2015 HCI H1 Biology Prelims P1&P2 Answers

HWA CHONG INSTITUTION / 2015 JC2 H1 BIOLOGY / PRELIMINARY EXAMINATION / MARK SCHEME 1 HWA CHONG INSTITUTION 2015 JC2 H1 BIOLOGY PRELIMINARY EXAMINATION PAPER 1 & 2 MARK SCHEME 1 D 11 D 21 D 2 D 12 D 22 C 3 B 13 A 23 A 4 D 14 B 24 A 5 B 15 C 25 A 6 A 16 A 26 D 7 C 17 B 27 D 8 D 18 B 28 C 9 C 19 C 29 D 10 A 20 B 30 A 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) (i) Account for the trends shown by the distribution of the two types of bacteria after six m o n t h s . [ 3 ] 1. Aerobic bacteria decrease with depthfrom mean number of bacteria per gram of stored soil of 12.5 x 107 to 0.9 x 107

HWA CHONG INSTITUTION / 2015 JC2 H1 BIOLOGY / PRELIMINARY EXAMINATION / MARK SCHEME 2 2. Anaerobic bacteria increase with depth from mean number of bacteria per gram of stored soil of 0.6 x 107 to 8.8 x 107 3. Oxygen content of soil decreases with depth as aerobic bacteria requires oxygen for respiration 4. Less oxygen available as the final electron acceptor in ETC, decrease in ATP synthesis (ii) Describe how aerobic bacteria are structurally adapted for cellular respiration. [2] 1. Presence of cytoplasmic membranes in aerobic bacteria. Hence, increase surface area 2. For more embedding of electron transport chain and ATP synthase to allow electron transfer / to drive oxidative phosphorylation of ADP to ATP (b) (i) State with evidence from Fig. 6.1 which depth, A or B, were samples taken from a greater depth. [2] 1. Depth B 2. Bacteria in samples taken from depth B shows a lowered mean dehydrogenase activity with 1.5 to 2.4 au compared to 3.2 to 5.7 au in samples taken from depth A (ii) Explain the roles of dehydrogenase in Krebs cycle of the aerobic bacteria. [2] 1. ref to substrate and ref oxidation / loss of protons and electrons 2. ref to reduction of coenzyme NAD and FAD and formation of NADH and FADH2 [Total: 9] Question 3 3 (a) State the structural features of DNA that make it a stable molecule. [2] [Any 2] 1. complementary bases / base pairing, hold(s) strands together 2. (because of) many hydrogen bonds 3. sugar-phosphate backbone / AW, with covalent / phosphodiester, bonds (b) Explain, in terms of mode of action of enzymes, the results of the investigation shown in Fig. 3.1. [3] 1. increasing concentration of ara-ATP (can be comparison between 0 and 5 / 20 or between 5 and 20) decreases enzyme activity, ref. to rate of DNA synthesis for enzyme activity 2. ara-ATP acting as an inhibitor 3. substrate unable to bind with active site / fewer enzyme-substrate complexes (formed) 4. further detail

HWA CHONG INSTITUTION / 2015 JC2 H1 BIOLOGY / PRELIMINARY EXAMINATION / MARK SCHEME 3 for either competitive inhibition e.g. competes with substrate for (binding to) the active site / similar, structure / shape, as substrate or complementary shape to active site O R f o r non-competitive inhibition e.g. binds to site other than active site / changes shape of active site (c) Define the term recessive. [1] allele which does not have its effect in heterozygote (d) Explain why females are less likely than males to have RGC. [2] 1. gene / allele, on X chromosome / sex linkage 2. female, needs 2 RGC alleles / homozygous recessive / can be heterozygous 3. male needs only 1 RGC allele to be affected (e) With reference to Fig. 3.2, and using the symbols R for the dominant allele and r for the recessive allele, state the genotypes of the individuals 1 and 6. [2] 1 X RXr 6 XrY [Total: 10] Question 4 (a) Explain what is meant by recombinant plasmid DNA. [2] 1a. DNA from two sources: bacteria plasmid, and 1b. (eukaryotic) foreign gene of interest 2. Where they are ligated together (b) Outline the roles of the enzymes used in the formation of the recombinant plasmid DNA. [2] 1. Restriction enzymes / endonucleases 2. Cleavage of phosphodiester bonds of the DNA 3. DNA ligase 4. Catalyses the formation of phosphodiester bond (c) One of the foreign genes inserted into the plasmid, codes for resistance to a particular antibiotic. Explain why it is important to include a gene for antibiotic resistance in the plasmid produced. [2] 1. To identify / select bacteria that have taken up the plasmid 2. After the new plasmids are incubated with bacteria, the bacteria is grown on media containing the antibiotics

HWA CHONG INSTITUTION / 2015 JC2 H1 BIOLOGY / PRELIMINARY EXAMINATION / MARK SCHEME 4 3. Only those bacteria that have taken up the plasmid will be able to grow / survive on the media (d) Explain why a farmer might choose to grow a crop that was genetically engineered to be resistant to insects, rather than spray the crop with insecticide. [2] 1. Insecticide may be harmful to humans and other living things; 2. Thus, it may kill useful insects; 3. GM crop may be cheaper to grow and manage; 4. Genetically engineered resistance may be more specific for a harmful insect; (e) Suggest why farmer Y might be concerned about the possibility of his crops being fertilised by pollen from farmer X’ s c r o p . [ 1 ] 1. Farmer Y may lose some of his markets if he cannot state that his crops are GM free (f) Deduce the conclusions that can be drawn about cross-pollination and the gap between crops. [ 3 ] 1. When the crops are adjacent to each other there is cross-pollination 2. Cross-pollination can be reduced by moving crops 5 m apart 3. The difference between separating 5 m or 7 m does not influence cross pollination at the edge, only 10 m into crop [Total: 12] FREE RESPONSE QUESTION Question 5 (a) Describe and explain the fluid mosaic model of membrane structure. [5] 1. A ‘mosaic’ / collage of proteins 2. randomly distributed 3. There are integral and peripheral proteins found in plasma membranes 4. A fluid phospholipid bilayer which is free to move about laterally 5. Due to the weak hydrophobic interactions between the hydrophobic fatty acid tails 6. Presence of cholesterol regulates membrane fluidity 7. Ref. to membranes being asymmetrical / characteristics of fluid mosaic model exist due to the composition of the components of the membrane: proteins and lipids 8. Ref. to membranes being amphipathic / consisting of phospholipids will a hydrophilic phosphate head and hydrophobic hydrocarbon tails

HWA CHONG INSTITUTION / 2015 JC2 H1 BIOLOGY / PRELIMINARY EXAMINATION / MARK SCHEME 5 (b) Outline the roles and functions of membranes within cells and at the surface of cells. [8] Within cells: 1. Organelles can be bound by membranes 2. Maintenance of characterist