2015 RI H2 Biology Prelims Paper 2 Answers

© RI 2015 Preliminary Examination 9648/02 [Turn over RAFFLES INSTITUTION 2015 Year 6 Preliminary Examination Higher 2 BIOLOGY 9648/02 Paper 2 Core paper 16th SEPTEMBER 2015 2 hours Additional materials: Answer Sheet READ THESE INSTRUCTIONS FIRST Write your index number, CT group & name on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Section A Answer all questions. Section B Answer either ONE question. At the end of the examination, hand in your essay SEPARATELY. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 24 printed pages. CIVICS GROUP CANDIDATE NAME INDEX NUMBER 5 S 0 3 1 For Examiner’s Use Section A 1 / 9 2 /10 3 /12 4 /10 5 /10 6 / 9 7 / 9 8 /11 Section B 9 or 10 /20 Total /100

2 © RI 2015 Preliminary Examination 9648/02 For Examiner’s Use Section A Answer all the questions in this section. 1 Meristematic root tissue from a barley seedling was prepared and its chromosomes are observed under a microscope. Fig. 1.1 shows a cell from the root tissue at the metaphase stage of mitosis. Fig. 1.1 Fig. 1.2 shows the changes in amount of DNA at different stages of the barley life cycle. Fig. 1.2 (a) Mark out with an arrow clearly on Fig 1.2 which part of the graph corresponds to the Raffles Institution Internal Examination

3 © RI 2015 Preliminary Examination 9648/02 [Turn over For Examiner’s Use stage shown in Fig. 1.1. [1] Accept all part of line except the corners (b) From stages A to D in Fig. 1.2, state all stages (i) that has/have the same number of chromosomes as shown in Fig. 1.1; [1] A and B; (ii) has/have a different number of chromosomes as shown in Fig. 1.1. [1] C and D; (c) Explain how stages in Y lead to variation. [4] 1. Crossing over * between non-sister chromatids * of homologous chromosomes/bivalents/homologous pair takes place during prophase I*; Or where equivalent portions of non-sister chromatids* of homologous chromosomes break and rejoin during prophase I* 2. gives rise to new combination of alleles* / mixing of alleles from both parental chromosomes which creates genetic variation in gametes; A: new linkage groups in place of new combination of alleles 3. Independent assortment* of homologous chromosomes/bivalents/homologous pair at metaphase plate during metaphase I * and their subsequent separation during anaphase I OR Homologous chromosomes are arranged independently of other homologous pairs at metaphase plate during metaphase I * and their subsequent separation during anaphase I 4. results in 2n possible (types of ) gametes where n is the number of homologous pairs OR Gametes with different combinations of parental (maternal and paternal) chromosomes (d) Explain the significance of the event occurring at X. [2] 1. X refers to fertilization*; (point 1 is essential) 2. random fusion of gametes* results in greater variation/varied offspring with different genotypes and phenotypes; 3. Restoration of the diploid number of chromosomes; [Total : 9]

4 © RI 2015 Preliminary Examination 9648/02 For Examiner’s Use 2 Fig. 2.1 shows DNA replication. (a) (i) Use an arrow to show the direction of replication of the leading strand in the box provided in Fig. 2.1. [1] (ii) What do 5’ and 3’ on the DNA molecule represent? [2] 1. 5’ represents the end (of strand of nucleotide) with carbon 5 on deoxyribose/pentose sugar having free phosphate group 2. 3’ represents end (of strand of nucleotide) with carbon 3 on deoxyribose/pentose sugar having free hydroxyl group (iii) Name the following molecules. [1] V: DNA polymerase W: Primase R: RNA primase Note: RNA primase forms DNA primer DNA primase forms RNA primer (iv) Describe the role of two named enzymes that are required for DNA replication. [2]. (role is needed, not description of how) 1. Helicase Unzips the DNA double helix/ separates the two DNA strands by breaking hydrogen bonds between the complementary base pairs. 2. Topoisomerase Breaking and rejoining DNA strands to relieve overwinding strain ahead of (a) (i) V W Fig. 2.1

5 © RI 2015 Preliminary Examination 9648/02 [Turn over For Examiner’s Use replication fork 3. DNA Polymerase Addition of free deoxyribonucleotides/elongation of the new DNA strand by formation of phosphodiester bond between nucleotides. 4. DNA ligase form phosphodiester bonds to join the Okazaki fragments sealing the nicks. 5. Primase to synthesise the RNA primers to provide free 3’OH for DNA Polymerase to elongate the new DNA strand (b) Fig. 2.2 shows transcription. Describe how the structure of molecule Z is adapted to its role in transcription. [2] 1. molecule Z = RNA polymerase, 2. which has a specific active site* which is complementary in shape/conformation* and charge to substrate such as DNA template and ribonucleotides; / DNA binding site* that is complementary to nucleotide sequences at the promoter. 3. catalytic amino acids capable of catalyzing the formation of phosphodiester bond* elongating the RNA (c) Describe how a silent mutation can result in no change in protein structure. [2] 1. Single base substitution mutation (involves a replacement of a DNA nucleotide with a different nitrogenous base) 2. Change in codon resulting in same amino acid incorporated in polypeptide chain due to degeneracy of the genetic code (R: wobble) OR 3. Any mutation (e.g. insertion, deletion, substitution) in introns 4. which will be spliced out in post-transcriptional modification (1 and 2 OR 3 and 4) 5. Same primary structure and hence no change in secondary and tertiary structure [Total : 10] Z Fig. 2.2

6 © RI 2015 Preliminary Examination 9648/02 For Examiner’s Use 3 The Jacob-Monod hypothesis describes lactose metabolism in the bacterium Escherichia coli. An investigation of this reaction in E. coli at 25 °C was carried out as described below. • 100 cm3 of gel beads coated with E. coli were placed into each of seven identical funnels fitted with outlet taps. • 100 cm3 of solution containing 2 grams of lactose was poured into each funnel at 0 min. • At each time shown in the table, the solution from the respective funnel was released and collected. • The mass of lactose in each solution was measured. The results are shown in the table below. With reference to Table 3.1 Funnel Time (min) Mass of lactose collected in the solution (g) 1 0 2.00 2 10 2.00 3 20 1.48 4 30 0.92 5 40 0.40 6 50 0.12 7 60 0.04 (a) (i) calculate the average mass of lactose broken down per minute in funnel 5. [1] (2.00 – 0.40)/40 = 0.04 g per minute (ii) explain the results from funnels 3 to 7. [4] 1. As from 20 to 60 min/ time passes/ over 40 min, lactose digested increased from 0.52 g to 1.96 g A: 1a: from 20 to 60 min, lactose collected decreas