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Form 5 Biology Practice Paper: Respiration, Photosynthesis, Genetics and Ecology (12 Questions, 100 Marks, 75 Minutes)
Form 5 Biology – 12 questions – 100 marks – 75 minutes
Form 5 Biology paper: 12 questions, 100 marks, 75 minutes on respiration, enzyme Q10, photosynthesis rates, chromatography Rf values, genetics ratios and ecology estimates.
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Paper details Form 5 Biology practice paper with 12 structured questions on respiration, enzyme kinetics, photosynthesis, pigment chromatography, monohybrid and dihybrid crosses, sex-linked inheritance, DNA base pairing, mark-recapture, energy transfer, quadrat sampling and eutrophication.
Section A: Respiration, Photosynthesis and Enzymes (Q1-Q4, 34 marks)
Answer all questions; show the formula, substitution and unit for every calculation.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | 1. The efficiency of aerobic and anaerobic respiration is a key comparison. (a) Calculate the amount, in moles, of glucose in 0.90 g of glucose. (2 marks) (b) If one glucose molecule yields 38 ATP in aerobic respiration, find the moles of ATP produced. (2 marks) (c) If the same glucose is respired anaerobically and yields only 2 ATP per molecule, find the moles of ATP. (2 marks) (d) Find how many times more ATP aerobic respiration produces than anaerobic respiration. (2 marks) | Answer: (a) Amount = 0.90 / 180 = 0.01 mol, that is 5 mmol. (b) Aerobic respiration yields 0.01 x 38 = 0.19 mol ATP, that is 190 mmol. (c) Anaerobic respiration yields only 0.01 x 2 = 0.01 mol ATP, that is 10 mmol. (d) The aerobic yield is 0.19 / 0.01 = 18 times greater, so aerobic respiration is far more efficient. |
| 2 | 2. An enzyme catalyses at a rate of 12.0 units per minute at 20 degrees and 24.0 units per minute at 30 degrees. (a) Calculate the temperature coefficient Q10 of the enzyme. (3 marks) (b) Assuming the rate rises by the same coefficient, predict the theoretical rate at 40 degrees. (2 marks) (c) The measured rate at 40 degrees is only 30.0 units per minute. Explain the difference and state the optimum temperature range. (4 marks) | Answer: (a) Q10 = rate at 30 degrees / rate at 20 degrees = 24.0 / 12.0 = 2. (b) Predicted rate at 40 degrees = 24.0 x 2 = 48 units per minute. (c) The measured rate is only 30.0 units per minute because 40 degrees is above the optimum: the tertiary structure of the enzyme is disrupted, the active site changes shape and some enzyme molecules are permanently denatured, so the rate no longer follows Q10; the optimum is about 35 to 37 degrees. |
| 3 | 3. A leaf at a light intensity of 20 kilolux shows a net carbon dioxide uptake of 18.0 units, while its respiration rate is 6.0 units. (a) Find the gross rate of photosynthesis at this light intensity. (3 marks) (b) Explain the meaning of the light compensation point and state the net uptake at that point. (3 marks) (c) State two external factors that may become limiting at high light intensity. (2 marks) | Answer: (a) Gross photosynthesis = net uptake + respiration = 18 + 6 = 24 units. (b) The light compensation point is the light intensity at which gross photosynthesis exactly equals respiration, so net carbon dioxide uptake is 0; the plant accumulates no net organic matter and would weaken if kept at that intensity. (c) Low carbon dioxide concentration, high temperature or restricted water supply can close stomata or reduce enzyme activity and become limiting. |
| 4 | 4. Chloroplast pigments are separated by paper chromatography. The solvent front moves 9.0 cm and four pigment spots move 1.8, 3.6, 5.4 and 7.2 cm. (a) Calculate the Rf value of the pigment that moved 3.6 cm. (3 marks) (b) Calculate the Rf values of the other three pigments. (4 marks) (c) Identify which pigment is most likely carotene and explain why. (2 marks) | Answer: (a) Rf = 3.6 / 9.0 = 0.4. (b) The other three are 1.8 / 9.0 = 0.2, 5.4 / 9.0 = 0.6 and 7.2 / 9.0 = 0.8. (c) The pigment that moved furthest, with the highest Rf of 0.8, is carotene, because it is the most soluble in the non-polar solvent and is least adsorbed on the paper. |
Section B: Inheritance and Molecular Genetics (Q5-Q8, 33 marks)
Data: use 180 as the molar mass of glucose, energies are in kJ, and give percentage answers to one decimal place.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | 5. Two pea plants, both of genotype Tt, are crossed and produce 640 offspring. (a) Use a Punnett square to give the genotypes and their ratio in the offspring. (3 marks) (b) Find the expected numbers of the dominant and recessive phenotypes. (3 marks) (c) Explain how a test cross can determine the genotype of a plant showing the dominant phenotype. (2 marks) | Answer: (a) The genotypic ratio is TT : Tt : tt = 1 : 2 : 1 and the phenotypic ratio is 3 : 1. (b) Dominant phenotype = 640 x 3 / 4 = 480 plants; recessive phenotype = 640 x 1 / 4 = 160 plants. (c) Cross the unknown plant with a homozygous recessive (tt) plant: if all offspring show the dominant phenotype the parent is TT, while a ratio near 1 : 1 of dominant to recessive offspring shows the parent is Tt. |
| 6 | 6. Two individuals, both of genotype AaBb, are crossed and produce 1 600 offspring; the two gene loci are on different chromosome pairs. (a) State the types of gametes produced by each parent. (2 marks) (b) Find the expected ratio and numbers of the four phenotypes. (4 marks) (c) Name the genetic law used and state its premise. (3 marks) | Answer: (a) Each parent produces four gamete types, AB, Ab, aB and ab, in equal proportions. (b) The phenotypic ratio is 9 : 3 : 3 : 1: A_B_ = 1600 x 9 / 16 = 900, A_bb = 1600 x 3 / 16 = 300, aaB_ = 1600 x 3 / 16 = 300 and aabb = 1600 x 1 / 16 = 100 plants. (c) The calculation uses the law of independent assortment, which assumes the two loci are on different chromosome pairs so that their segregation during meiosis is independent. |
| 7 | 7. Haemophilia is caused by a recessive allele on the X chromosome. A carrier mother (X^H X^h) marries a normal father (X^H Y). (a) Give the possible genotypes and ratio of their sons and daughters. (3 marks) (b) If the couple has four sons, find the probability that all four are normal. (3 marks) (c) Explain why none of the daughters is affected. (2 marks) | Answer: (a) Sons receive one X chromosome from the mother: half are X^H Y (normal) and half are X^h Y (affected), a ratio of 1 : 1; daughters receive X^H from the father, so half are X^H X^H (normal) and half are X^H X^h (carriers), also 1 : 1. (b) Each son has probability 0.5 of being normal, so all four are normal with probability 0.5 to the power 4 = 0.06, about 6.25%. (c) The father contributes an X chromosome carrying the dominant allele X^H, which masks the recessive allele from the mother, so daughters are not affected but may be carriers who pass the allele to the next generation. |
| 8 | 8. A DNA double helix contains 1 200 base pairs, and adenine makes up 30% of all bases. (a) Find the total number of bases in this DNA. (2 marks) (b) Find the number of adenine and thymine bases. (3 marks) (c) Find the number of guanine bases and state the base pairing rule used. (3 marks) | Answer: (a) Total bases = 1 200 x 2 = 2400. (b) Adenine = 2400 x 30% = 720; thymine pairs with adenine in equal number, so there are 720 thymine bases. (c) Guanine plus cytosine = 2400 – 1440 = 960, and the two are equal in number, so guanine = 960 / 2 = 480 bases, using the complementary base pairing rule of adenine with thymine and guanine with cytosine. |
Section C: Ecology and Environment (Q9-Q12, 33 marks)
Write explanations in full sentences and support each point with the calculated values or data given.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | 9. Researchers catch 120 fish in a pond, mark them and release them. A week later they catch 80 fish, of which 12 are marked. (a) Estimate the total fish population using the Lincoln index. (3 marks) (b) State two assumptions that make this estimate valid. (3 marks) (c) If some marks fall off during the week, will the estimate be too high or too low? Explain. (3 marks) | Answer: (a) Lincoln index = marked x second catch / marked in second catch = 120 x 80 / 12 = 800 fish. (b) Assumptions: marking does not affect survival or the chance of being caught, marked fish have mixed evenly with the population before the second catch, and the population shows no significant immigration, emigration, birth or death. (c) If some marks fall off, the number of marked fish in the second catch is too small, the denominator shrinks and the estimate is too high. |
| 10 | 10. Producers in an ecosystem fix 250 000 kJ of energy per square metre per year, and the transfer efficiency between trophic levels is 10%. (a) Find the energy available to primary consumers. (2 marks) (b) Find the energy available to the third trophic level. (2 marks) (c) Explain where the rest of the energy goes and why food chains usually have only four or five trophic levels. (4 marks) | Answer: (a) Primary consumers receive 250 000 x 10% = 25000 kJ. (b) The third trophic level receives 25000 x 10% x 10% = 250 kJ. (c) About nine tenths of the energy is released by respiration and lost as heat, and some passes to decomposers in faeces and remains; only about one tenth reaches the next level, so by the fourth or fifth level too little energy remains to support a viable population. |
| 11 | 11. The plants in a meadow are sampled with 10 quadrats, each of area 0.25 square metres, and the counts total 75 plants. The survey area is 2 500 square metres. (a) Find the mean number of plants per quadrat. (2 marks) (b) Find the plant density per square metre. (3 marks) (c) Estimate the total number of plants in the survey area and state how sampling accuracy can be improved. (3 marks) | Answer: (a) Mean = 75 / 10 = 7.5 plants per quadrat. (b) Density = 7.5 / 0.25 = 30 plants per square metre. (c) Total = 30 x 2500 = 75000 plants; use random sampling with quadrat positions chosen from random numbers and increase the number of quadrats to reduce sampling error. |
| 12 | 12. Dissolved oxygen in a river is 9.2 mg per litre upstream and falls to 2.3 mg per litre downstream after a sewage discharge. (a) Find the percentage fall in dissolved oxygen. (3 marks) (b) Describe in order the three steps from nutrient enrichment to fish death. (3 marks) (c) Explain why better sewage treatment allows dissolved oxygen to recover. (2 marks) | Answer: (a) Percentage fall = (9.2 – 2.3) / 9.2 x 100% = 75%. (b) Sewage adds nitrates and phosphates, so algae and aquatic plants grow rapidly; when they die, decomposers break them down and use up dissolved oxygen; oxygen levels then fall so low that fish and other aerobes suffocate. (c) Better treatment removes most nutrients, limiting algal growth, so the oxygen demand of decomposers falls and dissolved oxygen recovers through diffusion from the air and photosynthesis by aquatic plants. |
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