NYJC Prelim Paper 2 MS

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 9744 / H2 Biology / 02

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 causative 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. 1. Prokaryotic cell have no membrane-bound organelles but C.albicans has membrane bound organelle/ give example of membrane bound organelle (nucleus/mitochondria/Golgi apparatus/ER) that is absent in prokaryotes but present in C.albicans ; 2. Prokaryotes have plasmid(s) but C.albicans have no plasmids; 3. Prokaryotes has a peptidoglycan cell wall but C.albicans cell wall is made from other compounds (glucan and chitin) ; 4. Prokaryotes have 70S ribosomes but C.albicans has 80S ribosome ; 5. Prokaryotes have circular DNA but C.albicans have linear DNA [2] 9744 / H2 Biology / 02

3 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. Any 4 1. Nucleus contains gene that is transcribed to pre-mRNA/mRNA sequence/AW ; 2. nuclear pore has protein complexes that facilitates exit of mature mRNA to, cytoplasm 3. ribosome on rER translates mRNA sequence to protein; 4. RER, transports protein to Golgi (apparatus / body) / modifies protein ; 5. Golgi apparatus further modifies protein by adding carbohydrates / sugars, to proteins ; A glycosylation/ any other example of post-translational modification [max 1 of the following] 6. Golgi apparatus packages protein / makes vesicle(s) ; 7. Secretory vesicle fuses with cell surface membrane to embed TMP1 on CSM; 8. 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. Any 2 1. Globular proteins are s oluble but fibrous proteins are insoluble ; 2. Globular proteins have hydrophilic amino acids on the exterior and hydrophobic amino acids in the interior but fibrous protein hydrophobic amino acids on the exterior; 3. Globular proteins have spherical 3D conformation/compact conformation but fibrous proteins usually have linear/chain-like confomation; 4. Globular proteins such as haemoglobin has a primary, secondary, tertiary and quaternary structure but fibrous proteins such as collagen has a primary, secondary and quaternary structure/ haemoglobin has a tertiary structure but collagen has no tertiary structure; [2] 9744 / H2 Biology / 02

5 Fig. 2.1 shows part of the base sequence of the mRNA that codes for the 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. 1. R group interactions occur between the 8 helices which further folds into specific 3D shape; Any 2 2. Hydrogen bond forms between polar groups; 3. ionic bond between positive and negative group ; 4. Hydrophobic interactions between non-polar side chains ; 5. (Idea of) no Disulphide/ covalent bonds present; [3] 9744 / H2 Biology / 02

6 [Total: 11] 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. 1. anything within range 0.6 to 0.8 ; 2. units of µmol dm –3 min –1 / µmol per dm 3 per min ; A µmol per dm 3 / min or µmol dm –3 / min initial rate of reaction = ......................................................... [2] 9744 / H2 Biology / 02

7 9744 / H2 Biology / 02

8 (b) The procedure was repeated to find the effects on the activity of the pepsin using a N-acet