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HKDSE Biology Paper 2 Elective Units Data Practice Paper (10 Questions, 100 Marks)
HKDSE Biology Paper 2 Elective Data
HKDSE Biology Paper 2 elective units data practice paper: 10 original questions worth 100 marks in 60 minutes, covering glomerular filtration and ADH, heat loss by sweating.
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Paper details Level: S6 (DSE) | Subject: Biology | Time: 60 minutes | Marks: 100 | Questions: 10
Section A E1 Human Physiology and E2 Applied Ecology (50 marks)
Questions 1-5
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | An adult has a glomerular filtration rate of 125 cm3 min-1 and passes 1.5 dm3 of urine per day. (a) Calculate the volume of filtrate produced each day in dm3. (3 marks) (b) Calculate the percentage of the filtrate water reabsorbed by the kidney tubules. (4 marks) (c) Explain how ADH changes the permeability of the collecting duct to water and state the effect on urine volume during dehydration. (3 marks) |
Answer: (a) 125 x 60 x 24 = 180000 cm3 = 180 dm3. (b) (180 – 1.5) / 180 x 100% = 99.2%. (c) ADH increases the number of water channels in the membranes of collecting duct cells, so more water is reabsorbed and a smaller volume of more concentrated urine is produced. Marking notes: filtrate volume = filtration rate x time; the denominator for reabsorption is the filtrate, not the urine; ADH acts on the collecting duct. |
| 2 | An athlete sweats 0.8 dm3 per hour at 32 degrees C. The latent heat of vaporisation of sweat is 2.4 kJ g-1. (a) Calculate the mass of sweat lost per hour in grams. (3 marks) (b) Calculate the heat lost per hour and per minute by sweating, in kJ. (4 marks) (c) Explain why the same sweat loss removes heat less effectively on a very humid day. (3 marks) |
Answer: (a) 0.8 dm3 = 800 g. (b) 800 x 2.4 = 1920 kJ per hour; 1920 / 60 = 32 kJ per minute (about 533 W). (c) The high water vapour pressure of humid air slows evaporation, so less latent heat is removed and body temperature tends to rise. Marking notes: 1 dm3 of water is 1000 g; energy is latent heat times mass; cooling depends on evaporation, not on sweating itself. |
| 3 | In an oral glucose tolerance test a subject drinks 75 g of glucose (relative molecular mass 180) dissolved in 250 cm3 of water. (a) Calculate the mass concentration of the solution in g dm-3. (3 marks) (b) Calculate its molar concentration in mol dm-3. (3 marks) (c) Explain how blood glucose returns close to the fasting level two hours after drinking. (4 marks) |
Answer: (a) 75 / 250 = 0.30 g cm-3 = 300 g dm-3. (b) 300 / 180 = 1.67 mol dm-3. (c) The rise in blood glucose stimulates beta cells of the pancreas to secrete insulin, which promotes uptake of glucose by liver and muscle cells and its conversion to glycogen, while glycogen breakdown in the liver is inhibited. Marking notes: convert volume units first; molar concentration = mass concentration / molar mass; both uptake and storage must be mentioned. |
| 4 | A researcher catches 40 fish in a pond, marks them and releases them. A week later 50 fish are caught, of which 8 carry a mark. (a) Estimate the total number of fish in the pond by the mark-recapture method, showing the calculation. (4 marks) (b) State two assumptions of this method. (3 marks) (c) If marked fish learn to avoid the traps, will the estimate be too high or too low? Explain. (3 marks) |
Answer: (a) population = 40 x 50 / 8 = 250 fish. (b) Marking does not affect survival or the chance of being caught; marks do not fall off; there is no birth, death, immigration or emigration between the two samples (any two). (c) Too high. The proportion of marked fish in the second sample falls, so the denominator in the formula is smaller and the estimate is inflated. Marking notes: the formula is marked in the first sample x total in the second / marked in the second; assumptions concern random mixing; the direction of bias follows the denominator. |
| 5 | A grassland has a gross primary production of 48000 kJ m-2 yr-1 and producer respiration of 22000 kJ m-2 yr-1. Primary consumers assimilate 2600 kJ m-2 yr-1. (a) Calculate the net primary production. (4 marks) (b) Calculate the efficiency of energy transfer from producers to primary consumers. (3 marks) (c) Explain why energy decreases greatly between trophic levels. (3 marks) |
Answer: (a) NPP = 48000 – 22000 = 26000 kJ m-2 yr-1. (b) 2600 / 26000 x 100% = 10%. (c) Energy is lost as heat in respiration and in excretory and undigested material; only a small fraction is stored as new biomass. Marking notes: NPP = GPP – respiration; transfer efficiency uses NPP as the denominator; energy flow is one-way and is not recycled. |
Section B E3 Microorganisms and Humans, E4 Biotechnology (50 marks)
Questions 6-10
| # | Question | Answer and marking notes |
|---|---|---|
| 6 | A bacterium divides every 30 minutes during exponential growth, starting from 2000 cells. (a) Calculate the number of generations in 4 hours. (3 marks) (b) Calculate the number of cells after 4 hours. (4 marks) (c) Explain why the cell number hardly changes at first after transfer to fresh medium. (3 marks) |
Answer: (a) 4 x 60 / 30 = 8 generations. (b) 2000 x 2^8 = 2000 x 256 = 512000 cells. (c) This is the lag phase: the cells are synthesising enzymes and other cell components to adapt to the new medium and have not yet begun to divide actively. Marking notes: generations = time / doubling time; growth is by powers of two; the lag phase is not a death phase. |
| 7 | When milk is heated at 63 degrees C, the number of viable bacteria falls by 90% every 2 minutes. The initial count is 1000000 per cm3. (a) Calculate the logarithmic reduction in viable count after 10 minutes. (3 marks) (b) Calculate the number of viable bacteria per cm3 after treatment. (4 marks) (c) Explain why pasteurised milk still has to be refrigerated. (3 marks) |
Answer: (a) 10 / 2 = 5 log units. (b) 1000000 / 10^5 = 10 per cm3. (c) Pasteurisation kills most vegetative cells but is not sterilisation; surviving cells and spores can multiply at room temperature, so cooling slows their growth. Marking notes: each 90% reduction is one log unit; divide by ten to the power of the reduction; pasteurisation is not sterilisation. |
| 8 | A 0.1 cm3 sample of a 10^-4 dilution is spread on an agar plate and, after incubation, 68 colonies are counted. (a) Calculate the number of viable cells per cm3 of the original culture (cfu cm-3). (4 marks) (b) Explain why only plates with 30 to 300 colonies are counted. (3 marks) (c) State one precaution that reduces error in this experiment. (3 marks) |
Answer: (a) 68 / 0.1 = 680 cfu cm-3 of the dilution; dividing by 10^-4 gives 6.8 x 10^6 cfu cm-3. (b) Too few colonies give a large random error; too many colonies overlap and cannot be distinguished, so the count is too low. (c) Plate three replicates per dilution and take the mean; mix the suspension before sampling; use the same medium and incubation time. Marking notes: divide by the plated volume, then by the dilution factor; 30 to 300 is the countable range; replicates reduce random error. |
| 9 | A circular plasmid is 4800 bp long and is cut by a restriction enzyme at 3 sites. (a) State the number of fragments produced. (3 marks) (b) Gel electrophoresis shows fragments of 2100 bp and 1500 bp. Find the length of the third fragment. (4 marks) (c) State the function of primers in PCR and explain why a pair of primers is needed. (3 marks) |
Answer: (a) Three cut sites in a circular plasmid give 3 fragments. (b) 4800 – 2100 – 1500 = 1200 bp. (c) A primer binds to a complementary sequence on the template strand and provides the 3-prime end from which DNA polymerase starts synthesis; a pair allows both strands to be copied and defines the amplified region. Marking notes: n cuts in circular DNA give n fragments; fragment lengths add up to the plasmid length; primers set the start point and the region amplified. |
| 10 | Yeast ferments 2 dm3 of medium containing 20 g dm-3 glucose under anaerobic conditions and produces 18 g of ethanol. (a) Calculate the total mass of glucose in the medium. (3 marks) (b) Using C6H12O6 -> 2 C2H5OH + 2 CO2, calculate the theoretical maximum mass of ethanol. (4 marks) (c) Calculate the actual yield as a percentage of the theoretical maximum. (3 marks) |
Answer: (a) 20 x 2 = 40 g of glucose. (b) 180 g of glucose gives 92 g of ethanol, so 40 x 92 / 180 = 20.4 g. (c) 18 / 20.4 x 100% = 88%. Marking notes: the 1 to 2 mole ratio becomes a 180 to 92 mass ratio; theoretical yield comes from stoichiometry; the percentage uses the theoretical value as denominator. |
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