NYGH 2017-S3EOY-IP Bio P1 and P2 Ans

2017 S3 IP BIOLOGY EOY MARK SCHEME PAPER 1 Qn Ans Explanation 1 D Fact. Water is the main constituent of protoplasm. 2 D In cold climates, hibernating animals will store fats for warmth. Vacuoles are the storage organelles of the cell. 3 C Potato has live cells which have partially permeable membranes. Thus osmosis will occur from the water into the potato cells and move from cell to cell to enter the sugar solution, down a water potential gradient. 4 C The diagram shows substance in a vesicle fusing with the membrane and releasing the substance outside the cell. This is exocytosis. 5 B only B shows positive food test results for two nutrients (reducing sugar and fats). The others show positive observations for 3 nutrients, or all negative results 6 A Sugars are preferentially used by the body as a quick source of energy. Fats and proteins are not metabolised as quickly. The substance with the highest amount of reducing sugar is shown by brick-red ppt (because Benedict’s test is a semi-quantitative test) 7 A digestion of starch by amylase gives maltose, which are disaccharides. Maltose is then digested by maltase into glucose (basic units). 8 C Using the lock and key hypothesis, Z is the lock (enzyme) and X and Y are the keys (substrates: phenylalanine and hydroxyl group). W has an active site that does not fit the substrates, and so must be the defective enzyme. 9 A PAH is a human enzyme, so optimum activity should be around 37oC. Slow increase in activity at low temp (due to inactivation of enzyme) and sharp decrease in activity after 37oC (due to denaturation) 10 C Light energy is used in the photolysis of water molecules into protons and oxygen. 11 B During the day, rate of photosynthesis exceeds that of respiration. Hence, the plant appears to give out oxygen and take in carbon dioxide. 12 C Plant photosynthesises in light and oxygen is produced. Bubbles of oxygen displaces the water. 13 A Dilute sodium bicarbonate solution supplies the plant with additional carbon dioxide, hence photosynthetic rate increases, thereby increasing the rate of oxygen production. 14 B Q and S. Increasing light intensity results in an increase in rate of photosynthesis. 15 A (y-x) g: change in mass is due to photosynthesis. (y-z) g: plant respires instead of photosynthesise in the dark and respiration requires breakdown of starch to provide energy. 16 A Water is able to form hydrogen bonds with other molecules (cohesion) and with the walls of xylem (adhesion). Water being polar means it cannot pass through phospholipid bilayer easily. Not 4: Low molecular mass is a result of adding the mass of individual atoms that make up water. 17 D both 1 and 2 are xylem, as seen by the striped patterns of lignin. Therefore, they transport water and mineral ions. 18 D radioactive carbon dioxide is taken into leaves and used for photosynthesis, to make radioactive sugar. Radioactive sugar enters the phloem and reaches the root. A shows the position of STEM phloem. B shows STEM xylem. C shows ROOT xylem. D shows ROOT phloem.

2017 End-of-Year Examination Secondary Three 2 Nanyang Girls’ High School Biology PAPER 2 19 C Buccal cavity, P: ingestion Stomach, Q: digestion Small intestines, R: absoprtion 20 B Gall bladder stores bile produced by liver. Bile emulsifies fats; so removal of gall bladder affects fat digestion. Bilirubin in bile cause faeces to have a dark color. The absence of bile would cause faeces to be a lighter color. Amino acids absorption does not depend on bile/gall bladder. Liver is the organ that produces bile. 21 C Hepatic portal vein delivers absorbed nutrients e.g. glucose from small intestines to liver for storage, hence less glucose exits the liver through hepatic vein. 22 D Only small soluble molecules can be absorbed without further digestion. 23 C Heart muscles will be affected during a heart attack. 24 C Blood cells are forced to move in a single file due to the narrow lumen of the capillaries, not because of low blood pressure and velocity. 25 C On the graph, regions of smaller cross-sectional area correspond to higher blood velocities. 26 D Both elastic recoil of vessel wall and pumping action of heart create high pressure due to the force generated. 27 B Carbon dioxide produced by cells travel in the blood vessel to the lungs where it exits the capillary into the alveolus. The alveolus leads into the bronchiole, bronchus, trachea and then larynx. 28 C Alveoli are the sites of gaseous exchange in the lungs. The lesser the alveoli, the smaller the surface area available for gaseous exchange. 29 A Breathing is a process that involves inhalation and exhalation. Out of all the time periods, V to W is the only one that registers a complete inhalation and exhalation. 30 A Cilia helps to sweep dust and mucus up the trachea. The sweeping motion of cilia requires energy produced by mitochondria. 1ai Guard cells [1] 1aii The guard cells control the opening and closing of the stomata, thereby regulating the passage of oxygen and carbon dioxide in and out of the the leaf. OR The guard cells control the opening and closing of the stomata, allowing the exchange of oxygen during respiration and carbon dioxide during photosynthesis. REJECT × gaseous exchange only (incomplete) × exchange of carbon dioxide and oxygen only (incomplete) [1]

2017 End- of-Year Examination Secondary Three 3 Nanyang Girls’ High School Biology [Turn over 1bi Reject × plural form of chloroplast if student only labels one chloroplast × cell wall (without mentioning “thicker”, “inner”) [2] 1bii During the day, the guard cells photosynthesise, converting light energy to chemical energy. The chemical energy is used to pump potassium ions (K+) from neighbouring epidermal cells into the guard cells. As the result, the water potential of the guard cells is lowered. Water from neighbouring epidermal cells then enters guard cells by osmosis. This causes the guard cells to swell and become turgid. The guard cells have a thicker cell wall on one side of the cell (the side around the stomatal pore). This causes the swollen guard cells to curve more than the thinner outer cell wall, and results in the openng of the stomatal pore. [4] [Total: 8] 2ai amino group acidic group [1] 2aii peptidase [1] aiii inability to digest dipeptide to amino acids leads to less amino acids absorbed and eventually malnutrition / protein deficiency REJECT dipeptide not digested to amino acids (stating the obvious) [1] thicker inner cell wall [1] chloroplast [1]

2017 End-of-Year Examination Secondary Three 4 Nanyang Girls’ High School Biology 2b Structure of aa1 drawn; Structure of aa2 drawn; [2] Total [ 5 ] 3a Mean change in volume = (1.3-0.1 + 2.6-1.3 + 3.7-2.6) / 3 = (1.2 + 1.3 + 1.1) / 3 = 1.2 ml Mean transpiration rate = 1.2 / 30 = 0.04 ml / min [2] 3b volume of water into plant is assumed to be water lost by transpiration rate of transpiration is the same as rate of absorption It is assumed that plant does not use water for photosynthesis It is assumed that plant does not retain more water taken in in its cells REJECT the rate of transpiration of the plant was the same for all the 3 trials (this is not necessary (Trial 1 is 0.04 ml/min Trial 2 is 0.0433 ml/min Trial 3 is 0.0367 ml/min) [2] 3ci Set-up Transpiration rate/ arbitrary units B 2 A 8 [1] 3cii states & explains any 2 factors that affect the rate of transpiration Set-up A has higher transpiration rate because • plant has more leaves which increases surface area for transpiration • air conditioner decreases humidity of the computer lab which increases rate of diffusion of water vapour • there is wind from the air-conditioner, which blows away the water vapour that accumulates outside the stomata • there is light from the lamp, which causes the stomata