Biology SBI4U Practice Test Suggested Answers PDF

TVO ILC SBI4U Practice Test Suggested Answers Biology SBI4U Practice Test Suggested Answers Total Marks: 95 TVO ILC...

TVO ILC SBI4U Practice Test Suggested Answers Biology SBI4U Practice Test Suggested Answers Total Marks: 95 TVO ILC SBI4U Practice Test Suggested Answers Part A: Population Dynamics (15 marks) 1. a) (2 marks, allocated as follows) The size of the global human population stayed very low until about 2000 to 4000 years ago. It then began to increase gradually, mostly due to agriculture (and related technological and cultural advances). (1 mark) Over the last 150 to 200 years, the population increase has been exponential due to several factors, including the Industrial Revolution, advances in medicine, and the Green Revolution in agriculture. (1 mark) b) (1 mark - You could choose any one of these three for the mark) Greater social equality for women (for example, better education, more economic opportunities, and so on), leading women to have fewer children Improved living standards and better retirement benefits and health care mean that people don’t have to have large families in order to ensure that they are cared for by their children when they become elderly. Greater urbanization: Fewer children are required, when families move away from the farm, as their labour is no longer needed. 2. (3 marks) DE = N/SE = 350 deer/(530 km2 – 80 km2) = 350 deer/450 km2 = 0.78 deer/km2 The ecological density of the white-tailed deer in the valley is 0.78 deer/km2. 3. a) (1 mark) Logistic growth (or S-shaped or sigmoid growth) Copyright © 2020 The Ontario Educational Communications Authority. All rights reserved. ii TVO ILC SBI4U Practice Test Suggested Answers b) (3 marks) White-tailed deer population on Farlane Island 700 Stationary 600 500 400 L o 300 g L 200 a 100 g 0 0 5 10 15 20 Time (years) c) (1 mark) dN/dt = rN [(K – N) / K] d) (1 mark) The deer population increases because the population has not yet reached the carrying capacity of the habitat. Births are greater than deaths so there is a net gain of individuals in the population. e) (2 marks) There would be a decline in deer numbers as the competitor began to consume the same resources as the deer. This interspecific competition would lower the carrying capacity of the deer so that, after year 16, the stable number of deer would be lower than the 600 it was before. Copyright © 2020 The Ontario Educational Communications Authority. All rights reserved. iii TVO ILC SBI4U Practice Test Suggested Answers f) (1 mark) White-tailed deer population on Farlane Island 700 Stationary 600 500 400 L o 300 g L 200 a 100 g 0 0 5 10 15 20 Time (years) Copyright © 2020 The Ontario Educational Communications Authority. All rights reserved. iv TVO ILC SBI4U Practice Test Suggested Answers Part B: Biochemistry (20 marks) 4. (7 marks). a) Condensation b) Redox c) Neutralization d) Redox e) Neutralization f) Condensation g) Hydrolysis 5. (1 mark) When two atoms share their valence electrons, a covalent bond is formed. This sharing of valence electrons enables each atom to obtain a “full valence shell”. This type of bond exists between atoms in molecules. 6. (3 marks, allocated as follows) Both are forms of active transport that require the cellular membrane to assist in the movement of the material. (1 mark) Exocytosis involves the removal of materials such as wastes from the cell. (1 mark) Endocytosis involves bringing materials such as water (pinocytosis) and solid particles (phagocytosis) into the cell. (1 mark) 7. (2 marks) The salt created a hypertonic environment relative to the interior of the roast turkey’s cells (hypotonic environment). This caused water to migrate out of the roast by osmosis, making it dry. Copyright © 2020 The Ontario Educational Communications Authority. All rights reserved. v TVO ILC SBI4U Practice Test Suggested Answers 8. (5 marks, allocated as follows) 2 marks for correct structure of the triglyceride molecule 1 mark each for the following labels: Reactants: Glycerol and 3 Fatty acids Product: Triglyceride Functional groups: Any hydroxyl group (OH) and any carboxyl group (H-C=O). Main linkage: Ester linkage Ester linkage O H H H H H H H O H H H H H H C C C C C C C H H C O C C C C C C C H H Hydroxyl HO H H H H H H H H H H H H H C OH O H H H H H H O H H H H H H H C OH C C C C C C C H H C O C C C C C C C H H C OH HO H H H H H H H H H H H H H O H H H H H H O H H H H H H H C O C C C C C C C H C C C C C C C H H H H H H H H HO H H H H H H Glycerol 3 Fatty acids Triglyceride 9. (2 marks) Enzymes lower the activation energy required for the reaction to occur by attaching to the reactants and positioning them so that they are in the optimal orientation to break or make chemical bonds between them. Copyright © 2020 The Ontario Educational Communications Authority. All rights reserved. vi TVO ILC SBI4U Practice Test Suggested Answers Part C: Metabolic Processes (19 marks) 10. a) (2 marks, allocated as follows) ATP has three phosphate tails at one end of the molecule. This is the site of energy storage. Each phosphate group has a negative charge, so they repel each other. This means that the three groups hold potential energy. (1 mark) When the last bond between phosphate groups is broken (ATP → ADP), this potential energy is released to enable cell work. (1 mark) I f energy is added and a source of atoms in a phosphate group is present, then ATP can be formed again. b) (2 marks) ADP + Pi + energy → ATP → ATP ADP + Pi + energy ATP is converted to ADP in order to make energy for the cell. However, the foods that we eat can provide the matter and energy needed to add a third phosphate to produce ATP again. 11. a) (2 marks) 6 CO2 + 6 H2O + energy → C6H12O6 + 6 O2 b) (2 marks) C6H12O6 + 6 O2 → 6 CO2 + 6 H2O + ATP energy c) (1 mark) You would expect there to be a strong odour of alcohol. Copyright © 2020 The Ontario Educational Communications Authority. All rights reserved. vii TVO ILC SBI4U Practice Test Suggested Answers 12. (5 marks) Type of reaction Light-depentant reactions Light-indepentant reactions Location in chloroplast Thylakoid membrane Stroma Purpose Converts light energy to Uses ATP and NADPH to make chemical energy stored in ATP glucose and NADPH Products Oxygen Glucose 13. (1 mark) The dark reaction receives energy captured from sunlight that is stored in the form of ATP and NADPH molecules. 14. (4 marks) Any four of the following: a) Photosystem 680 (also called photosystem II) b) NADPH c) Calvin cycle d) Krebs cycle e) Glycolysis f) Pyruvate oxidation g) Glucose (C6H12O6) Copyright © 2020 The Ontario Educational Communications Authority. All rights reserved. viii TVO ILC SBI4U Practice Test Suggested Answers Part D: Molecular Genetics (20 marks) (approximate time: 22 minutes) 15. (2 marks) 3'TAACGTAACCCCAT5' 16. (3 marks, allocated as follows) Watson and Crick described DNA as a double helix (twisted ladder), made up of repeating units called nucleotides. (1 mark) Each nucleotide contains a nitrogen base, a sugar, and a phosphate. (1 mark) The sugars and phosphates of the repeating nucleotides make up the sides of the ladder, while the nitrogen bases make up the rungs. (1 mark) 17. (5 marks) The answers are shown in bold. Enzyme Function DNA gyrase Prevents helix from super-coiling DNA helicase Unwinds and unzips DNA helix (breaks hydrogen bonds between nitrogen bases) DNA polymerase III It is the enzyme responsible for “stringing” nucleotides together to make the newly synthesized DNA. It catalyzes the formation of phosphodiester bonds between adjacent deoxyribonucleoside triphosphates. DNA ligase Glues together Okazaki fragments RNA primase Lays down RNA primers 18. (4 marks: ½ mark each) DNA codon (5-3) TAC CAG GTA GAC mRNA codon AUG GUC CAU CUG Amino acid Methionine Valine Histidine Leucine Copyright © 2020 The Ontario Educational Communications Authority. All rights reserved. ix TVO ILC SBI4U Practice Test Suggested Answers 19. a) i) (1 mark) CCG ii) (1 mark) Proline b) (1 mark) A missense mutation would result. Instead of leucine, the codon now codes for proline, thus changing the sequence of amino acids. This can change the resulting protein. 20. (3 marks: ½ mark for each term) 1. g 2. f 3. a 4. e 5. c 6. h Copyright © 2020 The Ontario Educational Communications Authority. All rights reserved. x TVO ILC SBI4U Practice Test Suggested Answers Part E: Homeostasis (21 marks) (approximate time: 20 minutes) 21. (6 marks: 2 marks for each step, with 1 mark within each step allocated for the description and 1 mark for the active/passive transport assignment) Step 1: Filtration The movement of fluids from the blood into Bowman’s capsule is called filtration. The dissolved solutes leave the afferent arteriole and enter Bowman’s capsule, which is porous. Plasma protein, blood cells, and platelets are too large to move through the pores of the glomerulus, while smaller molecules pass through the walls and enter the nephron. It only uses passive transport. Step 2: Reabsorption The transfer of essential solutes and water from the nephron back into the blood is called reabsorption. It uses active transport to move sodium (N + ) ions, glucose, and amino acids back into the blood and it uses passive transport to move water from the nephron back into the blood. Step 3: Secretion The movement of materials, such as nitrogen-containing waste, excess H + ions, and some minerals and drugs are secreted into the urine. It only uses passive transport. Copyright © 2020 The Ontario Educational Communications Authority. All rights reserved. xi TVO ILC SBI4U Practice Test Suggested Answers 22. (6 marks) The answers are shown in bold. Hormone Gland Target Action Estrogen Ovary Developing follicle Any one of: Thickens the lining of the uterus Causes the egg to mature (undergo meiosis) Causes female secondary sex characteristics to develop Cortisol Adrenal Muscles, liver Any one of: cortex Increases the amount of amino acids in the blood, which the liver converts to glucose for energy Converts fats to fatty acids to produce energy for the body 23. (4 marks, allocated as follows) After depolarization, the charge on the inside of the nerve membrane becomes positive. This causes the sodium gates to close, stopping the influx of sodium. (1 mark) A sodiumpotassium pump, located in the cell membrane, restores the condition of the resting membrane by transporting sodium ions out of the neuron, while moving potassium ions into the neuron, at a ratio of 3 Na+ to 2 K+ ions. (1 mark) The energy for the pump comes from ATP. (1 mark) The process of restoring the original polarity of the nerve membrane is called repolarization. (1 mark) Copyright © 2020 The Ontario Educational Communications Authority. All rights reserved. xii TVO ILC SBI4U Practice Test Suggested Answers 24. (5 marks) Example Community interaction term 1. Certain species of ants eat aphids that land on the a) Realized niche plant under which the ants are living. 2. Some species of worms obtain their food by living in b) Interspecific competition the intestines of other animals and eating the digesting material before it can be absorbed by the host animal. 3. Remora fish attach themselves to whales or sharks to c) Fundamental niche catch a free ride and obtain opportunistic meals, while causing no harm to their host. 4. Certain species of ants protect aphids from d) Intraspecific competition predation by ladybugs. In turn, the ants obtain a meal of sugary “honeydew” released by the aphids. 5. Two species of parrot both eat the same types of fruit. e) Predation f) Mutualism g) Commensalism h) Parasitism Write alphabetical letter of answer: 1. Predation 2. Parasitism 3. Commensalism 4. Mutualism 5. Interspecific competition Copyright © 2020 The Ontario Educational Communications Authority. All rights reserved. xiii