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HKDSE Biology Paper 1B Data and Practical Analysis Practice Paper (10 Questions, 100 Marks)
HKDSE Biology Paper 1B Data Analysis
HKDSE Biology Paper 1B data and practical analysis practice paper: 10 original questions worth 100 marks in 60 minutes, covering magnification and actual size.
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Paper details Level: S6 (DSE) | Subject: Biology | Time: 60 minutes | Marks: 100 | Questions: 10
Section A Cells, Enzymes and Physiology (50 marks)
Questions 1-5
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | In a micrograph a cell measures 45 mm and the magnification is x1500. (a) Calculate the actual length of the cell in um. (4 marks) (b) A second cell has an actual diameter of 60 um. What magnification would show it as 90 mm? (3 marks) (c) Explain why units must be made consistent before calculating magnification. (3 marks) |
Answer: (a) Actual length = 45 mm / 1500 = 0.03000 mm = 30 um. (b) Magnification = 90 mm / 60 um = 90 mm / 0.060 mm = x1500. (c) Magnification has no units, so image and object must be in the same unit; mm and um differ by a factor of 1000. Marking notes: actual size is image size divided by magnification; 1 mm = 1000 um; the answer needs the um label. |
| 2 | In an enzyme experiment the gas volume read at 2 minutes is 24 cm3 and at 4 minutes is 36 cm3. (a) Calculate the mean rate over the first 2 minutes in cm3 min-1. (3 marks) (b) Calculate the mean rate between 2 and 4 minutes. (3 marks) (c) Calculate the percentage fall in rate and explain the cause. (4 marks) |
Answer: (a) 24 / 2 = 12 cm3 min-1. (b) (36 – 24) / (4 – 2) = 6 cm3 min-1. (c) (12 – 6) / 12 x 100% = 50%; substrate is used up and active sites become less saturated, so the rate falls. Marking notes: a rate is a change in volume over the matching time interval, not total volume over total time. |
| 3 | Three identical potato strips each of mass 2.40 g are placed in distilled water, 0.30 mol dm-3 sucrose and 0.60 mol dm-3 sucrose. After 30 minutes the strip in distilled water weighs 2.88 g and the strip in the 0.60 mol dm-3 solution weighs 1.92 g. (a) Calculate the percentage mass change in distilled water. (3 marks) (b) Calculate the percentage mass change in the 0.60 mol dm-3 solution. (3 marks) (c) Explain the difference and predict the result in the 0.30 mol dm-3 solution. (4 marks) |
Answer: (a) (2.88 – 2.40) / 2.40 x 100% = +20%. (b) (1.92 – 2.40) / 2.40 x 100% = -20%. (c) Distilled water has a higher water potential than the cell sap, so water enters by osmosis; the concentrated sucrose solution has a lower water potential and the cells lose water; the 0.30 mol dm-3 solution is close to isotonic so the mass barely changes. Marking notes: the initial mass is the denominator; water moves from high to low water potential. |
| 4 | A cross between two parents showing the dominant trait gives 400 offspring: 305 with the dominant phenotype and 95 with the recessive one. (a) State the expected dominant to recessive ratio and the two expected numbers. (3 marks) (b) Calculate the chi-square value. (4 marks) (c) The critical value at 5% with 1 degree of freedom is 3.84; decide whether the result fits the expected ratio. (3 marks) |
Answer: (a) 3:1, i.e. 300 dominant and 100 recessive. (b) chi-square = 0.0833 + 0.2500 = 0.3333. (c) 0.33 is below 3.84, so the deviation is not significant and the 3:1 ratio is accepted. Marking notes: each group contributes (observed – expected)2 / expected; compare with the critical value, not with the observed counts directly. |
| 5 | In a 200 m2 meadow, 10 random quadrats of 0.5 m2 each contain 24 plants of a species in total, and 7 quadrats contain the species. (a) Calculate the total sampled area and the population density in plants m-2. (4 marks) (b) Estimate the population size of the whole meadow. (3 marks) (c) Calculate the percentage frequency and state why random sampling matters. (3 marks) |
Answer: (a) Area sampled = 10 x 0.5 = 5 m2; density = 24 / 5 = 4.8 m-2. (b) 4.8 x 200 = 960 plants. (c) Frequency = 7 / 10 x 100% = 70%; random placement avoids bias from choosing convenient spots. Marking notes: density is per unit area; the total is density times total area; frequency uses the number of quadrats as denominator. |
Section B Genetics, Ecology and Data Analysis (50 marks)
Questions 6-10
| # | Question | Answer and marking notes |
|---|---|---|
| 6 | A pondweed 10 cm from a lamp releases 30 bubbles in 2 minutes; at 20 cm it releases 18 bubbles in 2 minutes. (a) Calculate the bubble rate at each distance in bubbles min-1. (4 marks) (b) Calculate the percentage fall in rate. (3 marks) (c) If the distance doubles, what fraction of the original light intensity remains? Explain what the bubbles show. (3 marks) |
Answer: (a) 10 cm: 30 / 2 = 15 min-1; 20 cm: 18 / 2 = 9 min-1. (b) (15 – 9) / 15 x 100% = 40%. (c) About 0.25 of the intensity remains (inverse square law); the bubbles are oxygen and their rate reflects the net rate of photosynthesis. Marking notes: compare rates over equal time; the original rate is the denominator. |
| 7 | A 2.5 g animal uses 0.42 cm3 of oxygen in 30 minutes in a respirometer. (a) Calculate the hourly oxygen uptake in cm3 h-1. (3 marks) (b) Calculate the uptake per gram per hour in cm3 g-1 h-1. (4 marks) (c) Explain why metabolic rate is usually expressed per unit body mass and state one precaution when comparing species. (3 marks) |
Answer: (a) 0.42 x (60 / 30) = 0.84 cm3 h-1. (b) 0.84 / 2.5 = 0.336 cm3 g-1 h-1. (c) Per unit mass removes the effect of body size; comparisons must use the same temperature, activity level and measuring period. Marking notes: convert to an hourly figure first, then divide by mass; keep the units complete. |
| 8 | An athlete at rest has a heart rate of 72 beats min-1 and a stroke volume of 70 cm3; during vigorous exercise the values are 120 beats min-1 and 100 cm3. (a) Calculate the cardiac output at rest and during exercise in dm3 min-1. (4 marks) (b) Calculate how many times greater the exercising output is. (3 marks) (c) Explain the benefit of the higher output to the muscles. (3 marks) |
Answer: (a) Rest: 72 x 70 = 5040 cm3 min-1 = 5.04 dm3 min-1; exercise: 120 x 100 = 12000 cm3 min-1 = 12 dm3 min-1. (b) 12000 / 5040 = 2.38 times. (c) Oxygen and glucose reach the muscles faster and carbon dioxide is removed faster, supporting a higher respiration rate. Marking notes: cardiac output = heart rate x stroke volume; the resting value is the denominator. |
| 9 | A potometer shows 3.6 cm3 of water taken up in 20 minutes by a leafy shoot with a total leaf area of 60 cm2. (a) Calculate the uptake rate in cm3 min-1. (3 marks) (b) Calculate the transpiration per square centimetre per hour in cm3 cm-2 h-1. (4 marks) (c) Explain the effect of stronger wind on the reading. (3 marks) |
Answer: (a) 3.6 / 20 = 0.18 cm3 min-1. (b) 3.6 x (60 / 20) / 60 = 0.18 cm3 cm-2 h-1. (c) Wind removes the saturated layer of air next to the leaf, steepening the water vapour gradient so transpiration and the movement of the air bubble both speed up. Marking notes: convert to an hourly value before dividing by leaf area; a potometer measures uptake, which approximates transpiration. |
| 10 | A double-stranded DNA fragment contains 2000 bases, of which 30% are adenine. (a) State the percentage of thymine. (2 marks) (b) Calculate the percentages of guanine and cytosine. (4 marks) (c) Calculate the number of guanine bases in the fragment. (4 marks) |
Answer: (a) A pairs with T, so thymine is 30%. (b) A + T = 60%, leaving 40% shared equally between G and C, i.e. 20% each. (c) 2000 x 20% = 400 bases. Marking notes: complementary base pairing fixes equal percentages; all four percentages must total 100; the count is the percentage of the total number of bases. |
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