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HKDSE Biology Practice Paper: Cells, Genetics and Ecology (12 questions, 100 marks, 75 minutes)
HKDSE Biology: Cells, Genetics and Ecology
Twelve questions on magnification, osmosis, enzymes, inheritance, protein synthesis, mitosis, energy flow and mark-recapture, with full answers and marking points.
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Paper details Level: S6 | Subject: Biology | Time: 75 minutes | Marks: 100 | Questions: 12
Section A (32 marks)
Multi-step calculations and data analysis; show working and units.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A specimen is viewed with a x10 eyepiece and a x40 objective. The image measures 2.4 mm in length. (a) Calculate the total magnification. (b) Calculate the actual length of the specimen in micrometres. (c) If the x10 objective is used instead, what will the image length be in mm? |
(a) Total magnification = 10 x 40 = 400. [2] (b) Actual length = 2.4 mm / 400 = 0.006 mm = 6 micrometres. [3] (c) Image length = 2.4 x 10 / 40 = 0.6 mm. [3] |
| 2 | Two cube-shaped cells have sides of 1 cm and 2 cm. (a) Calculate the surface area to volume ratio of each cell in cm-1. (b) Which cell can obtain enough materials by diffusion alone? Explain. |
(a) 1 cm cell: 6 x 1 squared / 1 cubed = 6 cm-1; 2 cm cell: 6 x 2 squared / 2 cubed = 3 cm-1. [4] (b) The 1 cm cell, because its surface area to volume ratio is larger (6 against 3), so diffusion across its membrane supplies its volume fast enough. [4] |
| 3 | 20 g of sucrose is dissolved in 250 cm3 of distilled water. (a) Calculate the concentration in g dm-3. (b) A potato strip of mass 4.0 g is placed in the solution. After 30 minutes its mass is 3.6 g. Calculate the percentage change in mass. (c) Explain the change in mass. |
(a) Concentration = 20 g / 0.25 dm3 = 80 g dm-3. [2] (b) Percentage change = (3.6 – 4.0) / 4.0 x 100 percent = -10 percent. [3] (c) The solution is more concentrated than the cell sap, so water leaves the cells by osmosis and the strip loses mass. [3] |
| 4 | An enzyme produces 36 bubbles in 3 minutes at 30 degrees C, but only 12 bubbles in 3 minutes at 60 degrees C. (a) Calculate the average rate of gas production at 30 degrees C in bubbles per minute. (b) Calculate the rate at 60 degrees C as a percentage of the rate at 30 degrees C. (c) Explain why the rate is lower at 60 degrees C. |
(a) Rate = 36 / 3 = 12 bubbles per minute. [2] (b) 12 / 36 x 100 percent = 33.3 percent, about one third. [3] (c) At 60 degrees C the enzyme is denatured, the active site changes shape and fewer enzyme-substrate complexes form. [3] |
Section B (36 marks)
Inheritance and cell division; use diagrams or ratios.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | Two heterozygous tall pea plants (Tt) are crossed and produce 640 offspring. (a) Use a genetic diagram to find the phenotypic ratio of the offspring. (b) Calculate the expected number of tall plants. (c) Only 470 tall plants are counted. Suggest one reason for the difference. |
(a) Tt x Tt gives TT : Tt : tt = 1 : 2 : 1, so tall : short = 3 : 1. [4] (b) 640 x 3 / 4 = 480 tall plants. [3] (c) Fertilisation is random and the sample is finite, so the observed number usually differs from the expected number. [2] |
| 6 | Colour blindness is caused by a recessive allele (b) on the X chromosome. A carrier mother (XBXb) marries a normal father (XBY). (a) State the ratio of phenotypes among the children. (b) If they have 4 sons, how many are expected to be colour-blind? (c) Explain why none of the daughters is colour-blind. |
(a) Daughters are XBXB or XBXb, all normal (half are carriers); sons are XBY or XbY, half colour-blind; overall normal : colour-blind = 3 : 1. [4] (b) 4 x 1/2 = 2 sons are expected to be colour-blind. [3] (c) A daughter receives the normal XB from her father, which is dominant to b, so she cannot be affected. [2] |
| 7 | A mRNA molecule has 90 nucleotides. The first three form the start codon and the last three form the stop codon. Each amino acid is coded by three bases. (a) Calculate the number of amino acids coded by this mRNA. (b) A protein has 150 amino acids, excluding the stop codon. How many nucleotides are needed in its coding region at least? (c) State two differences in base composition between DNA and mRNA. |
(a) (90 – 3) / 3 = 29 amino acids. [3] (b) 150 x 3 = 450 nucleotides. [3] (c) DNA contains thymine while mRNA contains uracil; DNA is double-stranded while mRNA is usually single-stranded. [3] |
| 8 | A fertilised egg divides by mitosis 4 times. (a) Calculate the number of cells produced. (b) A human body cell has 46 chromosomes. State the chromosome number of each cell after the divisions. (c) Explain the importance of mitosis for growth. |
(a) Number of cells = 2 to the power 4 = 16 cells. [3] (b) 46 chromosomes in each cell. [3] (c) Mitosis produces genetically identical diploid cells, so the organism gains cells for growth and repair without changing its genetic make-up. [3] |
Section C (32 marks)
Ecology, health and application; justify your reasoning.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | Producers in an ecosystem fix 50,000 kJ of energy. The efficiency of energy transfer is 10 percent at each trophic level. (a) Calculate the energy available to primary consumers. (b) Calculate the energy available to secondary consumers. (c) Explain why food chains rarely have more than four or five trophic levels. |
(a) Primary consumers receive 50,000 x 10 percent = 5,000 kJ. [2] (b) Secondary consumers receive 5,000 x 10 percent = 500 kJ. [2] (c) About 90 percent of the energy is lost at each level through respiration, heat and undigested material, so too little remains to support higher levels. [4] |
| 10 | In a mark-recapture study, 40 animals are caught and marked. Later, 50 animals are caught and 5 of them are marked. (a) Estimate the population size. (b) State two assumptions of this method. (c) If marking makes animals easier to catch again, will the estimate be too high or too low? Explain. |
(a) N = 40 x 50 / 5 = 400 animals. [3] (b) The population is closed (no migration, births or deaths between samples); marks do not affect survival or catchability and are not lost. [3] (c) The estimate is too low, because the proportion of marked animals in the second sample is overestimated, which makes N smaller. [2] |
| 11 | Two unaffected parents are both carriers (Aa) of a recessive disorder and already have one affected child (aa). (a) State the probability that the next child is affected. (b) State the probability that the next child is a carrier. (c) If they plan to have 4 children, how many are expected to be affected? (d) Explain why marriage between close relatives increases the risk to the children. |
(a) The probability is 1/4 (25 percent). [2] (b) The probability is 1/2 (50 percent). [2] (c) 4 x 1/4 = 1 child is expected to be affected. [2] (d) Close relatives are more likely to carry the same rare recessive allele, so two carriers are more likely to marry and produce aa children. [2] |
| 12 | A leaf absorbs carbon dioxide at 0.6 cm3 per minute in the light and releases carbon dioxide at 0.2 cm3 per minute in the dark. (a) Calculate the true rate of photosynthesis in cm3 per minute. (b) Calculate the net carbon dioxide absorbed in 30 minutes in the light. (c) Explain the effect of increasing light intensity on the two rates. |
(a) True rate of photosynthesis = 0.6 + 0.2 = 0.8 cm3 per minute. [3] (b) Net absorption = 0.6 x 30 = 18 cm3. [2] (c) The rate of photosynthesis rises with light intensity until another factor becomes limiting, while the respiration rate stays roughly constant. [3] |
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