中文（繁體）
中文（简体）
English
HKDSE Biology Paper 1B Structured Mock Paper (10 questions, 100 marks)
HKDSE Biology Paper 1B
HKDSE Biology Paper 1B structured mock paper: 10 questions, 100 marks, 2 hours 30 minutes, covering monohybrid genetics, microscopy, surface area to volume ratio, photosynthesis, cardiac output, osmosis, enzyme activity, quadrat sampling, yeast respiration and DNA base composition, with full answers and marking notes.
Download Question paper PDF | Answer key PDF
Paper details Level: HKDSE | Subject: Biology | Time: 2 hours 30 minutes | Marks: 100 | Questions: 10
Section A: cells, genetics and plant physiology (50 marks)
Structured questions: show calculations, genetic diagrams and full explanations; numerical answers must carry units. 10 questions, 100 marks, 2 hours 30 minutes, calculator allowed.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Two tomato plants are crossed (Tt x Tt), where T (tall) is completely dominant to t (dwarf). 400 offspring are obtained. (a) State the gametes produced by each parent and their ratio. (3 marks) (b) Use a Punnett square to give the genotypic and phenotypic ratios. (4 marks) (c) How many of the 400 offspring are expected to be dwarf? (3 marks) |
Answer: (a) Both parents are Tt, so each produces T and t gametes in a 1:1 ratio. (b) TT:Tt:tt = 1:2:1; with complete dominance tall:dwarf = 3:1 (300:100). (c) Dwarf plants are tt, one quarter of the total: 400 x 1/4 = 100. (Marking: gametes 1, square 2, ratio 1, calculation and conclusion 2) |
| 2 | A student uses a x10 eyepiece and a x40 objective to observe a guard cell; the image measures 60 mm long. (a) Calculate the total magnification. (3 marks) (b) Calculate the actual length of the cell in micrometres. (4 marks) (c) The field of view is 1.8 mm at x100. What is its diameter at x400? (3 marks) |
Answer: (a) Total magnification = 10 x 40 = 400. (b) Actual length = 60 / 400 = 0.15 mm = 150 micrometres. (c) Magnification rises fourfold so the field diameter falls fourfold: 1.8 x 100 / 400 = 0.45 mm. |
| 3 | Two cube models of side 3.0 cm and 1.0 cm represent cells of different sizes. (a) Calculate the surface area, volume and surface area to volume ratio of each. (4 marks) (b) How many times larger is one ratio than the other? (3 marks) (c) Which model can supply its whole interior by diffusion? Explain. (3 marks) |
Answer: (a) 3.0 cm cube: surface area 6 x 3^2 = 54 cm2, volume 3^3 = 27 cm3, ratio 2.0; 1.0 cm cube: 6 cm2, 1 cm3, ratio 6.0. (b) 6.0 / 2.0 = 3 times. (c) The 1.0 cm model: a larger surface area to volume ratio gives more surface per unit volume for diffusion, and the diffusion distance to the centre is shorter. |
| 4 | An aquatic plant at 25 degrees C and light intensity 20 units releases 36 oxygen bubbles per minute. In complete darkness it still releases 8 bubbles per minute, which represents the respiration rate. (a) State the net rate of photosynthesis in bubbles per minute. (3 marks) (b) Calculate the gross rate of photosynthesis. (4 marks) (c) At light intensity 40 units the rate rises to 52 bubbles per minute. Suggest what limits the rate now. (3 marks) |
Answer: (a) Net rate = 36 bubbles per minute. (b) Gross rate = net + respiration = 36 + 8 = 44 bubbles per minute. (c) Doubling the light adds only 16 bubbles, so light is no longer the only limiting factor; carbon dioxide concentration or temperature (enzyme activity) is now limiting. |
| 5 | At rest an athlete has a stroke volume of 70 mL and a heart rate of 72 beats per minute. (a) Calculate the resting cardiac output in litres per minute. (4 marks) (b) During vigorous exercise the stroke volume is 110 mL and the heart rate is 150 beats per minute. Calculate the cardiac output. (3 marks) (c) How many times the resting value is this? (3 marks) |
Answer: (a) Cardiac output = 70 x 72 = 5040 mL per minute = 5.04 L per minute. (b) 110 x 150 = 16500 mL per minute = 16.5 L per minute. (c) 16.5 / 5.04 = 3.27 times. |
Section B: human physiology, ecology and molecular genetics (50 marks)
Structured questions: show calculations, genetic diagrams and full explanations; numerical answers must carry units. 10 questions, 100 marks, 2 hours 30 minutes, calculator allowed.
| # | Question | Answer and marking notes |
|---|---|---|
| 6 | Five potato strips each of mass 5.0 g are placed in sucrose solutions of different concentrations for 30 minutes. Final masses: 0.0 M 5.6 g; 0.2 M 5.3 g; 0.4 M 5.0 g; 0.6 M 4.7 g. (a) Calculate the percentage change in mass in the 0.0 M solution. (3 marks) (b) Which concentration is isotonic to the cell sap? Explain. (3 marks) (c) Explain why the mass falls in the 0.6 M solution. (4 marks) |
Answer: (a) (5.6 – 5.0) / 5.0 x 100% = +12.0%. (b) 0.4 M: no change in mass means no net movement of water, so the solution and the cell sap have the same water potential. (c) The 0.6 M solution has a lower water potential, so water leaves the cells by osmosis through the cell membrane; the cells lose turgor and may become plasmolysed, so mass falls by 6.0%. |
| 7 | A student investigates the effect of temperature on amylase. Times for complete digestion of 10 cm3 of starch solution: 20 C 40 s; 30 C 16 s; 40 C 8 s; 50 C 20 s; 60 C no reaction. Activity is taken as 100 divided by time in seconds. (a) Calculate the activity at 20 C and at 40 C, and the ratio between them. (3 marks) (b) State the optimum temperature. (3 marks) (c) Explain why the enzyme is completely inactive at 60 C. (4 marks) |
Answer: (a) 20 C: 100 / 40 = 2.5; 40 C: 100 / 8 = 12.5; a ratio of 5 times. (b) The optimum is 40 C, where activity is highest and the time shortest. (c) The high temperature denatures the protein: hydrogen and ionic bonds break, the shape of the active site changes, so starch can no longer bind to form enzyme-substrate complexes. The loss is permanent and is not reversed on cooling. |
| 8 | A 25 cm by 25 cm quadrat is used to survey a plant in a grassland; ten random quadrats contain 80 plants in total. (a) Calculate the area of one quadrat in square metres. (3 marks) (b) Calculate the total area sampled. (2 marks) (c) Calculate the density in plants per square metre. (2 marks) (d) Estimate the population if the grassland is 2,500 square metres. (3 marks) |
Answer: (a) 0.25 x 0.25 = 0.0625 m2. (b) 0.0625 x 10 = 0.625 m2. (c) 80 / 0.625 = 128 plants per m2. (d) 128 x 2,500 = 320,000 plants (an estimate, since quadrat sampling has random error). |
| 9 | At 25 C yeast produces 30 cm3 of carbon dioxide in 10 minutes; at 35 C, with all other conditions the same, it produces 45 cm3. (a) Calculate the gas production rate at each temperature in cm3 per minute. (4 marks) (b) Calculate the percentage increase in rate. (3 marks) (c) Describe how to show that the gas is carbon dioxide. (3 marks) |
Answer: (a) 25 C: 30 / 10 = 3.0 cm3 per minute; 35 C: 45 / 10 = 4.5 cm3 per minute. (b) (4.5 – 3.0) / 3.0 x 100% = 50%. (c) Bubble the gas through limewater: a clear solution turns milky, which shows carbon dioxide. Bicarbonate indicator turning from red to yellow gives the same evidence. |
| 10 | A double-stranded DNA sample contains 1,200 bases, of which adenine (A) is 30%. (a) State the percentage of thymine (T). (3 marks) (b) Calculate the percentage of guanine (G) and of cytosine (C). (4 marks) (c) Calculate the number of cytosine bases in the sample. (3 marks) |
Answer: (a) A pairs with T, so T is also 30%. (b) A + T = 30% + 30% = 60%, leaving 40% shared equally between G and C, so G = C = 20%. (c) 1,200 x 20% = 240 cytosine bases. |
A TutorZone practice paper; original questions and passages, not publisher material.
