2015 HCI H2 Biology Prelims Paper 2 Answers

HWA CHONG INSTITUTION / 2015 JC2 H2 BIOLOGY / PRELIMINARY EXAMINATION / MARK SCHEME 1 HWA CHONG INSTITUTION 2015 JC2 H2 BIOLOGY PRELIMINARY EXAMINATION PAPER 2 MARK SCHEME STRUCTURED QUESTIONS Question 1 (a) Describe the role of haem group in haemoglobin. [2] 1. Haem group in haemoglobin consists of an iron ion held in a porphyrin ring structure 2. Ref to iron ion bind with oxygen (b) Discuss the advantages of having four subunits in haemoglobin. [3] 1. Binding of four oxygen molecules per haemoglobin results in increased oxygen carrying capacity 2. Ref to cooperative binding 3. Change in 3D conformation in one subunit results in changes in 3D conformation of the other subunits (c) Explain how the differences in haemoglobin structure of the Greylag and Andean geese contribute to their different oxygen affinities. [4] 1. Ref to different amino acid sequence in the for Greylag goose and Andean goose haemoglobin 2. Different R-groups interactions results in the different specific 3D conformation 3. Haem groups are more exposed to bind to oxygen in Andean goose 4. Ref to different subunit interactions and different extend of cooperativity [Total: 9] Question 2 (a) Distinguish between the helical structures shown in regions X and Y. [2] 1. The helix in region X is single-stranded but the helix in region Y is double-stranded 2. Hydrogen bonds are formed between amine (-NH) and carbonyl (-CO) groups along the polypeptide backbone in region X but hydrogen bonds are formed between complementary bases on the two DNA strands in region Y 3. Helix in region X makes one complete turn every 3.6 amino acids but helix in region Y makes one complete turn every 10 base-pairs (b) Describe the roles of DNA polymerase in the S phase of the cell cycle. [2] 1. DNA polymerase holds the DNA template and deoxyribonucleoside triphosphates / deoxyribonucleotides close together in the correct orientation 2. DNA polymerase catalyses the formation of phosphodiester bond between 5’ phosphate group of an incoming dNTP and the 3’ hydroxyl group of the growing daughter strand 3. DNA polymerase catalyses the formation of a daughter DNA strand that is complementary to the parental DNA strand

HWA CHONG INSTITUTION / 2015 JC2 H2 BIOLOGY / PRELIMINARY EXAMINATION / MARK SCHEME 2 4. DNA polymerase removes the RNA primer and replaces it with dNTPs. 5. DNA polymerase proofreads and removes mismatched bases / nucleotides on the newly-synthesised daughter DNA strands (c) ddCTP was added to a DNA replication reaction in large excess over the concentration of deoxycytidine triphosphates (dCTP). Explain how the addition of ddCTP would affect DNA replication. [4] 1. Daughter strands would be synthesised until the first guanine base (G) encountered in parental / template strand 2. ddCTP is incorporated into daughter strands instead of dCTP and extension of daughter strand terminated / stops 3. Since ddCTP has a similar shape / 3D conformation to dCTP and ddCTP is added to DNA replication reaction in large excess over dCTP 4. ddCTP outcompetes dCTP for the active site of DNA polymerase and forms phosphodiester bond with 3’ hydroxyl group of daughter strand 5. Since ddCTP lacks 3’ hydroxyl group on the pentose, the incorporated ddC nucleotide cannot form phosphodiester bond with incoming dNTPs (d) ddCTP is used in DNA sequencing reactions to determine the DNA base sequence. In such reactions, ddCTP is added at 10% the concentration of the dCTP. Suggest how ddCTP facilitates the determination of DNA base sequence. [2] 1. 1 in 10 chance of ddCTP being incorporated whenever a G is encountered on the template strand 2. A population of DNA fragments of different sizes will be synthesised 3. From the lengths of DNA fragments, one can deduce the position of the G nucleotides on the template strand [Total: 10] Question 3 (a) With reference to Fig. 3.1, (i) suggest why structure C, the first viral enzyme involved in the reproductive cycle of HTLV- 1, is crucial in its classifica tion as a retrovirus. [1] Structure C which is enzyme reverse transcriptase synthesizes a DNA intermediate (ii) describe the roles of structures A and B in the reproductive cycle of HTLV-1 in CD4 + T c e l l s . [ 2 ] 1. Structure A binds to CD4 + receptor to facilitate fusion of membranes for the release of viral contents into host T cell 2. Structure B serves as a template to form viral DNA for integration into host T cell genome (b) Suggest how HTLV-1 infection could lead to the onset of T-cell leukemia. [1] Idea of insertional mutagenesis

HWA CHONG INSTITUTION / 2015 JC2 H2 BIOLOGY / PRELIMINARY EXAMINATION / MARK SCHEME 3 (c) With reference to Fig. 3.2 and 3.3, (i) identify the time period(s) which correspond(s) to this late stage of HIV reproductive cycle shown in Fig. 3.3. [1] X and Z (ii) describe what happens during this stage of HIV reproductive cycle. [3] 1. Newly assembled immature HIV are budding off from the host T cell, acquiring host cell membrane as viral envelope 2. This is evident from the sharp increase of HIV RNA copies per ml plasma from Week 0 to 6 and Year 9 to 11 (iii) suggest how infection by HIV could cause diseases in time period Z. [2] Failure of HIV patients’ immune system as evident from the mass destruction of CD4 + T cells by Year 11 could potentially leads to life-threatening infections [Total: 10] Question 4 (a) (i) State whether the AFP gene is found in the euchromatic or heterochromatic region in the nucleus of a foetal cell. [1] Euchromatin / euchromatic region (ii) Explain your answer in (a)(i). [ 3 ] 1. Euchromatin is highly de-condensensed 2. allows RNA polymerase to bind 3. resulting in expression of AFP protein (b) (i) Explain how chromatin remodelling can lead to a significant decrease in cellular AFP mRNA upon birth. [2] 1. Deacetylation of histone tails catalysed by histone deacetylases 2. hence, the chromatin structure becomes more compact 3. prevents access of RNA polymerase to AFP gene (ii) Explain how cellular AFP proteins declines in the absence of AFP mRNA. [2] 1. Ubiquitin binds to and marks existing AFP proteins 2. Proteasome binds to ubiquitinated protein and degrades it (c) State one limitation of using AFP as a biomarker in the detection of liver cancer. [1] AFP protein is found in blood serum during liver regeneration and this may result in misdiagnosis for liver cancer [Total: 9]

HWA CHONG INSTITUTION / 2015 JC2 H2 BIOLOGY / PRELIMINARY EXAMINATION / MARK SCHEME 4 Question 5 (a) Describe how structures P differ from structures Q. [2] 1. structures Pare genetically identical while structures Q are not genetically identical 2. structures P are derived from a single parent where structures Q are derived from both parents / ref. to maternal and paternal (b) (i) Draw a genetic diagram to explain the results of the test cross for plant B. [4] F1 phenotype normal leaves, few spines on fruit heart-shaped leaves, many spines on fruit F1 genotype H f h F x h f h f Gametes H f H F h f h F h f male gametes H f h F H F h f female gametes h f H f h f h F h f H F h f h f h f phenotypes normal leaves, many spines heart- shaped leaves, few spines normal leaves, few spines heart- shaped leaves, many spines ii) Account for the different test cros