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DSE Biology Practice Paper: Cells, Genetics and Ecology
DSE Biology: Cells, Genetics and Ecology
DSE Biology practice paper: 12 questions, 100 marks, 75 minutes on magnification, enzymes, osmosis, genetics, DNA and ecology, with a chi-squared test.
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Paper details Level: S6 | Subject: Biology | Time: 75 minutes | Marks: 100 | Questions: 12
Section A: Cells, enzymes and transport (Questions 1-4, 24 marks)
Answer all questions. In calculations show the formula, the substitution and the units, giving 3 significant figures.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A student observes onion epidermal cells with a compound microscope. (a) The eyepiece is 10x and the objective is 40x; find the total magnification. (b) At this setting a cell has an image diameter of 30 mm; find the actual diameter in µm. (c) Another cell has an image length of 45 mm at a magnification of 1 500x; find its true length. (d) Explain why the eyepiece graticule must be calibrated before actual sizes are calculated. (6 marks) | Answer: (a) Total magnification = 10 x 40 = 400x (1 mark). (b) Actual diameter = 30 mm / 400 = 0.075 mm = 75 µm (2 marks). (c) 45 mm / 1 500 = 0.030 mm = 30 µm (2 marks). (d) The real length of each graticule division changes with the objective, so it must be calibrated against a stage micrometer before divisions can be converted into true lengths (1 mark). Marking: 1 mm = 1 000 µm; no marks for a count of divisions without conversion. |
| 2 | Two cube-shaped model cells have sides of 20 µm and 10 µm. (a) Find the surface area and volume of each. (b) Find the surface area to volume ratio of each in µm⁻¹. (c) Compare the two ratios and explain why small cells absorb nutrients more readily. (6 marks) | Answer: (a) 20 µm cube: surface area 6 x 20 squared = 2 400 µm squared, volume 20 cubed = 8 000 µm cubed; 10 µm cube: 600 µm squared and 1 000 µm cubed (2 marks). (b) The ratios are 2 400 / 8 000 = 0.30 µm⁻¹ and 600 / 1 000 = 0.60 µm⁻¹ (2 marks). (c) Halving the side doubles the ratio from 0.30 to 0.60, so each unit of volume has more membrane area for diffusion and active transport and nutrients are absorbed faster (2 marks). Marking: the answer must refer to the ratio, not to surface area alone. |
| 3 | Yeast and glucose solution are kept in a 25 °C water bath and the gas collected in a syringe every 30 seconds: 12 cm³ in 2 minutes. The same apparatus in a 60 °C bath gives gas at 1.5 cm³ per minute. (a) Find the average rate at 25 °C in cm³/min. (b) Find the percentage decrease in rate at 60 °C. (c) Explain why the rate falls at the higher temperature. (6 marks) | Answer: (a) Rate = 12 cm³ / 2 min = 6.0 cm³/min (2 marks). (b) Decrease = (6.0 – 1.5) / 6.0 x 100% = 75% (2 marks). (c) 60 °C is above the optimum, so the three-dimensional shape of the enzyme and its active site is destroyed (denaturation); the substrate can no longer bind and the rate falls sharply (2 marks). Marking: the terms denaturation and change in active site shape are required. |
| 4 | Three potato strips each of mass 4.00 g are placed in distilled water, 0.3 mol/dm³ sucrose solution and 1.0 mol/dm³ sucrose solution. After 30 minutes their masses are 4.60 g, 3.60 g and 3.30 g. (a) Find the percentage change in mass in distilled water. (b) Find the percentage change in mass in 0.3 mol/dm³ sucrose. (c) Explain the difference between (a) and (b). (6 marks) | Answer: (a) (4.60 – 4.00) / 4.00 x 100% = +15.0% (2 marks). (b) (3.60 – 4.00) / 4.00 x 100% = -10.0% (2 marks). (c) Distilled water has a higher water potential than the cell sap, so water enters by osmosis and the cells become turgid and heavier; the 0.3 mol/dm³ sucrose solution has a lower water potential, so the cells lose water and become lighter (2 marks). Marking: water potential and osmosis must both be named; the sign of each percentage is required. |
Section B: Genetics and molecular biology (Questions 5-8, 32 marks)
Use standard symbols for genotypes in genetics questions and give probabilities as fractions or percentages.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | In pea plants the allele for tall stems T is dominant to the allele for short stems t. Two heterozygous tall plants are crossed and produce 400 offspring. (a) Write the genotypes of the parents and the expected phenotypic ratio. (b) Estimate how many tall and how many short plants are expected. (c) Find the number of homozygous tall plants and their proportion among the tall offspring. (d) If a heterozygous tall plant is test-crossed with a short plant, what ratio and numbers are expected? (8 marks) | Answer: (a) The parents are Tt x Tt, giving 3 tall : 1 short (genotypes 1 TT : 2 Tt : 1 tt) (2 marks). (b) Tall = 400 x 3/4 = 300 and short = 400 x 1/4 = 100 (2 marks). (c) Homozygous tall TT = 400 x 1/4 = 100, which is 100/300 = 1/3 of the tall offspring (2 marks). (d) Tt x tt gives 1 Tt : 1 tt, that is 200 tall and 200 short (2 marks). Marking: the 1:2:1 genotypic ratio and the 3:1 phenotypic ratio must not be confused. |
| 6 | In peas, round R is dominant to wrinkled r and yellow Y is dominant to green y; the two genes are on different chromosomes. Two round yellow heterozygous plants are crossed and produce 320 seeds. (a) State the expected phenotypic ratio. (b) Calculate the expected number of each phenotype. (c) Explain the cytological basis of independent assortment. (d) If 200 round yellow seeds are actually observed, is this above or below expectation? (8 marks) | Answer: (a) Round yellow : round green : wrinkled yellow : wrinkled green = 9 : 3 : 3 : 1 (2 marks). (b) 320 x 9/16 = 180 round yellow, 60 round green, 60 wrinkled yellow and 320 x 1/16 = 20 wrinkled green (4 marks). (c) The two genes are on different chromosomes; in meiosis I the homologous pairs line up randomly at the equator so the non-homologous chromosomes assort independently and the alleles are distributed independently (1 mark). (d) 200 is above the expected 180, so the observation is high, but a chi-squared test is needed before calling the difference significant (1 mark). |
| 7 | Red-green colour blindness is caused by a recessive allele b on the X chromosome. A carrier mother (X^B X^b) marries a man with normal vision (X^B Y). (a) Deduce the chance of colour blindness in their sons and daughters. (b) Find the probability that all four of their children have normal vision. (c) Explain why colour blindness is far more common in sons than in daughters. (8 marks) | Answer: (a) Sons: with a X^B X^b mother and a X^B Y father, half of the sons are X^b Y and are colour blind (50%); all daughters are X^B X^B or X^B X^b and have normal vision, half of them being carriers (4 marks). (b) Each child has a 3/4 chance of normal vision, so all four are normal with probability 0.75 to the fourth power = 0.316, about 31.6% (4 marks). Marking: sons and daughters must be treated separately. |
| 8 | Analysis of a DNA sample shows that adenine (A) makes up 30% of the bases. In a population at Hardy-Weinberg equilibrium a recessive condition affects 1 person in 10 000. (a) Find the percentage of guanine (G) and cytosine (C). (b) Find the recessive allele frequency q and the dominant allele frequency p. (c) Find the percentage of carriers (heterozygotes) in the population. (8 marks) | Answer: (a) A pairs with T and G with C, so T = 30% and G = C = (100% – 60%) / 2 = 20% (2 marks). (b) q squared = 1 / 10 000 = 0.0001, so q = 0.01 and p = 1 – 0.01 = 0.99 (3 marks). (c) Carriers = 2pq = 2 x 0.99 x 0.01 = 0.0198, that is 1.98% (3 marks). Marking: no marks if p squared is used instead of 2pq. |
Section C: Ecology, energy flow and data analysis (Questions 9-12, 44 marks)
Use correct biological terminology in explanations; in experimental questions state the assumptions, sources of error and how each can be reduced.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A student estimates the number of fish in a pond by mark-recapture: 40 fish are caught, marked and released; a week later 50 fish are caught and 8 of them are marked. Separately, 5 quadrats of 1 m² in a meadow contain 12 plants each on average. (a) Estimate the total number of fish. (b) If the meadow is 200 m², estimate the total number of plants. (c) State two assumptions of the mark-recapture method, and say whether the estimate is too high or too low if marking makes fish easier to catch. (10 marks) | Answer: (a) N = (40 x 50) / 8 = 250 fish (3 marks). (b) Density is 12 plants per m², so the total is 12 x 200 = 2 400 plants (3 marks). (c) Assumptions: marking does not affect survival or the chance of being caught; the population is closed with no immigration, emigration, births or deaths during the study; marks do not fall off (2 marks for two valid assumptions). If marking makes fish easier to catch, the proportion of marked fish in the second sample is too high and the estimate is too low (2 marks). |
| 10 | In a grassland the producers fix 50 000 kJ m⁻² yr⁻¹ and only 10% of the energy passes to each successive trophic level. (a) Find the energy available to the primary, secondary and tertiary consumers. (b) Find the transfer efficiency between the primary and secondary consumers. (c) Explain where the other 90% of the energy goes. (10 marks) | Answer: (a) Primary consumers 50 000 x 0.1 = 5 000, secondary 500 and tertiary 50 kJ m⁻² yr⁻¹ (4 marks). (b) Efficiency = 500 / 5 000 x 100% = 10% (3 marks). (c) The remaining energy is lost as heat from respiration, lost in excretory products and undigested material, and passes to decomposers in the parts not eaten, so less energy is available at each higher trophic level (3 marks). Marking: at least three routes are required. |
| 11 | Elodea is placed at different distances from a lamp and the number of bubbles per minute is used as a measure of the rate of photosynthesis: 60 bubbles at 0 cm, 30 at 20 cm and 15 at 40 cm. (a) List the three readings. (b) Describe the relationship between distance and bubble count. (c) Find the total number of bubbles in 10 minutes at 0 cm. (d) When the distance rises from 0 cm to 20 cm the light intensity falls to one quarter by the inverse square law, yet the rate only halves; explain why. (12 marks) | Answer: (a) 60 bubbles per minute at 0 cm, 30 at 20 cm and 15 at 40 cm (2 marks). (b) Every extra 20 cm of distance halves the bubble count, so the rate is inversely proportional to distance (2 marks). (c) 60 x 10 = 600 bubbles (3 marks). (d) From 0 to 20 cm the light intensity falls to one quarter, so if light were the only limiting factor the rate should fall to 15 bubbles per minute; the measured 30 shows that light is no longer the only limiting factor and that carbon dioxide concentration or temperature is now limiting, so the rate only halves (5 marks). Marking: the idea of a limiting factor is required. |
| 12 | 120 seeds are placed in a dish: 84 germinate and 36 do not. If the germination rate were 50%, 60 would germinate and 60 would not. Use a chi-squared test to decide whether the observations deviate significantly from the expectation (1 degree of freedom, 5% critical value 3.84). (a) State the expected values. (b) Calculate chi-squared. (c) State your conclusion. (d) Find the actual germination rate and suggest two reasons why it is higher than 50%. (12 marks) | Answer: (a) Expected: 60 germinating and 60 non-germinating seeds (2 marks). (b) Chi-squared = (84 – 60) squared / 60 + (36 – 60) squared / 60 = 9.6 + 9.6 = 19.2 (4 marks). (c) With 1 degree of freedom the 5% critical value is 3.84; 19.2 is greater, so the null hypothesis is rejected and the results deviate significantly from the 50% expectation (3 marks). (d) Germination rate = 84 / 120 x 100% = 70.0%; possible reasons: the seeds had high viability, the temperature and moisture in the dish were favourable, or the seeds had already been treated to break dormancy (3 marks). Marking: the conclusion must compare the value with the critical value and reject the null hypothesis. |
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