S5 Biology Practice Paper: Genetics, Ecology And Bioengineering (12 Questions, 100 Marks, 75 Minutes, Answers Included)

S5 Biology Practice Paper: Genetics, Ecology and Bioengineering (12 Questions, 100 Marks, 75 Minutes, Answers Included)

2026-11-06

S5 Biology: Genetics, Evolution, Ecology and Bioengineering

S5 Biology paper: 12 questions, 100 marks, 75 minutes on magnification, DNA base composition, monohybrid crosses, Hardy-Weinberg, chi-square, trophic levels and biomagnification.

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Section A: Cells and Metabolism (24 marks)

Each question carries 6 marks; show your working or reasoning.

#	Question	Answer and marking notes
1	A student observes onion epidermal cells. With a 10× eyepiece and a 15× objective, a cell in the photograph measures 45 mm while its actual length is 30 µm. (a) Calculate the magnification of the photograph. (b) The student then uses a total magnification of 400×; a second cell measures 60 mm in the photograph. Calculate its actual length in µm.	(a) Magnification = image length ÷ actual length = 45 mm ÷ 0.030 mm = 1500×. Note that 10× times 15× gives only 150×; the photograph is enlarged further, so the measured ratio must be used. (b) Actual length = 60 mm ÷ 400 = 0.15 mm = 150 µm. Marking points: correct magnification formula, conversion between mm and µm, answers with units.
2	A viral DNA molecule contains 1,200 base pairs, of which adenine (A) makes up 32% of all bases. (a) State the percentages of thymine (T), guanine (G) and cytosine (C). (b) Calculate the numbers of A-T and G-C base pairs. (c) Calculate the total number of hydrogen bonds.	(a) A = T = 32%, so G = C = 18%. (b) Of 1,200 base pairs, A-T pairs = 768 and G-C pairs = 432. (c) Hydrogen bonds = 2 × 768 + 3 × 432 = 2,832. Marking points: A pairs with T and G with C, two bonds per A-T pair and three per G-C pair.
3	Two agar cubes A and B with sides of 2 mm and 20 mm are placed in a dye solution. (a) Calculate the surface area to volume ratio of A and of B in mm⁻¹. (b) How many times larger is the ratio of A than that of B? (c) Explain why the smaller cube is stained through a greater percentage of its volume.	(a) A: surface area 24 mm², volume 8 mm³, ratio 3 mm⁻¹; B: surface area 2,400 mm², volume 8,000 mm³, ratio 0.3 mm⁻¹. (b) The ratio of A is 10 times that of B. (c) A larger surface area to volume ratio gives more surface per unit volume for dye entry, so a greater percentage of the small cube is stained. Marking points: formula, units, explanation linked to the ratio.
4	During photosynthesis a leaf in bright light absorbs 0.88 g of carbon dioxide in one hour. (a) Calculate the moles of CO₂ fixed (molar mass 44 g/mol). (b) Using 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂, calculate the mass of glucose formed (molar mass 180 g/mol). (c) Calculate the volume of oxygen released in the same hour in cm³ (1 mol of gas = 24 dm³ at room temperature).	(a) n(CO₂) = 0.88 ÷ 44 = 0.02 mol. (b) n(glucose) = 0.02 ÷ 6 = 0.00333 mol, so mass = 0.00333 × 180 = 0.60 g. (c) V(O₂) = 0.02 × 24 dm³ = 0.48 dm³ = 480 cm³. Marking points: 6:1:6 stoichiometry, separate mass and volume steps.

Section B: Genetics and Evolution (36 marks)

Each question carries 9 marks; note the marks for each part.

#	Question	Answer and marking notes
5	In peas the tall allele T is dominant to dwarf t. In a monohybrid cross two heterozygous tall plants produce 400 offspring. (a) State the parental genotypes and the gametes produced. (b) Predict the numbers of tall and dwarf offspring. (c) If one heterozygous tall plant is crossed with a dwarf plant, how many dwarf plants are expected among 400 offspring?	(a) The parents are Tt × Tt and each produces T and t gametes. (b) Genotypes are 1 TT : 2 Tt : 1 tt, so the phenotypic ratio is 3 tall : 1 dwarf, that is 300 tall and 100 dwarf. (c) Tt × tt gives 1 Tt : 1 tt, so about 200 dwarf plants are expected. Marking points: Punnett square, clear ratio, conversion to numbers.
6	In peas yellow cotyledons Y are dominant to green y, and round seeds R are dominant to wrinkled r; the two genes are on different chromosomes. In a dihybrid cross two heterozygous plants (YyRr) give 640 seeds. (a) State the phenotypic ratio. (b) Calculate the expected numbers of yellow round, yellow wrinkled, green round and green wrinkled seeds. (c) Calculate the expected number of double recessive (yyrr) seeds and their percentage of the offspring.	(a) The phenotypic ratio is 9:3:3:1. (b) Expected yellow round 360, yellow wrinkled 120, green round 120 and green wrinkled 40 seeds. (c) The double recessive yyrr is 1/16, that is 40 seeds, or 6.25% of the offspring. Marking points: independent assortment, all four phenotypes, percentage method.
7	A recessive disorder affects 1 person in 10,000 in a population at Hardy-Weinberg equilibrium. (a) Calculate the recessive allele frequency q and the dominant allele frequency p. (b) Calculate the carrier (heterozygote) frequency as a percentage. (c) Estimate the number of carriers in every 1,000 people.	(a) q² = 1/10,000, so q = 0.01 and p = 0.99. (b) Carrier frequency 2pq = 2 × 0.99 × 0.01 = 0.0198, that is 1.98%. (c) About 19.8 carriers per 1,000 people. Marking points: square root of q², the 2pq formula, conversion to a percentage.
8	A student classifies 400 offspring as 315 tall and 85 dwarf and tests the 3:1 expectation with a chi-square test. (a) State the expected numbers. (b) Calculate the chi-square value. (c) The critical value at p = 0.05 with 1 degree of freedom is 3.84; decide whether the 3:1 hypothesis is rejected and explain your answer.	(a) Expected 300 tall and 100 dwarf. (b) chi-square = (315 – 300)² ÷ 300 + (85 – 100)² ÷ 100 = 3.00. (c) 3.00 is smaller than the critical value 3.84, so the deviation is not significant and the 3:1 hypothesis is not rejected. Marking points: expected values, chi-square formula, comparison with the critical value.

Section C: Ecology and Bioengineering (40 marks)

Each question carries 10 marks; give full working or explanations.

#	Question	Answer and marking notes
9	Producers in an ecosystem fix 50,000 kJ per square metre per year and 10% of the energy passes to the next trophic level. (a) Calculate the energy available to the primary, secondary, tertiary and quaternary consumers. (b) If the quaternary consumers need 20 kJ per square metre per year, calculate the producer energy required. (c) Explain why food chains rarely contain more than four or five trophic levels.	(a) 5,000, 500, 50 and 5 kJ respectively. (b) The producers must supply 20 ÷ 10⁻⁴ = 200,000 kJ. (c) About 90% of the energy is lost at each transfer, so after four or five trophic levels too little energy remains to support another level. Marking points: repeated multiplication by 0.1, working backwards, explanation based on energy loss.
10	A mark-recapture study estimates the fish population of a pond: 40 fish are marked and released, and a week later 50 fish are caught of which 8 are marked. (a) Estimate the population using the Lincoln index. (b) What would the estimate be if only 5 of the 50 were marked? (c) State two assumptions of the method.	(a) N = (M × C) ÷ R = (40 × 50) ÷ 8 = 250 fish. (b) With 5 marked recaptures, N = (40 × 50) ÷ 5 = 400 fish. (c) Assumptions: marking does not affect survival or catchability; the population is closed (no births, deaths, immigration or emigration); marked animals mix evenly with the population. Marking points: formula, substitution, any two assumptions.
11	0.60 g of glucose is fermented anaerobically by yeast and the carbon dioxide is collected over water. The reaction is C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ and collection takes 8 minutes. (a) Calculate the moles of glucose (molar mass 180 g/mol). (b) Calculate the theoretical volume of carbon dioxide in cm³ (1 mol = 24 dm³). (c) Calculate the mean rate of gas production in cm³ per minute.	(a) n(glucose) = 0.60 ÷ 180 = 0.00333 mol. (b) n(CO₂) = 2 × 0.00333 = 0.00667 mol, so volume = 0.00667 × 24,000 cm³ = 160 cm³. (c) Rate = 160 ÷ 8 = 20 cm³ per minute. Marking points: 1:2 ratio, conversion between dm³ and cm³, rate as volume divided by time.
12	The water of a lake contains 0.000002 ppm of DDT. DDT is fat soluble and its concentration rises tenfold at each trophic level. (a) Calculate the DDT concentration in phytoplankton, zooplankton, small fish and large fish in ppm, and (b) the concentration factor of the large fish relative to the water. (c) Explain why fish-eating birds at the top level are affected most severely; name the process involved.	(a) Phytoplankton 0.00002 ppm, zooplankton 0.0002 ppm, small fish 0.002 ppm and large fish 0.02 ppm. (b) The concentration factor is 0.02 ÷ 0.000002 = 10,000 times. (c) DDT is fat soluble and degrades very slowly, so it accumulates in fatty tissue along the food chain; birds at the top level eat large amounts and excrete it slowly, so they carry the highest concentration. The process is biomagnification. Marking points: tenfold steps, factor calculation, fat solubility and biomagnification.

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