IB Biology SL Paper 2 Practice Paper | 10 Questions, 100 Marks, 75 Minutes

2026-11-01

IB Biology SL

IB Biology SL Paper 2 practice paper: cardiac output and pulmonary ventilation; 10 questions, 100 marks, 75 minutes, with full answers.

Download　Question paper PDF　|　Answer key PDF

Section A: Cells, membranes and enzymes (Questions 1-4, 30 marks)

Answer all questions. Write your answers in the answer booklet and label each answer clearly; the mark allocation is shown in brackets.

#	Question	Answer and marking notes
1	1. A student observes onion epidermal cells with a light microscope. (a) A cell measures 45 mm across in the photograph and its actual diameter is 0.15 mm; find the magnification. (b) Convert 0.15 mm to micrometres. (c) Explain why an electron microscope has a higher resolution than a light microscope. (8 marks)	(a) magnification = image size / actual size = 45 / 0.15 = x300. (2 marks, unit required) (b) 0.15 mm x 1000 = 150 micrometres. (2 marks) (c) An electron microscope uses a beam of electrons instead of visible light; the electron wavelength is much shorter, so the diffraction limit is smaller and closer points can be resolved. A light microscope is limited by the wavelength of visible light, giving a resolution of about 200 nm, so organelle detail is not visible. (4 marks: wavelength 2, comparison 2)
2	2. Potato strips are left for 30 minutes in distilled water and in 0.5 mol dm-3 sucrose solution. A strip in the sucrose solution changes in mass from 2.4 g to 2.76 g. (a) Calculate the percentage change in mass. (b) Explain the net movement of water in each treatment and the role of water potential. (7 marks)	(a) percentage change = (2.76 – 2.4) / 2.4 x 100% = +15.0%, i.e. a gain of 15.0%. (3 marks: equation 1, value 1, sign 1) (b) Distilled water has the highest water potential (about zero), so water moves into the cells by osmosis and the strip gains mass. The 0.5 mol dm-3 sucrose solution has a lower water potential than the cell sap, so water leaves the cells and they become plasmolysed. Water always moves net from higher to lower water potential. (4 marks)
3	3. A yeast suspension releases 24 cm3 of gas in 40 s at 20 C, and 72 cm3 in 40 s at 30 C. (a) Calculate the rate at 20 C in cm3 s-1. (b) Calculate Q10. (c) Explain why the rate falls sharply at 60 C. (8 marks)	(a) rate = 24 / 40 = 0.6 cm3 s-1. (2 marks) (b) rate at 30 C = 72 / 40 = 1.8 cm3 s-1; Q10 = 1.8 / 0.6 = 3.0. (3 marks) (c) At 60 C the enzyme is above its optimum. Hydrogen and ionic bonds break, the tertiary structure and active site are distorted (denaturation), enzyme-substrate complexes cannot form and the rate falls sharply; denaturation is irreversible. (3 marks)
4	4. Two cube models have sides of 2 mm and 4 mm. (a) Calculate the surface area and volume of each. (b) Calculate the surface area to volume ratio. (c) Explain the significance of SA:V for gas exchange in a unicellular organism and in a large animal. (7 marks)	(a) Side 2 mm: surface area 6 x 2^2 = 24 mm2, volume 2^3 = 8 mm3. Side 4 mm: surface area 96 mm2, volume 64 mm3. (2 marks) (b) SA:V = 3.0 and 1.5; doubling the side halves the ratio. (2 marks) (c) A unicellular organism has a large SA:V, so diffusion across the surface supplies the whole cell. A large animal has a small SA:V, so surface diffusion cannot supply internal tissues and specialised surfaces (lungs, gills) plus a transport system are needed. (3 marks)

Section B: Genetics and genetic ratios (Questions 5-6, 20 marks)

Answer all questions. Write your answers in the answer booklet and label each answer clearly; the mark allocation is shown in brackets.

#	Question	Answer and marking notes
5	5. In pea plants, tall (T) is dominant to dwarf (t). Two Tt plants are crossed; 240 seeds give 185 tall and 55 dwarf plants. (a) Use a Punnett square to find the expected ratio and numbers. (b) Calculate chi-square. (c) The critical value at 1 degree of freedom and p = 0.05 is 3.84; state your conclusion. (11 marks)	(a) Tt x Tt gives TT : Tt : tt = 1 : 2 : 1, so tall : dwarf = 3 : 1; expected 180 tall and 60 dwarf. (4 marks) (b) chi-square = (185 – 180)^2 / 180 + (55 – 60)^2 / 60 = 0.556. (5 marks) (c) Since 0.556 is less than 3.84, the deviation is not significant and the 3:1 hypothesis is not rejected; the difference can be explained by chance. (2 marks)
6	6. Two AaBb plants are crossed (the two gene loci are on different chromosomes) and 320 offspring are obtained. (a) Calculate the expected numbers for a 9:3:3:1 ratio. (b) If the two loci were linked on the same chromosome, how would the ratio deviate, and why? (9 marks)	(a) A-B- : A-bb : aaB- : aabb = 9 : 3 : 3 : 1, i.e. 180 : 60 : 60 : 20 plants, total 320. (4 marks) (b) With linkage the parental gametes (AB, ab) are over-represented and recombinant gametes (Ab, aB) are under-represented, so the offspring shift towards a 3:1 pattern with more double dominant and double recessive individuals. The tighter the linkage, the fewer the recombinants and the greater the deviation from 9:3:3:1. (5 marks)

Section C: Energy flow in ecosystems (Questions 7-8, 20 marks)

Answer all questions. Write your answers in the answer booklet and label each answer clearly; the mark allocation is shown in brackets.

#	Question	Answer and marking notes
7	7. In an ecosystem producers fix 45,000 kJ m-2 yr-1 and primary consumers assimilate 4,500 kJ m-2 yr-1. (a) Calculate the efficiency of energy transfer. (b) Explain why most energy is lost at each level. (c) Estimate the energy available at the fourth trophic level. (11 marks)	(a) efficiency = 4,500 / 45,000 x 100% = 10%. (3 marks) (b) Most energy is lost as heat from respiration; some is not eaten (leaf litter, roots), some is not digested and is lost in faeces, and some is excreted. Only about 10% is transferred to the next level. (4 marks) (c) 45,000 x 0.1^3 = 45 kJ m-2 yr-1. (4 marks)
8	8. A grassland has a gross primary production of 12,800 kJ m-2 yr-1 and producer respiration of 6,400 kJ m-2 yr-1. (a) Calculate net primary production. (b) Describe the fate of NPP. (c) Explain why NPP is not the same as the net increase in biomass. (9 marks)	(a) NPP = GPP – respiration = 12,800 – 6,400 = 6,400 kJ m-2 yr-1, which is 50% of GPP. (3 marks) (b) NPP is used for growth (adding biomass), eaten by primary consumers, shed as litter and root exudates for decomposers, and partly stored as peat or wood. (3 marks) (c) Part of NPP is consumed by herbivores and part dies and is decomposed, so NPP is not equal to the net biomass increase; biomass also changes through death, emigration and fire. (3 marks)

Section D: Human physiology and gas exchange (Questions 9-10, 30 marks)

Answer all questions. Write your answers in the answer booklet and label each answer clearly; the mark allocation is shown in brackets.

#	Question	Answer and marking notes
9	9. An athlete at rest has a stroke volume of 70 cm3 and a heart rate of 75 beats per minute. (a) Calculate cardiac output in cm3 min-1. (b) Convert it to dm3 min-1. (c) During vigorous exercise the heart rate is 150 beats per minute and stroke volume 110 cm3; calculate cardiac output and the increase. (d) Explain how the rise in cardiac output increases oxygen supply to muscles. (15 marks)	(a) cardiac output = stroke volume x heart rate = 70 x 75 = 5,250 cm3 min-1. (4 marks) (b) 5,250 / 1000 = 5.25 dm3 min-1. (3 marks) (c) 110 x 150 = 16,500 cm3 min-1, which is 3.1 times the resting value. (4 marks) (d) More blood is pumped each minute, so more oxygenated blood reaches the muscle capillaries, the oxygen concentration gradient across the capillary wall increases and diffusion into muscle fibres speeds up; carbon dioxide and lactate are also removed faster. (4 marks)
10	10. An adult at rest has a tidal volume of 0.5 dm3 and a breathing rate of 15 per minute. (a) Calculate pulmonary ventilation in dm3 min-1. (b) During exercise the tidal volume is 2.0 dm3 and the rate 30 per minute; calculate the ventilation. (c) State three structural features of alveoli that suit gas exchange. (d) Calculate the volume of oxygen inhaled per minute at rest (air is 21% oxygen). (15 marks)	(a) pulmonary ventilation = tidal volume x breathing rate = 0.5 x 15 = 7.5 dm3 min-1. (4 marks) (b) 2.0 x 30 = 60 dm3 min-1, which is 8 times the resting value. (4 marks) (c) Very numerous with a total surface area of about 70 m2; walls one cell thick (squamous epithelium) so the diffusion distance is short; surrounded by a dense capillary network that maintains a steep concentration gradient; surfactant prevents collapse. (4 marks) (d) 7.5 x 0.21 = 1.575 dm3 min-1, about 1575 cm3 min-1. (3 marks)

Find a tutor — free Tutor Registration →

A TutorZone practice paper; original questions and passages, not publisher material.