2016 HCI H1 Biology Prelims P1&P2 Answers

1 HWA CHONG INSTITUTION 2016 JC2 H1 BIOLOGY PRELIMINARY EXAMINATION PAPER 1 & 2 MARK SCHEME MULTIPLE CHOICE QUESTIONS 1 B 11 B 21 B 2 B 12 A 22 D 3 D 13 D 23 A 4 D 14 A 24 A 5 C 15 A 25 D 6 D 16 A 26 D 7 C 17 D 27 B 8 D 18 C 28 B 9 A 19 C 29 A 10 B 20 C 30 C STRUCTURED QUESTIONS QUESTION 1 (a) Explain the role of helicase in DNA replication. [2] 1. separates/unwinds / unzips strands / helix / breaks H-bonds 2. so nucleotides can attach / parental DNA strands can act as templates (b) Explain how cytarabine prevents DNA replication. [2] 1. similar structure to cytosine added instead of cytosine 2. prevents complementary base pairing / prevents strand elongation With reference to Fig. 1.2, (c) (i) state the duration of metaphase in the cell. [1] 18min (ii) complete line Y on the graph. [1] horizontal until 18 minutes, then decreases as straight line to 0 μm at 28 minutes

2 (iii) account for your answer in (c)(ii). [ 3 ] 1. chromosomes / pairs of sister chromatids align singly at the metaphase plate during metaphase of mitosis 2. hence, from 0 minute to 18 minutes, the distance of each chromatid and the pole to which it is moving was constant at 20 μm 3a. sister chromatids separate at the centromere to become daughter chromosomes 3b. and migrate towards the opposite poles in anaphase 4. hence, the distance of each chromatid / daughter chromosome and the pole decreased from 20 μm to 0 μm from 18 minutes to 28 minutes at the end of anaphase 5. each chromatid / daughter chromosome did not move / remain at the pole in telophase 6. hence, the distance of each chromatid / daughter chromosome was constant at 0 μm [Total: 9]

3 QUESTION 2 (a) (i) State precisely where RuBP and GP are located in the chloroplast. [1] Stroma (ii) Explain why the concentration of RuBP changed between 200 and 275 seconds. [2] 1. during this period, CO 2 concentration is lower 2. since less CO 2 fixed by RuBP / CO2 combining with RuBP / RuBP converted to GP 3. moreover, RuBP is regenerated from triose phosphate (b) Suggest how the decrease in the concentration of GP leads to a decrease in harvest for commercial suppliers of Chlorella. [2] 1. a decrease in the concentration of GP will lead to less triose phosphate / glyceraldehyde-3-phosphate being produced 2. so less conversion of triose phosphate to carbohydrates / lipids / amino acids / proteins (c) In the light dependent stage, illumination of chloroplasts is important for maintaining the high pH in the stroma. Explain how the illumination of chloroplasts maintains the high pH in the stroma. [3] 1. illumination of chloroplasts excite and displace electrons from special chlorophyll a / photosystem / ref. to photoactivation of chlorophyll 2. electron is then passed down the electron transport chain 3. free energy is used to pump protons into thylakoid lumen from the stroma 4. protons also released from the photolysis of water 5. this leads to lower concentration of protons in stroma maintains pH [Total: 8]

4 QUESTION 3 (a) (i) Suggest and explain one such adaptation in the woolly mammoth. [2] 1. having thick fur / small ears 2. to reduce heat loss (ii) Explain how natural selection may have brought about the evolution of the woolly mammoth from the steppe mammoth. [4] 1. presence of much cooler conditions in the environment which serves as selection pressure 2. mammoths with favourable traits are at a selective advantage 3. differential survival and reproductive abilities 4. those selected for can pass down favourable allele to offspring 5. over time, the change in allelic frequency in the population (b) Explain the likely effect of these differences on a molecule of mammoth haemoglobin. [3] 1. difference in primary sequence of amino acids of the polypeptide results in presence of different side chains 2. this changes the bonds formed between the R groups, resulting in a change in the 3D conformation / tertiary structure of each haemoglobin chain 3. as 2 α and 2 β chains are required to form the quaternary structure of haemoglobin / ref. to effect of quaternary structure 4. greater effect on β chain 5. there will be a change in function of each haemoglobin molecule [Total: 9]

5 QUESTION 4 (a) Using the symbols A and a for colour of grain and B and b for texture of grain, draw a genetic diagram to explain these results. [4] F1 phenotype purple and smooth purple and smooth F1 genotype AaBb x AaBb Gametes AB Ab aB ab AB Ab aB ab Random fertilisation male gametes AB Ab aB ab female gametes AB AABB AABb AaBB AaBb Ab AABb AAbb AaBB Aabb aB AaBB AaBb aaBB aaBb ab AaBb Aabb aaBb aabb F2 genotypic ratio 9 A _ B _ : 3 A_ bb : 3 aa B_ : 1 aabb F2 phenotypic ratio 9 purple & : 3 purple & : 3 yellow & : 1 yellow & smooth shrunken smooth shrunken (b) Explain why yellow and shrunken grains breed true. [2] 1. the yellow shrunken grains are double hom ozygous recessive at the two gene loci 2. when it is self-fertilised, it will produce only double homozygous recessive offspring, which resemble the parental genotype (c) Explain why it is possible for rice plants to express genes from a bacterium. [1] 1. all organisms share the same genetic code (d) (i) Describe the role of the rice endosperm-specific promoter that was added to psy and crt 1 in step 1. [2] 1. role of promoter is to switch on genes 2. only in the rice endosperm and not expressed anywhere else would be a waste of energy 3. since it is the edible part of seed

6 (ii) Explain how a length of DNA can be inserted into a plasmid in step 2. [3] 1. use restriction enzyme to cut plasmid open 2. sticky ends anneal via complementary base pairing, A - T / C - G 3. role of DNA ligase, in joining sugar-phosphate backbone (e) (i) Suggest one possible risk to the environment of growing a genetically engineered crop. [1] 1. gene transfer to other rice 2. contamination of other crop (e.g. ref to organic crop) 3. gene transfer to wild relative (ii) Suggest one possible risk to human health of eating a genetically engineered crop. [1] 1. allergy 2. long term toxicity 3. antibiotic resistance of gut bacteria if antibiotic marker used [Total: 14]

7 FREE RESPONSE QUESTIONS QUESTION 5 (a) Describe how the molecular structure of cellulose is related to its support function. [6] 1. alternate inverted β-glucose units linked by (1,4) glycosidic bonds allow cellulose to form long, unbranched and straight chains 2. few organisms have enzymes to hydrolyse (1,4) glycosidic bonds, hence cellulose is stable 3. alternate inverted β-glucose units linked by (1,4) glycosidic bonds allow cellulose to form long, unbranched and straight chains 4. allow formation of linear chains of polysaccharides that can be packed tightly 5. many chains run parallel to each other and t heir hydroxyl group (OH) project outwards from each chain 6. extensive hydrogen bonds form between para llel chains/ Extensive hydrogen bonds form between the protruding OH groups of neighbouring chains 7. allowing establishment of rigid cross-links between chains 8. cross-linked cellulose chains associate in groups to form microfibrils 9. microfibrils associate with other, non-cellulose pol