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Form 5 Biology Practice Paper: Genetics, DNA and Ecology (12 questions, 100 marks, 75 minutes)
Form 5 Biology – 12 questions – 100 marks – 75 minutes
Form 5 biology practice paper: 12 questions, 100 marks, 75 minutes on DNA, inheritance, translation, energy flow and ecology, with worked answers and marking points.
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Paper details Form 5 biology practice paper with 12 original questions on DNA structure, genetics, energy flow, nutrient cycles and conservation, each with full worked answers.
Section A: Genetics and DNA (Q1-Q4)
Genetics answers must give genotypes and gametes, with a Punnett square or probability reasoning.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A double-stranded DNA sample contains adenine (A) making up 20% of all bases. (a) Calculate the percentage of thymine (T), guanine (G) and cytosine (C). (3 marks) (b) Explain the base-pairing rule used. (2 marks) (c) If the DNA has 2000 base pairs, calculate the number of G bases. (3 marks) | (a) A pairs with T, so T = 20%; A + T = 40%, leaving 60% shared equally, so G = C = 30%. (3 marks) (b) A pairs with T by two hydrogen bonds and G pairs with C by three, so A = T and G = C in double-stranded DNA. (2 marks) (c) Total bases = 4000; G = 30% of 4000 = 1200. (3 marks) |
| 2 | In peas, round seeds (R) are dominant to wrinkled seeds (r). A round-seeded plant is crossed with another round-seeded plant, giving 787 round and 263 wrinkled offspring. (a) State the genotype of both parents. (2 marks) (b) Calculate the ratio of round to wrinkled offspring and explain how it arises. (4 marks) (c) If a wrinkled plant is crossed with an Rr plant, what phenotypic ratio is expected? (2 marks) | (a) Both parents are Rr. (2 marks) (b) 787 divided by 263 = 2.99, close to 3:1. Each parent makes R and r gametes in equal numbers; random fusion gives RR:Rr:rr = 1:2:1 and, since R is dominant, a 3:1 phenotypic ratio. (4 marks) (c) rr x Rr gives half Rr and half rr, so round:wrinkled = 1:1. (2 marks) |
| 3 | The coding region of an mRNA molecule contains 300 nucleotides. (a) Calculate the number of codons formed. (2 marks) (b) If one codon is a stop codon, calculate the maximum number of amino acids in the polypeptide. (3 marks) (c) Explain why the number of amino acids measured in the cell may be smaller than your answer in (b). (3 marks) | (a) 300 divided by 3 = 100 codons. (2 marks) (b) 100 – 1 = 99, so at most 99 amino acids. (3 marks) (c) The initial methionine may be removed, a signal peptide may be cut off, and the protein may be modified or cleaved after translation, so the mature protein can be shorter. (3 marks) |
| 4 | On an island, 1 person in 10000 has an autosomal recessive disease, and the population is in Hardy-Weinberg equilibrium. (a) Calculate the frequency of the recessive allele q. (3 marks) (b) Calculate the dominant allele frequency p and the expected number of carriers. (3 marks) (c) State three assumptions required by this calculation. (3 marks) | (a) q squared = 1/10000 = 0.0001, so q = 0.01. (3 marks) (b) p = 0.99; carrier frequency 2pq = 0.0198, so 198 carriers. (3 marks) (c) Large population, random mating, no mutation, no selection and no migration. (3 marks) |
Section B: Energy flow and nutrient cycles (Q5-Q8)
Calculations need the formula, substitution and units; estimates must state sources of error.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | In an ecosystem, producers fix 50000 kJ per square metre per year, and the energy transfer efficiency between trophic levels is 10%. (a) Calculate the energy available to primary, secondary and tertiary consumers. (3 marks) (b) Explain two reasons for the large energy loss between levels. (4 marks) (c) Explain why food chains rarely have more than five trophic levels. (1 mark) | (a) Primary consumers 5000 kJ, secondary consumers 500 kJ and tertiary consumers 50 kJ. (3 marks) (b) Respiration oxidises part of the organic matter and releases energy as heat; some tissue is never eaten and part of the ingested food is indigestible and lost in faeces. (4 marks) (c) Only about one tenth of the energy passes to the next level, so by the fifth level too little energy remains to support a viable population. (1 mark) |
| 6 | A lake food web is: phytoplankton, water flea, small fish, large fish. A fat-soluble pesticide is present at 0.0004 ppm in the water and 2.0 ppm in large fish. (a) Calculate the biomagnification factor. (2 marks) (b) Explain why the concentration is highest in large fish. (3 marks) (c) Explain the risk to people who eat fish from this lake. (3 marks) | (a) 2.0 divided by 0.0004 = 5000 times. (2 marks) (b) The pesticide is fat-soluble and is not easily excreted, so it accumulates in fat tissue; each trophic level eats many prey and so gathers the contaminant from all of them. (3 marks) (c) People are top consumers, so long-term fish eating builds high concentrations that may affect the nervous and endocrine systems, with higher risk for pregnant women and children. (3 marks) |
| 7 | (a) Name three microbial processes in the nitrogen cycle and the group of organisms responsible for each. (3 marks) (b) Explain why leguminous plants grow well in nitrogen-poor soil. (2 marks) (c) Explain how heavy nitrogen fertiliser use causes eutrophication. (3 marks) | (a) Nitrogen fixation by Rhizobium and cyanobacteria, nitrification by nitrifying bacteria, and denitrification by denitrifying bacteria. (3 marks) (b) Legumes form root nodules with Rhizobium, which reduces atmospheric nitrogen to ammonia for making amino acids and proteins, so they do not depend on soil nitrate. (2 marks) (c) Excess nitrate washes into rivers and lakes, algae bloom, and when the algae die decomposers use up dissolved oxygen, so fish and other aquatic organisms suffocate. (3 marks) |
| 8 | An ecologist catches 40 beetles in a forest, marks them and releases them. A week later 50 beetles are caught, 8 of which are marked. (a) Use the Lincoln index to estimate the population size. (3 marks) (b) Give two reasons why the estimate may be too high. (4 marks) (c) Explain why a waiting period is needed before the second capture. (1 mark) | (a) Population size is about (40 x 50) divided by 8 = 250. (3 marks) (b) Marks may make beetles more visible to predators or reduce their mobility, so fewer marked beetles are recaptured; marks may also rub off. Both make the estimate too high. (4 marks) (c) To let marked and unmarked individuals mix so that the recapture sample is representative. (1 mark) |
Section C: Analysis, succession and conservation (Q9-Q12)
Data questions must quote specific values, and explanations must give cause and effect.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | The rate of oxygen release from an aquatic plant was measured at 25 degrees Celsius in arbitrary units. In 0.1% carbon dioxide the rates at 10, 20, 30 and 40 klx were 12, 24, 30 and 31; in 0.03% carbon dioxide the rates at the same light intensities were 10, 18, 20 and 20. (a) Explain why the rate stops rising above 20 klx in 0.03% carbon dioxide. (3 marks) (b) Identify the limiting factor at 40 klx and support your answer with data. (4 marks) (c) Predict the change in rate at 40 degrees Celsius and explain. (2 marks) | (a) Light is no longer limiting; carbon dioxide concentration becomes limiting, so carbon fixation saturates at a rate of 20. (3 marks) (b) At 40 klx the rate is 31 in 0.1% carbon dioxide but only 20 in 0.03%, so carbon dioxide is the limiting factor. (4 marks) (c) The rate falls because high temperature denatures the enzymes of photosynthesis. (2 marks) |
| 10 | An abandoned farmland undergoes succession: weeds in year 1, more herbs by year 3, shrubs by year 10 and a woodland by year 40. (a) Identify the type of succession and justify your answer. (3 marks) (b) Explain why species number rises during the shrub stage. (3 marks) (c) State two features of pioneer species. (3 marks) | (a) Secondary succession, because it starts on existing soil with surviving seeds and underground stems rather than on bare rock. (3 marks) (b) Shrubs add vertical niches, shelter and food, and make the microclimate more stable, so more species can live there. (3 marks) (c) Fast growth, tolerance of strong light and drought, many easily dispersed seeds, and the ability to bind soil. (3 marks) |
| 11 | In a family, unaffected parents have two daughters and one affected son; the maternal grandfather is also affected and other members are unaffected. (a) Deduce the mode of inheritance and explain. (3 marks) (b) Write the genotypes of the affected son and his parents. (3 marks) (c) If the affected son marries a woman with no family history, explain the chance of affected children. (3 marks) | (a) Autosomal recessive inheritance: unaffected parents have an affected child, so the allele is recessive; the affected grandfather shows the allele passing through the family. (3 marks) (b) The son is aa and both parents are Aa. (3 marks) (c) If the woman is AA all children are Aa carriers with no disease, a chance of 0; if she is a carrier, each child has a 1 in 2 chance, so counselling and testing are needed. (3 marks) |
| 12 | Two nature reserves of equal area are compared. Reserve A has 40 species with similar numbers of individuals; reserve B also has 40 species but 90% of individuals belong to one species. (a) Compare species richness and evenness. (3 marks) (b) Explain which reserve resists a disease outbreak better. (3 marks) (c) Suggest two management measures that maintain genetic diversity. (3 marks) | (a) Both reserves have the same species richness of 40, but evenness is higher in A while B is dominated by one species. (3 marks) (b) Reserve A resists better, because its species have similar abundance, so if one species is hit by disease the others still maintain ecosystem function. (3 marks) (c) Build wildlife corridors between fragmented habitats to allow gene flow, and keep effective population sizes large enough to retain genetic variation. (3 marks) |
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