S5 Biology Practice Paper: Inheritance, Gene Expression And Ecosystems (12 Questions, 100 Marks, 75 Minutes)

S5 Biology Practice Paper: Inheritance, Gene Expression and Ecosystems (12 questions, 100 marks, 75 minutes)

2026-11-06

S5 Biology: Inheritance, Gene Expression and Ecosystems

S5 biology practice paper: 12 questions on Mendelian inheritance, DNA base pairing, transcription, mutation, Hardy-Weinberg equilibrium and energy flow, with full answers and mark points.

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Section A: Mendelian Inheritance and Probability (32 marks)

Answer all questions in Section A and state genotypes and ratios clearly.

#	Question	Answer and marking notes
1	Mendelian inheritance: in peas the tall allele T is completely dominant to the short allele t. Two heterozygotes (Tt) are crossed and produce 480 offspring. (a) State the genotype of both parents and the gametes produced. (b) Use a Punnett square to list the offspring genotypes and their ratio. (c) Calculate the expected number of tall and short plants. (d) If only 90 short plants actually appear, give two possible reasons.	Answer: (a) Both parents are Tt and each produces T and t gametes in a 1:1 ratio. (b) Offspring genotypes TT:Tt:tt = 1:2:1. (c) Tall = 480 x 3/4 = 360 plants; short = 480 x 1/4 = 120 plants. (d) Random deviation due to a limited sample, or some short seeds died before germination or were misclassified. Mark points: genotypes and gametes (2), ratio (2), plant numbers (2), two reasons (2).
2	DNA structure: a double-stranded DNA molecule contains 3 000 bases, of which adenine (A) is 22%. (a) Calculate the number of guanine (G) bases. (b) Calculate the total number of hydrogen bonds in the molecule. (c) Explain why the percentage of G is 28% rather than 22%.	Answer: (a) G is 28%, so the number of G bases = 3 000 x 28% = 840. (b) There are 660 A-T pairs with 2 hydrogen bonds each and 840 G-C pairs with 3 hydrogen bonds each, giving 660 x 2 + 840 x 3 = 3 840 hydrogen bonds. (c) A pairs with T and G pairs with C, so the four bases total 100%; A + T = 44% and the remaining 56% is shared equally, giving 28% each. Mark points: G number (2), hydrogen bonds (3), complementary base explanation (3).
3	Transcription and translation: the coding region of an mRNA contains 900 nucleotides, including one stop codon. (a) Calculate the number of codons that can be translated. (b) Calculate the number of amino acids in the polypeptide. (c) Explain the effect on protein synthesis if the start codon mutates into a non-start codon.	Answer: (a) 900 / 3 = 300 codons. (b) 300 – 1 = 299 amino acids because the stop codon does not code for an amino acid. (c) The ribosome cannot initiate translation without a start codon, so the polypeptide is not made and gene expression fails at the translation stage. Mark points: codons (2), amino acids (3), start codon explanation (3).
4	Inheritance probability: an autosomal recessive disorder is caused by allele a. Two carriers (Aa) marry. (a) State the possible offspring genotypes and their ratio. (b) Calculate the probability that a child is affected. (c) If they plan three children, calculate the probability that all three are unaffected (3 decimal places).	Answer: (a) AA:Aa:aa = 1:2:1. (b) The probability of an affected child (aa) is 1/4 = 25%. (c) Each child has a 0.75 chance of being unaffected, so all three are unaffected with probability 0.75 to the power 3 = 0.422 (about 42.2%). Mark points: ratio (2), affected probability (3), three children (3).

Section B: Gene Expression, Splicing and Mutation (36 marks)

Answer all questions in Section B and show every step of working.

#	Question	Answer and marking notes
5	Gene expression: the pre-mRNA of a gene is 6 000 nucleotides long and its introns total 1 800 nucleotides. (a) Calculate the length of the mature mRNA. (b) Calculate the number of amino acids in the polypeptide, excluding the stop codon. (c) Explain the importance of splicing in eukaryotic cells.	Answer: (a) 6 000 – 1 800 = 4 200 nucleotides. (b) 4 200 / 3 = 1 400 codons, giving 1 399 amino acids after removing the stop codon. (c) Splicing removes introns and joins exons so that the mature mRNA is translated correctly; alternative splicing also lets one gene produce several different proteins, increasing protein diversity. Mark points: length (2), amino acids (3), significance of splicing (4).
6	Mutation: the second base of codon 250 of a gene changes from T to A. (a) Name this type of mutation. (b) Describe three possible consequences. (c) Calculate the position of the affected nucleotide, taking the first nucleotide of the coding region as position 1.	Answer: (a) A point mutation, specifically a base substitution. (b) A silent mutation leaves the amino acid unchanged; a missense mutation changes one amino acid and may damage protein function; a nonsense mutation creates a premature stop codon and shortens the protein. (c) Position = (250 – 1) x 3 + 2 = 749. Mark points: name (2), three consequences (4), position calculation (3).
7	Population genetics: 4% of a population shows the recessive phenotype. The population is in Hardy-Weinberg equilibrium and contains 2 500 individuals. (a) Calculate the recessive allele frequency q. (b) Calculate the dominant allele frequency p. (c) Calculate the number of carriers (heterozygotes).	Answer: (a) q squared = 0.04, so q = 0.2. (b) p = 1 – 0.2 = 0.8. (c) 2pq = 2 x 0.8 x 0.2 = 0.32, so the number of carriers = 2 500 x 0.32 = 800. Mark points: q (2), p (2), number of carriers (5).
8	Incomplete dominance: a red-flowered plant (C-R C-R) is crossed with a white-flowered plant (C-W C-W) and all F1 plants are pink. (a) Name this pattern of inheritance. (b) State the F1 genotype and phenotype. (c) F1 self-pollination gives 800 F2 plants; calculate the number of plants of each of the three phenotypes.	Answer: (a) Incomplete dominance, giving an intermediate phenotype. (b) The F1 genotype is C-R C-W and the phenotype is pink. (c) The F2 phenotypic ratio is 1:2:1, that is 200 red, 400 pink and 200 white plants. Mark points: name of pattern (2), genotype and phenotype (2), three plant numbers (4).

Section C: Populations, Energy Flow and Ecosystems (32 marks)

Answer all questions in Section C and give units with every numerical answer.

#	Question	Answer and marking notes
9	Energy flow: producers in a grassland fix 24 000 kJ m-2 of energy per year and the transfer efficiency is 10%. (a) Calculate the energy available to primary, secondary and tertiary consumers. (b) Calculate the energy of tertiary consumers as a percentage of the energy fixed by producers. (c) Explain why food chains rarely exceed four or five trophic levels.	Answer: (a) Primary consumers receive 2 400 kJ m-2 yr-1, secondary consumers 240 and tertiary consumers 24. (b) 24 / 24 000 x 100% = 0.1%. (c) About 90% of the energy is lost at each transfer through respiration, excretion and uneaten parts, so the remaining energy quickly becomes too small to support another trophic level. Mark points: three energy values (3), percentage (2), explanation (3).
10	Population estimate: fish in a pond are estimated by capture-recapture. 120 fish are caught and marked, then released; a second sample of 90 fish contains 18 marked fish. (a) Use the Lincoln index to calculate the population size. (b) State three assumptions of the method. (c) Explain whether the estimate is too high or too low if one assumption is violated.	Answer: (a) N = (120 x 90) / 18 = 600 fish. (b) Marking does not affect survival or the chance of recapture; marks do not fall off; the population is closed to births, deaths, immigration and emigration between samples; individuals mix randomly. (c) If marking makes fish easier to catch, the recapture count is too high and the estimate is too low; if marking increases mortality, the recapture count is too low and the estimate is too high. Mark points: calculation (3), three assumptions (3), direction of bias (2).
11	Biological magnification: a pesticide is present at 0.00005 ppm in water and at 5 ppm in fish from the same water. (a) Calculate the magnification factor. (b) Explain how biological magnification arises. (c) Explain why top consumers are at greatest risk.	Answer: (a) 5 / 0.00005 = 100 000 times. (b) Fat-soluble pollutants are not easily broken down or excreted, so they accumulate in fatty tissue over a lifetime of feeding and their concentration rises at each trophic level. (c) Top consumers live long and eat large amounts, so their total accumulated load is the greatest. Mark points: factor (2), cause (3), top consumers (3).
12	Carbon cycle: one hectare of woodland fixes 12 000 kg of carbon dioxide by photosynthesis each year, while respiration by plants, animals and decomposers releases 9 000 kg of carbon dioxide from the same area. (a) Calculate the net carbon fixation in kg of carbon dioxide. (b) Convert this to a mass of carbon (relative atomic mass: C = 12, CO2 = 44). (c) Describe two effects on the carbon cycle if the trees are cut down and burned.	Answer: (a) 12 000 – 9 000 = 3 000 kg of carbon dioxide. (b) 3 000 x 12/44 = 818 kg of carbon. (c) Photosynthesis fixes less carbon, reducing carbon uptake; burning releases the stored carbon as carbon dioxide very quickly, raising its concentration in the atmosphere. Mark points: net fixation (3), carbon conversion (3), two effects (2).

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