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IB Biology SL Paper 2 Practice Paper (structured questions) – 50 marks, 1 hour 15 minutes, answers and mark scheme
TutorZone original practice paper – IB Biology SL Paper 2 (structured questions)
7 questions, 20 parts – 50 marks – 1 hour 15 minutes – calculator allowed
Paper comparison
| Paper | Time | Marks | Nature |
|---|---|---|---|
| Paper 1 (multiple choice) | 45 min | 30 | other SL paper |
| Paper 2 (structured) | 1 h 15 min | 50 | this paper |
| Paper 3 (practical and data skills) | 1 h | 35 | other SL paper |
| Internal assessment | about 10 hours | 20% of grade | school assessed |
Data and formulae
| Item | Formula / value |
|---|---|
| Magnification | magnification = image size / actual size |
| Surface area : volume | cube of side s: 6s^2 : s^3 |
| Percentage change | (new – old) / old x 100% |
| Chi-squared test | chi-squared = sum (O – E)^2 / E; df = 1, critical value 3.84 at p = 0.05 |
| Energy transfer efficiency | energy in next level / energy in previous level x 100% |
Questions
Total: 50 marks · 1 h 15 min · answer all of Section A and one question from Section B
Section A (30 marks)
Q 1 · Data question: photosynthesis and light intensity [8]
A student used the same aquatic plant to measure the volume of oxygen released in 10 minutes at different light intensities. The results are shown below:
| Light intensity (klx) | Oxygen released in 10 min (cm³) |
|---|---|
| 0 | 2 |
| 5 | 14 |
| 10 | 28 |
| 20 | 42 |
| 30 | 46 |
| 40 | 47 |
(a) Calculate the percentage increase in oxygen released when light intensity rises from 5 klx to 10 klx. [2]
(b) Calculate the mean rate of oxygen release at 30 klx in cm3 per minute. [2]
(c) Describe how the curve changes above 30 klx and suggest one limiting factor. [2]
(d) At 0 klx a volume of 2 cm3 was still recorded. Suggest one possible explanation. [2]
Q 2 · Cell membranes and transport [6]
(a) Explain why large molecules such as proteins cannot enter a cell by diffusion alone. [2]
(b) Spinach leaf discs lost mass from 2.40 g to 2.10 g in 0.4 mol dm-3 sucrose solution. Calculate the percentage change in mass and state whether the solution is hypotonic, isotonic or hypertonic to the cells. [2]
(c) Explain why ATP is needed when root hair cells absorb mineral ions. [2]
Q 3 · DNA structure and protein synthesis [6]
(a) A DNA fragment has 240 base pairs and adenine makes up 30%. Calculate the number of cytosine bases. [2]
(b) A mRNA molecule has 183 bases including a stop codon. Calculate the maximum number of amino acids it can code for. [2]
(c) The same gene can produce different proteins in a muscle cell and a pancreatic cell. Explain why. [2]
Q 4 · Genetics and chi-squared test [5]
(a) Two heterozygotes (Tt x Tt) produced 400 offspring. Using a 3:1 ratio, state the expected numbers of dominant and recessive offspring. [2]
(b) The observed numbers were 310 dominant and 90 recessive. Calculate chi-squared and decide whether the result deviates significantly from 3:1 (df = 1, critical value at p = 0.05 is 3.84). [2]
(c) Explain why a recessive phenotype disappears in F1 and reappears in F2. [1]
Q 5 · Energy flow in ecosystems [5]
(a) Producers in an ecosystem contain 45,000 kJ m-2 yr-1 and primary consumers 4,500 kJ m-2 yr-1. Calculate the efficiency of energy transfer. [1]
(b) Tertiary consumers contain 45 kJ m-2 yr-1. Calculate the percentage of energy lost between producers and tertiary consumers. [2]
(c) Explain two reasons why so much energy is lost between trophic levels. [2]
Section B (20 marks)
Q 6 · Section B question 1: gas exchange and oxygen transport [20]
(a) Describe and explain three features of alveoli that make gas exchange efficient. [8]
(b) Explain the role of haemoglobin in oxygen transport and why the oxygen dissociation curve shifts right during exercise. [12]
Q 7 · Section B question 2: coordination and osmoregulation [20]
(a) Compare nervous and endocrine coordination (four points). [10]
(b) Explain ultrafiltration and selective reabsorption in the nephron, and how water balance is maintained after drinking a lot of water. [10]
Answers and mark scheme
| Q | part | Answer | Mark points | marks |
|---|---|---|---|---|
| Section A | ||||
| 1 | (a) | Percentage increase = (28 – 14) / 14 x 100% = 100% (1 for working, 1 for the answer with % sign). | working (28 – 14) / 14 x 100% answer 100% |
2 |
| 1 | (b) | 46 / 10 = 4.6 cm3 per minute (1 for dividing by 10 minutes, 1 for the unit). | 46 / 10 = 4.6 unit cm3 min-1 |
2 |
| 1 | (c) | Oxygen rises by only 1 cm3 (about 2.2%) from 30 to 40 klx, so the curve plateaus (1); limiting factor: carbon dioxide concentration or temperature (1). | only 1 cm3 rise from 30 to 40 klx, curve plateaus limiting factor: CO2 concentration or temperature |
2 |
| 1 | (d) | Ambient light was not fully excluded so a little photosynthesis still occurred (1); or residual bubbles / dissolved oxygen and reading error (1). | ambient light not fully excluded residual bubbles or reading error |
2 |
| 2 | (a) | Too large to pass through the phospholipid bilayer (1); charged or water-soluble molecules are repelled by the hydrophobic tails (1). | molecule too large for the bilayer charged or water-soluble, blocked by hydrophobic tails |
2 |
| 2 | (b) | (2.10 – 2.40) / 2.40 x 100% = -12.5% (1); the solution is hypertonic (1). | -12.5% hypertonic |
2 |
| 2 | (c) | Ion concentration in the soil is lower than inside the cell, so movement is against the gradient (1); active transport requires energy from ATP hydrolysis (1). | against the concentration gradient active transport needs ATP |
2 |
| 3 | (a) | A = 240 x 30% = 72, T = 72 (1); G + C = 96, so C = 48 (1). | A and T are 72 each G + C = 96, so C = 48 |
2 |
| 3 | (b) | (183 – 3) / 3 = 60 (1 for the codon calculation, 1 for the answer). | subtract the 3 stop codon bases (183 – 3) / 3 = 60 |
2 |
| 3 | (c) | Gene expression is regulated, so each cell type transcribes only the genes it needs (1); alternative splicing of mRNA gives different sequences (1). | gene expression is regulated, only needed genes transcribed alternative splicing |
2 |
| 4 | (a) | Dominant = 400 x 3/4 = 300 (1); recessive = 400 x 1/4 = 100 (1). | dominant 300 recessive 100 |
2 |
| 4 | (b) | chi-squared = (310 – 300)^2 / 300 + (90 – 100)^2 / 100 = 1.33 (1); 1.33 < 3.84 so the deviation is not significant (1). | chi-squared = 1.33 1.33 < 3.84, not significant, 3:1 accepted |
2 |
| 4 | (c) | All F1 are heterozygotes so the recessive allele is masked by the dominant allele (1); homozygous recessive individuals appear in F2 (1). | F1 heterozygous, recessive masked homozygous recessive in F2 |
1 |
| 5 | (a) | 4,500 / 45,000 x 100% = 10%. | 10% | 1 |
| 5 | (b) | (45,000 – 45) / 45,000 x 100% (1) = 99.9% (1). | (45,000 – 45) / 45,000 x 100% 99.9% |
2 |
| 5 | (c) | Respiration releases energy as heat (1); some biomass is not eaten, or energy remains in faeces and excretory products (1). | energy lost as heat in respiration uneaten biomass or energy in faeces |
2 |
| Section B | ||||
| 6 | (a) | Two marks each (any four): (i) large surface area from many alveoli (~75 m2), increasing diffusion; (ii) one-cell-thick squamous epithelium, short diffusion distance; (iii) dense capillary network and blood flow maintaining a gradient; (iv) moist surface so gases dissolve first; (v) ventilation-perfusion matching. | large surface area speeds diffusion thin single-layer wall, short distance dense capillaries keep a gradient moist surface dissolves gases ventilation matched to blood flow |
8 |
| 6 | (b) | Mark points (2 each): (i) haemoglobin has four haem groups, each binding one O2; (ii) oxyhaemoglobin forms at high partial pressure in the lungs; (iii) O2 is released at low partial pressure in tissues; (iv) cooperative binding gives the S-shaped curve; (v) during exercise CO2 rises and pH falls (Bohr effect), shifting the curve right; (vi) higher temperature also shifts it right, unloading more oxygen. | four haem groups each bind one O2 oxyhaemoglobin forms at high ppO2 in lungs O2 released at low ppO2 in tissues cooperative binding gives an S-shaped curve CO2 up and pH down (Bohr effect) shifts right higher temperature shifts right, more O2 unloaded |
12 |
| 7 | (a) | Two marks each: (i) signal: nerve impulse vs hormone; (ii) route: neurone vs blood; (iii) speed: fast vs slow; (iv) duration and target: brief and local vs longer and widespread; (v) specificity of target cells (any five). | signal: impulse vs hormone route: neurone vs blood speed: fast vs slow duration: brief vs long range: local vs widespread |
10 |
| 7 | (b) | Two marks each: (i) high pressure in the glomerulus forces water and small molecules into the capsule; (ii) podocytes and basement membrane keep cells and large proteins in the blood; (iii) the proximal tubule reabsorbs glucose, amino acids and most Na+ by active transport; (iv) water follows solutes by osmosis; (v) after drinking a lot, less ADH is secreted so the collecting duct is less permeable and dilute urine is produced. | ultrafiltration by high glomerular pressure podocytes and basement membrane block large molecules active transport of glucose and Na+ in proximal tubule water follows by osmosis less ADH, large volume of dilute urine |
10 |
Topic breakdown
| Q | Topic | Marks |
|---|---|---|
| 1 | Data question: photosynthesis and light intensity | 8 |
| 2 | Cell membranes and transport | 6 |
| 3 | DNA structure and protein synthesis | 6 |
| 4 | Genetics and chi-squared test | 5 |
| 5 | Energy flow in ecosystems | 5 |
| 6 | Section B question 1: gas exchange and oxygen transport | 20 |
| 7 | Section B question 2: coordination and osmoregulation | 20 |
Practice grade boundaries set by TutorZone (not official)
| Grade | Marks |
|---|---|
| 7 | 40–50 |
| 6 | 34–39 |
| 5 | 28–33 |
| 4 | 22–27 |
| 3 | 16–21 |
| 2 | 10–15 |
| 1 | 0–9 |
These boundaries are set by TutorZone for practice only and are not official IB boundaries; the final IB Biology grade combines Paper 1, Paper 2 and the internal assessment.
Common mistakes
- Copying readings in the data question without doing the percentage or unit conversion, losing easy marks.
- Giving a chi-squared value without comparing it with the critical value, or omitting the conclusion.
- Section B answers that define terms but never describe and explain, so only one mark is credited per point.
- Vague answers on gas exchange or osmoregulation that never link structure to function.
FAQ
Does this paper match the official IB format?
The question style and timing follow the public IB Biology SL Paper 2 specification (1 h 15 min, 50 marks, structured questions), but every question and data set is original.
How should the grade boundaries be read?
The boundaries on this page are set by TutorZone for practice only. The final IB grade combines Paper 1, Paper 2 and the internal assessment.
Do I answer both Section B questions?
No. Choose one of the two. Plan about 45 minutes for Section A and 30 minutes for Section B within the 75 minutes.
Can I download and print it?
Yes. Question and answer PDFs are linked on this page, and the answer paper includes the mark scheme for self-marking or peer marking.
Written by TutorZone tutors from public IB Biology material. All questions and data are original and not endorsed by the IB.
The answer paper includes mark points and is for study use only; please credit the source when sharing.
