2017 RVHS H2 Biology Prelims Paper 2 Questions

River Valley High School Pg 1 of 24 Year 6 H2 Biology 9744 Paper 2 2017 Preliminary Examination II RIVER VALLEY HIGH SCHOOL YEAR 6 PRELIMINARY EXAMINATION II CANDIDATE NAME CENTRE NUMBER S CLASS INDEX NUMBER H2 BIOLOGY 9744/02 Paper 2 Structured Questions 11 Sep 2017 2 hours Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your Centre number, index number and name in the spaces at the top of this page. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, 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 not use appropriate units. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 / 9 2 / 13 3 / 8 4 / 10 5 / 12 6 / 15 7 / 10 8 / 11 9 / 12 Total / 100 _____________________________________________________________________ This document consists of 24 printed pages.

River Valley High School Pg 2 of 24 Year 6 H2 Biology 9744 Paper 2 2017 Preliminary Examination II Answer all questions. 1 Fig. 1.1 shows an electron micrograph of part of a plant cell. Fig. 1.1 (a) Identify region A and state its function. [2] (b) Describe how the structure of the membrane at B allows it to perform its function. [3]

River Valley High School Pg 3 of 24 Year 6 H2 Biology 9744 Paper 2 2017 Preliminary Examination II Cyanobacteria are prokaryotic cells that are capable of carrying out photosynthesis. The structure of a cyanobacteria is shown in Fig. 1.2. Fig. 1.2 (c) With reference to Fig. 1.1 and Fig. 1.2, compare the visible structures of cyanobacteria with that of C. [2] Cyanobacteria are considered to be the ancestors of structure C. They continued to function after being engulfed by primitive euk aryotic cells and evolved over time. This theory is known as the endosymbiont hypothesis. (d) State two features of structure C that provide support for this hypothesis. [2] [Total: 9]

River Valley High School Pg 4 of 24 Year 6 H2 Biology 9744 Paper 2 2017 Preliminary Examination II 2 Fig. 2.1 shows the structure of a G-protein coupled receptor (GPCR). Fig. 2.1 (a) Describe how the structure of GPCR is adapted to its function. [3] One of the cellular events resulting from glucagon binding to GPCR, shown in Fig. 2.1, is the activation of glycogen phosphorylase which breaks down glycogen to glucose. (b) (i) Describe how binding of glucagon leads to activation of glycogen phosphorylase. [3]

River Valley High School Pg 5 of 24 Year 6 H2 Biology 9744 Paper 2 2017 Preliminary Examination II (ii) Explain why liver cells store glucose in the form of glycogen. [3] The binding of glucagon to GPCR leads to an increase in blood glucose level partly due to the action of glucose transporters. Glucose transporters transport glucose via facilitated diffusion. (c) (i) Explain what is meant by facilitated diffusion. [2] (ii) Explain why glucose transporters are necessary to facilitate this process. [2] [Total: 13]

River Valley High School Pg 6 of 24 Year 6 H2 Biology 9744 Paper 2 2017 Preliminary Examination II 3 Fig. 3.1 shows DNA replication in an Escherichia coli (A) and in a mammalian cell ( B). Diagrams are not shown to scale. A B Fig. 3.1 (a) State one way in which the DNA replication in these two organisms differs and explain the advantage of this to the mammalian cell. [2] (b) Explain why DNA replication is said to be semi-conservative. [2]

River Valley High School Pg 7 of 24 Year 6 H2 Biology 9744 Paper 2 2017 Preliminary Examination II End replication problem is a fundamental problem associated with replicating DNA in eukaryotes. Some cells contain telomerase, which is responsible for extending the ends of DNA in eukaryotes. Fig. 3.2 shows the action of a telomerase enzyme. Fig. 3.2 (c) Explain how the end-replication problem arises. [2] (d) With reference to Fig. 3.2, state two differences between transcription and the process of lengthening of DNA ends. [2] [Total: 8]

River Valley High School Pg 8 of 24 Year 6 H2 Biology 9744 Paper 2 2017 Preliminary Examination II 4 Huntington’s disease is a rare neurodegenerative disorder targeting the central nervous system. Transcriptional dysregulation is one of the commonly observed molecular abnormalities affected in this disease. Recent evidence suggests the involvement of a mutant Huntingtin protein in the processes regulating condensation of DNA, leading to activation of DNA damage response and death of nerve cells. DNA in various levels of condensation can be observed in the nerve cell nucleus. Fig. 4.1 shows one of the levels of condensation of chromatin. Fig. 4.1 (a) It is postulated that mutant Huntingtin protein facilitates packing of DNA into structure shown in Fig. 4.1. Describe how the DNA double helix is condensed into this structure. [2] (b) The chromosomal condensation in (a) is the main reason for the commonly observed transcriptional dysregulation in Huntington’s disease. Explain how transcription is affected. [3]

River Valley High School Pg 9 of 24 Year 6 H2 Biology 9744 Paper 2 2017 Preliminary Examination II It is observed that nerve cells could remove Huntingtin proteins via ubiquitination of specific amino acids. However, the mechanism that triggers ubiquitination is unclear. In a study to determine the mechanism for degradation of Huntingtin proteins, selected amino acids were investigated and the results are shown in Table 4.1. Table 4.1 13th amino acid: serine 16th amino acid: serine 6th amino acid: lysine 9th amino acid: lysine 15th amino acid: lysine Fate of Huntingtin protein Trial 1 de- phosphorylated de- phosphorylated ubiquitin not attached ubiquitin not attached ubiquitin not attached remains active Trial 2 phosphorylated de- phosphorylated ubiquitin not attached ubiquitin not attached ubiquitin not attached remains active Trial 3 de- phosphorylated phosphorylated ubiquitin not attached ubiquitin not attached ubiquitin not attached remains active Trial 4 phosphorylated phosphorylated ubiquitin attached ubiquitin attached ubiquitin attached degraded (c) With reference to Table 4.1, (i) state the level of control for Huntingtin gene expression. [1] (ii) describe the events at the selected amino acids that triggers the degradation of Huntingtin proteins. [2] (iii) describe how ubiquitination results in the removal of mutant Huntingtin protein. [2] [Total: 10]

River Valley High School Pg 10 of 24 Year 6 H2 Biology 9744 Paper 2 2017 Preliminary Examination II 5 Fig. 5.1 shows the structure of a T4 virus. Fig. 5.1 (a) Identify structure Y. [1] The T4 virus cannot reproduce by itself and relies upon a host cell for reproduction. (b) State specifically why T4 viruses rely on host cells for their reproduction. [2]