IB Biology SL Paper 2 Practice Paper (6 Questions, 50 Marks, 75 Minutes)

2026-11-02

IB Biology SL Paper 2

IB Biology SL Paper 2 practice paper: 6 original structured questions in 75 minutes on osmosis and water potential, dihybrid crosses with chi-squared.

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Paper details　Level: IB SL (Grade 11-12) | Subject: Biology | Time: 75 minutes | Marks: 50 | Questions: 6

Section A Cell transport, genetics and energy flow (25 marks)

Questions 1-3 (structured and data-based, marks per part as shown)

#	Question	Answer and marking notes
1	Osmosis practical: five potato cylinders, each 2 g, are placed in 0, 0.2, 0.4, 0.6 and 0.8 mol dm^-3 sucrose solutions. After 30 minutes they are blotted and reweighed; the masses are 2.24, 2.11, 2.02, 1.91 and 1.82 g respectively. (a) Calculate the percentage change in mass for the 0.2 and 0.6 mol dm^-3 treatments. (2 marks) (b) Describe the trend in percentage change in mass as sucrose concentration increases. (2 marks) (c) Estimate the sucrose concentration at which there is no net movement of water, and state the method used. (2 marks) (d) Explain, in terms of water potential, why the mass of the potato cylinders falls in 0.8 mol dm^-3 sucrose. (2 marks)	(a) 0.2 mol dm^-3: (2.11 – 2.00) / 2.00 x 100 = +5.5 per cent; 0.6 mol dm^-3: (1.91 – 2.00) / 2.00 x 100 = -4.5 per cent (1 mark each). (b) From 0.0 to 0.8 mol dm^-3 the percentage change falls steadily from +12.0 per cent to -9.0 per cent, a negative correlation (1 mark trend, 1 mark data). (c) Interpolate between 0.4 and 0.6 mol dm^-3: 0.4 + 0.2 x 1.0 / (1.0 + 4.5) = 0.44 mol dm^-3 (1 mark method, 1 mark answer). (d) The external solution has a lower water potential than the cell sap, so water leaves the cells by osmosis, turgor is lost and mass falls (1 mark comparison, 1 mark direction).
2	In a plant species, flower colour and stem height are each controlled by one pair of alleles: purple (P) is completely dominant to white (p) and tall (T) to dwarf (t). A PpTt plant is crossed with a pptt plant, giving 200 offspring: 48 purple tall, 45 purple dwarf, 52 white tall and 55 white dwarf. (a) State the expected phenotypic ratio and the expected number in each class. (2 marks) (b) Calculate the chi-squared value, showing the formula and the substitution. (3 marks) (c) State the degrees of freedom and the conclusion at p = 0.05, where the critical value is 7.81. (2 marks) (d) Explain why independent assortment of the two gene pairs gives this ratio. (2 marks)	(a) Expected purple tall : purple dwarf : white tall : white dwarf = 1 : 1 : 1 : 1, 50 plants in each class (1 mark ratio, 1 mark numbers). (b) chi-squared = sum of (O – E)^2 / E = (4 + 25 + 4 + 25) / 50 = 58 / 50 = 1.16 (1 mark formula, 1 mark substitution, 1 mark answer). (c) Degrees of freedom = 4 – 1 = 3; the value 1.16 is below 7.81, so the independent assortment hypothesis is not rejected and the deviation is not significant (1 mark df, 1 mark conclusion). (d) The two genes are on non-homologous chromosomes, so during meiosis I the homologous chromosomes segregate randomly to the poles and the four gamete types are produced in equal numbers (1 mark location, 1 mark meiosis).
3	In a grassland ecosystem the gross primary productivity (GPP) is 87000 kJ m^-2 yr^-1 and respiration by producers is 43500 kJ m^-2 yr^-1. Primary consumers ingest 8700 kJ m^-2 yr^-1, of which 5220 kJ m^-2 yr^-1 is egested in faeces. (a) Calculate the net primary productivity (NPP). (2 marks) (b) Calculate NPP as a percentage of GPP. (2 marks) (c) Calculate the assimilated energy of the primary consumers as a percentage of NPP (ecological efficiency). (2 marks) (d) Explain two reasons why so much energy is lost between trophic levels. (2 marks)	(a) NPP = 87000 – 43500 = 43500 kJ m^-2 yr^-1 (1 mark formula, 1 mark answer). (b) 43500 / 87000 x 100 = 50 per cent (1 mark substitution, 1 mark answer). (c) Assimilated = 8700 – 5220 = 3480; 3480 / 43500 x 100 = 8 per cent (1 mark subtraction, 1 mark percentage). (d) Respiration oxidises organic matter and releases energy as heat; some energy remains in undigested faeces and in uneaten plant parts (1 mark each).

Section B Membranes, codominant inheritance and the carbon cycle (25 marks)

Questions 4-6 (structured and extended response, marks per part as shown)

#	Question	Answer and marking notes
4	(a) Outline the fluid mosaic model of the cell membrane, referring to the arrangement of phospholipids, proteins and cholesterol. (3 marks) (b) Distinguish facilitated diffusion from active transport, giving one example of each. (3 marks) (c) Beetroot tissue is kept in water at different temperatures for 15 minutes; the absorbance of the bathing solution rises from 0.12 at 25 C to 0.48 at 60 C. Calculate the percentage increase in absorbance and explain why high temperature increases membrane permeability. (3 marks)	(a) The phospholipid bilayer has hydrophilic heads facing the aqueous phases and hydrophobic tails facing inwards; proteins are embedded in, span or are attached to the surface; cholesterol sits between phospholipids and regulates fluidity (1 mark each). (b) Facilitated diffusion follows the concentration gradient through channel or carrier proteins and needs no ATP (e.g. glucose uptake by red blood cells); active transport moves substances against the gradient using carrier proteins and ATP (e.g. the sodium-potassium pump) (1 mark each, 3 marks total). (c) (0.48 – 0.12) / 0.12 x 100 = 300 per cent (1 mark); high temperature denatures membrane proteins and loosens the phospholipid packing, damaging membrane integrity so pigment leaks out (2 marks)
5	Sickle cell anaemia is caused by the HbS allele, which is codominant with the normal HbA allele: HbAHbA is normal, HbAHbS is a carrier with mild symptoms and HbSHbS is affected. Two carriers marry. (a) Use a Punnett square to give the expected genotype ratio of their children. (2 marks) (b) Calculate the probability that a child is affected. (1 mark) (c) Calculate the probability that two children are both unaffected (normal or carriers). (2 marks) (d) Explain why the HbS allele remains at a relatively high frequency in regions where malaria is common. (3 marks)	(a) HbAHbA : HbAHbS : HbSHbS = 1 : 2 : 1 (1 mark square, 1 mark ratio). (b) Probability of being affected (HbSHbS) = 1/4 = 25 per cent (1 mark). (c) Each child has a 3/4 chance of being unaffected, so two unaffected children = 0.75^2 = 0.5625, about 56 per cent (1 mark method, 1 mark answer). (d) Carriers resist malaria better because infected red blood cells are cleared more readily, so they survive and reproduce more successfully; natural selection therefore keeps the HbS allele in the population (1 mark each).
6	(a) Outline three main pathways by which carbon moves between the atmosphere, producers and decomposers. (3 marks) (b) Explain how the greenhouse effect arises and how human activity strengthens it. (3 marks) (c) Name one ecosystem in which a pyramid of biomass may be inverted, and state one limitation of a pyramid of numbers. (2 marks)	(a) Photosynthesis fixes atmospheric carbon dioxide into organic compounds; respiration by producers and consumers releases organic carbon as carbon dioxide; decomposers release carbon dioxide when breaking down dead bodies and excretory products (1 mark each). (b) Greenhouse gases absorb long-wave infrared radiation emitted by the Earth surface and re-radiate it back, trapping heat; burning fossil fuels and deforestation raise carbon dioxide and methane concentrations, strengthening the effect (1 mark each). (c) In marine ecosystems the biomass of phytoplankton can be smaller than that of zooplankton, giving an inverted pyramid; a pyramid of numbers ignores differences in individual size, since one large tree can support many insects (1 mark each).

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