200 µm D 0.8 mm Stage Micrometer Eyepiece Graticule 100 μm Distance of one small division of eyepiece graticule = 100 μm / 40 divisions = 2.5 μm

6 ©EJC 2021 9744/01/J1H2PROMO/2021 8 The electron micrograph shows a cell structure in a eukaryotic cell. Which statement(s) about this cell structure is/are correct? 1 ATP is synthesised in this cell structure. (False) 2 The cell structure is made of protein molecules. (True – centrioles contain microtubules which are made of tubulin subunits and tubulin is a protein) 3 The cell structure replicates during interphase of the cell cycle. (True – duplication of centrioles occurs in inter