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IB Biology SL Paper 2 Structured Questions Practice Paper | 6 questions, 50 marks, 1 hour 15 minutes
IB Biology SL Paper 2
IB Biology SL Paper 2 practice paper: 6 questions, 50 marks, 75 minutes; Section A data-based questions on enzymes, osmosis, genetics and ecology, Section B extended questions on gas exchange and protein synthesis, with worked solutions.
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Paper details Level: IB SL | Subject: Biology | Time: 1 hour 15 minutes | Marks: 50 | Questions: 6
Section A: data-based questions (30 marks)
Structured questions: show your working and give correct units; answer every part of each question. Total 50 marks.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | An experiment used catalase of the same concentration to break down 2% hydrogen peroxide; the volume of oxygen released was read every 20 s: 40 s: 6.0 cm3 80 s: 12.0 cm3 120 s: 18.0 cm3 160 s: 20.0 cm3 (a) Calculate the mean rate of reaction for 0-40 s and for 80-120 s (cm3 min-1). (b) Explain why the rate for 80-120 s is the same as that for 0-40 s. (c) Suggest one reason why the rate falls during 120-160 s. (d) State one variable that must be kept constant and describe how enzyme activity is measured. |
(a) 0-40 s: 6.0 cm3 / (40 / 60) min = 9.0 cm3 min-1; 80-120 s: 6.0 cm3 / 40 s = 40 cm3 min-1. (2 marks) (b) Both are 9.0 cm3 min-1 because oxygen released increases in direct proportion to time: substrate is in excess, all active sites are occupied (saturated) and the reaction runs at Vmax; rate = change in volume / change in time. (3 marks) (c) Substrate concentration falls with time so active sites are no longer all occupied and the rate drops below Vmax; denaturation of some enzyme over time is also accepted. (2 marks) (d) Keep temperature, pH or enzyme concentration constant; measure oxygen released per unit time (initial rate). (1 mark) |
| 2 | Five groups of potato cylinders of equal length and mass (initial mean mass 2.4 g, 3 cylinders per group) were kept in sucrose solutions for 30 minutes, then blotted and reweighed: 0.00 mol dm-3: 2.76 g 0.20 mol dm-3: 2.52 g 0.40 mol dm-3: 2.4 g 0.60 mol dm-3: 2.28 g 0.80 mol dm-3: 2.16 g (a) Calculate the percentage change in mass for each group. (b) Estimate the osmotic concentration of the potato cells in mol dm-3 and explain your method. (c) Explain why the cylinders in 0.00 mol dm-3 gained mass. (d) State why the cylinders were blotted before each weighing. |
(a) Percentage change = (final – initial) / initial x 100: 0.00 = 15.0%; 0.20 = 5.0%; 0.40 = 0.0%; 0.60 = -5.0%; 0.80 = -10.0%. (3 marks) (b) About 0.40 mol dm-3 (2 marks): read off the concentration at which the change in mass is zero, where the external solution and the cell sap have the same water potential and there is no net water movement (1 mark). (c) Distilled water has a higher water potential than the cell sap, so water enters the vacuole by osmosis across the partially permeable membrane; the cell wall limits expansion, producing turgor pressure and so mass increases. (1 mark) (d) To remove solution on the surface, which would make the final mass too high and overestimate the percentage change. (1 mark) |
| 3 | In tomato, fruit colour is controlled by alleles R/r and fruit shape by alleles S/s. Two homozygous parents were crossed and all F1 were red and round; selfing the F1 gave 1600 F2 plants: 900 red round, 300 red long, 300 yellow round and 100 yellow long. (a) State the genotypes of the two homozygous parents. (b) Calculate the phenotypic ratio from the F2 numbers and state whether it supports independent assortment. (c) If the F1 is test-crossed with a yellow long plant, how many of 800 offspring would you expect of each phenotype? (d) Explain how independent assortment produces so many gamete combinations. |
(a) RRSS (red round) x rrss (yellow long); the F1 is RrSs. (2 marks) (b) Ratio = 900 : 300 : 300 : 100 = 9 : 3 : 3 : 1 (1 mark). This is the expected ratio when the two gene pairs lie on different homologous chromosomes and assort independently, so the data support independent assortment (1 mark). (c) RrSs x rrss gives RrSs (red round), Rrss (red long), rrSs (yellow round) and rrss (yellow long) in equal numbers, i.e. 200 of each phenotype out of 800 offspring. (2 marks) (d) In meiosis I the two pairs of homologous chromosomes line up independently on the equator, so R/r and S/s can combine freely to give equal numbers of RS, Rs, rS and rs gametes. (1 mark) |
| 4 | An ecologist caught 120 frogs in a pond, marked them on the hind leg and released them. One week later 100 frogs were caught, 15 of which were marked. (a) Estimate the population size using the Lincoln index and state one assumption of the method. (b) Gross primary production (GPP) of the pond is 48,600 kJ m-2 yr-1 and respiration loses 36,200 kJ m-2 yr-1; calculate net primary production (NPP). (c) If 10% of the energy is transferred to primary consumers, calculate the energy available to them. (d) Explain why food chains rarely have more than four or five trophic levels. |
(a) N = (M x C) / R = (120 x 100) / 15 = 800 frogs (2 marks). Assumption: marks do not affect survival or the chance of recapture, the population is closed (no births, deaths, immigration or emigration between samples) and marked individuals have mixed randomly by the second sample (1 mark). (b) NPP = GPP – R = 48,600 – 36,200 = 12,400 kJ m-2 yr-1 (2 marks). (c) 12,400 x 10% = 1,240 kJ m-2 yr-1 (1 mark). (d) Every trophic level loses much energy through respiration and excretion, so transfer efficiency is only about 10%; energy falls sharply at each step and cannot support further levels. (1 mark) |
Section B: extended questions (20 marks)
Structured questions: show your working and give correct units; answer every part of each question. Total 50 marks.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | (a) A student at rest has a tidal volume of 0.45 dm3 and a breathing rate of 16 min-1; calculate the pulmonary ventilation rate. (b) Inhaled air contains 21% oxygen and exhaled air 16%; calculate the oxygen consumption per minute in cm3 min-1. (c) During exercise tidal volume rises to 1.1 dm3 and breathing rate to 24 min-1; calculate the ventilation rate and the factor by which it increases. (d) Explain the physiological reason for the faster breathing rate during exercise and why gas exchange must occur on a moist surface. |
(a) Ventilation = tidal volume x breathing rate = 0.45 x 16 = 7.2 dm3 min-1. (2 marks) (b) Oxygen in = 7.2 x 21% = 1.512 dm3; oxygen out = 7.2 x 16% = 1.152 dm3; consumed = 1.512 – 1.152 = 0.36 dm3 min-1 = 360 cm3 min-1. (3 marks) (c) 1.1 x 24 = 26.4 dm3 min-1; 26.4 / 7.2 = 3.67 times. (2 marks) (d) Respiration in muscles increases during exercise, so blood carbon dioxide rises and pH falls; this stimulates the breathing centre in the medulla, which increases nerve impulses so breathing becomes deeper and faster to remove carbon dioxide and supply oxygen. (2 marks) Oxygen and carbon dioxide must dissolve in a film of water before diffusing across the alveolus and capillary walls, so a moist surface keeps them dissolved and diffusing quickly. (1 mark) |
| 6 | A structural gene is 1,200 base pairs long, of which 900 bases are in protein-coding exons; the average relative molecular mass of an amino acid is about 110. (a) Calculate the number of codons and the number of amino acids in the final protein (assume the stop codon codes for no amino acid). (b) Calculate the approximate molecular mass of the protein. (c) If one extra base is inserted after the fourth base of the coding region, explain the effect on the structure of the protein. (d) Explain why the same gene can produce very different amounts of protein in liver cells and muscle cells. |
(a) Number of codons = 900 / 3 = 300 (1 mark); minus one stop codon, amino acids = 300 – 1 = 299 (2 marks). (b) Approximate molecular mass = 299 x 110 = 32,890 (2 marks). (c) Inserting one base causes a frameshift: from that point every codon is read in a different frame, so most codons specify different amino acids and a stop codon usually appears early, giving a shortened, non-functional protein. (3 marks) (d) The two cell types contain different combinations of transcription factors that bind the promoter and enhancer in different ways, so the gene is transcribed at different rates; mRNA stability and translation efficiency may also differ, although the DNA sequence is the same. (2 marks) |
A TutorZone practice paper; original questions and passages, not publisher material.
