2024 TMJC Prelim H2 Bio Paper 1 (A)

1 CANDIDATE NAME CIVICS GROUP H2 BIOLOGY 9744/01 Paper 1 Multiple Choice Questions 19 September 2024 1 hour Additional material: Multiple Choice Answer Sheet ANSWERS WITH EXPLANATION QUESTION ANSWER QUESTION ANSWER 1 B 16 A 2 C 17 B 3 C 18 C 4 D 19 A 5 A 20 D 6 C 21 D 7 A 22 C 8 D 23 C 9 D 24 B 10 A 25 C 11 B 26 D 12 B 27 D 13 D 28 D 14 C 29 B 15 A 30 C ________________________________________________________________________________ This document consists of 24 printed pages. TAMPINES MERIDIAN JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION

2 Tampines Meridian Junior College 2024 JC2 Preliminary Examination H2 Biology [Cell 1/2, KU-1] 1. Six organelles found in eukaryotic cells are shown. They are not drawn to scale. Which organelles are involved in the synthesis and secretion of a glycoprotein? A. 1, 2, 3 and 4 B. 1, 2, 4 and 6 C. 2, 3 and 5 D. 3, 4, 5 and 6 Explanation • 1 = nucleus, 2 = mitochondria, 3 = centrioles, 4 = RER, 5 = chloroplast, 6 = Golgi apparatus • Nucleus carries gene coding for the protein. It is also where transcription takes place. • RER is where translation takes place. The protein enters the ER lumen where folding and glycosylation occurs. • Golgi further modifies the glycoprotein and packages them into a secretory vesicles. • Mitochondria synthesizes ATP needed for protein synthesis and movement of vesicles along microtubules. [Cell 2/2, KU-1] 2. What is present in all viruses, prokaryotes and eukaryotes? A. thymine B. deoxyribose C. adenine D. phospholipid Explanation • Thymine is present in DNA. Some viruses are RNA viruses (hence no thymine). • Deoxyribose is present in DNA. Some viruses are RNA viruses (hence no DNA). • Some viruses are non-enveloped, hence no phospholipids.

3 Tampines Meridian Junior College 2024 JC2 Preliminary Examination H2 Biology [BioMol 1/2, KU-1] 3. The diagrams show the structures of two amino acids, aspartic acid and lysine. Which groups form the bonds that maintain the primary, secondary and tertiary structures of proteins? primary structure secondary structure tertiary structure A. 2, 6 1, 3, 4, 5 1, 3, 4, 5 B. 1, 3, 4, 5 2, 6 2, 6 C. 1, 3, 4, 5 1, 3, 4, 5 2, 6 D. 2, 6 2, 6 1, 3, 4, 5 Explanation: • 1 and 4 are the amino group, and 3 and 5 are the carboxylic acid group. They form peptide bonds that link amino acids together to form the primary structure (sequence of amino acids). • The C=O group and ─NH groups of the peptide bond regions are needed to form hydrogen bonds in alpha helices and beta pleated sheets. • The R-groups, which are CH2COOH in aspartic acid and CH2CH2CH2CH2NH2 in lysine forms bonds with other R-groups to maintain the globular (tertiary) structure of proteins.

4 Tampines Meridian Junior College 2024 JC2 Preliminary Examination H2 Biology [BioMol 2/2, HI-2] 4. DNA can form a triple helix, but such structures are extremely rare in nature. Base pairing within a normal double-stranded DNA is termed complementary base pairing. Base pairing of the third DNA strand with one of the two strands of the double-stranded DNA is termed Hoogsteen base pairing. The diagrams illustrate • how a third strand can bind to a double-stranded DNA molecule • how bases from the third strand can interact with the double-stranded DNA molecule. Which statements are correct regarding the formation of DNA triple helix? 1. Both major grooves and minor grooves of the double-stranded DNA are accessible to the third DNA strand for binding. 2. Complementary base pairing between adenine and thymine and Hoogsteen base pairing between guanine and cytosine are equally strong. 3. DNA triple helices can exhibit greater thermal stability than DNA double helices. 4. Synthetic oligonucleotides can be designed to bind to the target gene through triple helix formation, leading to gene silencing. A. 1 and 2 B. 1 and 4 C. 2 and 3 only D. 2, 3 and 4

5 Tampines Meridian Junior College 2024 JC2 Preliminary Examination H2 Biology Explanation 1. Only the bases along the major grooves are exposed for interaction with the third DNA strand, as shown in the diagram. In fact, all DNA -binding proteins interact with DNA via the major grooves as well. 2. Both CBP between A and T and HBP between G and C form two hydrogen bonds, hence equally strong. 3. Because there are more hydrogen bonds in triple helix, it takes a higher temperature to denature the three strands into single strands, hence exhibits greater thermal stability. 4. When a third strand is bound to a gene, RNA polymerase will be unable to unwind the original double helix for transcription, terminating transcription prematurely, leading to gene silencing. [Enzymes 1/1, HI-2] 5. CYP3A4 is an important enzyme in the human digestive system where it is needed to break down a range of different toxins. The activity of CYP3A4 has been shown to be reduced by substances called furanocoumarins. Furanocoumarins are found in some fruits and so dangerous concentrations of toxins may develop in the human digestive system when fruits containing furanocoumarins are eaten. From the information provided, what can be concluded about molecules of the enzyme CYP3A4? A. They lower the activation energy of the toxin breakdown reactions. B. They bind specifically through the active site to a substrate found in some fruits. C. They change permanently when acted upon by furanocoumarin molecules. D. They resume normal activity when concentrations of furanocoumarins decrease. Explanation A. The substrates of CYP3A4 are the toxin molecules. Hence, the activation energy of the breaking down of toxin is lowered by the enzyme CYP3A4. B. The toxins (substrates) are not specifically stated to be found in fruits. C. This option suggests that furanocoumarin binds permanently to and change the conformation of CYP3A4 . How furanocoumarin binds and inhibits CYP3A4 is not stated in the question stem. D. This option suggests that furanocoumarin binds reversibly to CYP3A4 . How furanocoumarin binds and inhibits CYP3A4 is not stated in the question stem. [Transport 1/1, KU-1] 6. When a small quantity of phospholipid is added to a test -tube of water and then shaken vigorously, an emulsion is formed by small droplets called liposomes. Which diagram shows the arrangement of phospholipid molecules in a cross -section of a liposome? ANSWER: (C) Explanation Water is present outside the liposome and within the liposome. The hydrophilic phosphate heads must face the water molecules.

6 Tampines Meridian Junior College 2024 JC2 Preliminary Examination H2 Biology [Transport 2/2, KU-1] 7. Which letter in the diagram represents the first step in cell signalling for a protein ligand molecule? ANSWER: (A) Explanation • Protein ligands are too big to enter the cell via a channel (C) or directly through the bilayer (D). They bind to transmembrane protein receptor (A) at the extracellular domain (signal reception), transducing the signal to the inside of the cell. [Transformation of Energy 1/2, HI-2] 8. Cyanobacteria are photosynthetic prokaryotes. A scientist exposed cyanobacteria to light of different colours and intensities and made the following observations: • Most cyanobacteria are blue in colour. • At low light intensities, glucose production in cyanobacteria is low. • When light intensity reaches a certain level the rate of glucose production in