2016 HCI H1 Biology Prelims P1 Questions

HWA CHONG INSTITUTION JC2 Preliminary Examination Higher 1 CANDIDATE NAME CT GROUP 15S____ CENTRE NUMBER INDEX NUMBER BIOLOGY Paper 1 Multiple Choice Additional Materials: Optical Mark Sheet 8875/01 22 September 2016 1 hour INSTRUCTIONS TO CANDIDATES 1. Write your name, CT group, Centre number and index number in the spaces provided at the top of this cover page. 2. Fill in your particulars on the Optical Mark Sheet. Write your NRIC number and shade accordingly. 3. There are thirty 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 Optical Mark Sheet. 4. At the end of the paper, you are to submit only the Optical Mark Sheet. INFORMATION FOR CANDIDATES 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 23 printed pages and 1 blank page.

2 © Hwa Chong Institution 2016 8875 H1 Biology / JC2 Preliminary Exams / Paper 1 1 The electronmicrograph shows part of an animal cell with structures W to Z labelled. Which statements correctly describe the activities of structures W to Z? 1 Structure X is involved in the synthesis and processing of membrane proteins. 2 Translation occurs in structures W and X. 3 DNA replication occurs only in structure Y. 4 Condensation of chromosomes occur in structure Z.   A 1 B 1 and 2 C 3 and 4 D 1, 2 and 4 W X Z Y

3 © Hwa Chong Institution 2016 8875 H1 Biology / JC2 Preliminary Exams / Paper 1 2 Which row shows the correct match between the descriptions of different types of biological molecules and where they are found? 1 a linear polymer of 1,4 linked -glucose molecules 2 an amphipathic, phosphate containing molecule 3 a highly branched polymer of 1,4 and 1,6 linked -glucose molecules 1 2 3 A eukaryote and prokaryote cell walls cell surface membranes of both eukaryotes and prokaryotes forming storage granules in the cells of prokaryotes B eukaryote cell walls cell surface membranes of both eukaryotes and prokaryotes forming storage granules in the cells of some eukaryotes C eukaryote cell walls cell surface membranes of both eukaryotes and prokaryotes forming starch grains in the cells of all eukaryotes D forming storage granules in the cells of eukaryotes prokaryote cell walls eukaryote cell walls 3 The diagrams show different types of bond found in biological molecules. Which combination of bonds could not be found in a protein with a tertiary structure? A 1, 2, 3 and 4 B 1, 2 and 4 only C 3 and 5 only D 5 only 1 2 3 4 5

4 © Hwa Chong Institution 2016 8875 H1 Biology / JC2 Preliminary Exams / Paper 1 4 Which levels of protein structure are always involved when competitive and non-competitive inhibitors bind to enzymes? competitive non-competitive A primary, secondary and tertiary secondary B quaternary and tertiary quaternary and tertiary C secondary primary and tertiary D tertiary tertiary A active site : complementary to ATP and hexose phosphate allosteric site : absent B active site : complementary to hexose phosphate allosteric site : complementary to glucose C active site : complementary to ATP and hexose phosphate allosteric site : complementary to ATP D active site : complementary to hexose bisphosphate allosteric site : absent 5 The enzyme phosphofructokinase (PFK) is involved in phosphorylation of hexose phosphate during glycolysis as shown in the diagram. It is involved in controlling the rate of glycolysis and thus respiration, by end-product inhibition. Which statement correctly describes the binding site(s) of PFK? hexose phosphate hexose bisphosphate ATP phosphofructokinase ADP

5 © Hwa Chong Institution 2016 8875 H1 Biology / JC2 Preliminary Exams / Paper 1 7 What is the correct description of centrioles, nuclear envelope and spindle during mitosis in animal cells? phase centrioles nuclear envelope spindle A anaphase replicate absent present B metaphase present reforms present C prophase move apart breaks up forms D telophase replicate breaks up breaks up 6 Which diagrams show the correct relationships? A B C D

6 © Hwa Chong Institution 2016 8875 H1 Biology / JC2 Preliminary Exams / Paper 1 8 A space probe brought back some living material from a distant planet. Proteins very similar to those found on Earth were present. DNA of this living material, which exists as double helices, contains six different nucleic acid bases, identified as H, I, J, K, L and M. Among these, J, L and M were found to be pyrimidine bases. Studies of the DNA from this living material gave the following results: ratio of bases in double-stranded DNA numerical value H / J 1.00 (H + I) / (J + M) 1.02 (H + K) / (J + M) 1.14 (H + K) / (J + L) 1.01 Assuming that the base-pairing follows Chargaff’s rule, what can be concluded from these data? 1 H base pairs with J. 2 H, I, K are likely to be purine bases. 3 It is probable that I base pairs with M, and K base pairs with L. A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2 and 3

7 © Hwa Chong Institution 2016 8875 H1 Biology / JC2 Preliminary Exams / Paper 1 9 In an investigation to study the mode of DNA replication, Escherichia coli (E. coli) cells were grown in a nutrient medium containing heavy isotope of nitrogen ( 15N) for an extended period of time until all the DNA was labelled. These E. coli cells were then transferred to a nutrient medium containing only light isotope of nitrogen ( 14N) and were allowed to multiply over three generations. The DNA of the E. coli cells was then harvested at nine different time intervals. Subsequently, density gradient centrifugation of these E. coli DNA using caesium chloride was performed. The diagram shows the results obtained. Which statements are consistent with the results observed? 1 There was no evidence of semi-conservative DNA replication. 2 In the 1 st generation, only hybrid 14N/15N DNA was produced. 3 In the 3 rd generation, 75% hybrid 14N/15N DNA and 25% light 14N/14N DNA were produced. 4 In the subsequent 4 th generation, only light 14N/14N DNA would be produced. A 2 B 1 and 4 C 2 and 3 D 1, 3 and 4 time elapsed since switching to 14N: 15N 1st generation 2nd generation 3rd generation 15N 14N 14N

8 © Hwa Chong Institution 2016 8875 H1 Biology / JC2 Preliminary Exams / Paper 1 10 A polypeptide has the amino acid sequence glycine-arginine-lysine-serine. The table gives possible tRNA anticodons for each amino acid. amino acid tRNA anticodons arginine UCC GCG glycine CCA CCU lysine UUC UUU serine AGG UCG Which DNA sequences would code for this polypeptide? A CCACGCAAGAGC B CCTTCCTTCTCG C GGAAGGAAAAGC D GGTTGGTTGTGC 11 A segment of the DNA sequences of normal and mutant β-globin genes are shown. A base substitution from A to G results in the disease β-thalassemia. What is the most plausible explanation as to why the indicated mutation results in β-thalassemia? A A restriction site is generated, so that the gene is cut into two. B A new splice site