IB DP Biology: Climate change: causes, feedbacks, ecological impacts and phenology (DP Biology) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover IB Diploma Programme Biology section D4.3 (Climate change) from the IB Biology guide, first assessment 2025. Questions 10, 11 and 12 draw on the additional higher level statements and are HL only. All data are supplied for the questions. Related pages: study guide, revision notes, IB DP Biology hub, checklist and free diagnostics.

Questions

1. Ice cores show that temperature and carbon dioxide concentration rose and fell together over hundreds of thousands of years. State the type of correlation, and explain why this alone does not show that carbon dioxide causes warming. [3]

2. A station records carbon dioxide at 379 ppm in 2005 and 418 ppm in 2022. Methane rises from 1,775 ppb to 1,911 ppb over the same years.

(a) Calculate the percentage change in carbon dioxide. [2] (b) Calculate the mean annual increase in carbon dioxide, in ppm per year, to two decimal places. [1] (c) Show that the percentage rise in methane was smaller than that for carbon dioxide. [1]

3. Outline how melting permafrost can lead to a positive feedback in global warming. [5]

4. A boreal region has the net carbon balance shown (g C m⁻² year⁻¹, positive means net gain): 1980s +52, 1990s +31, 2000s −8. Mean area burned rose from 0.8 to 2.0 million hectares per year.

(a) Define tipping point. [1] (b) Calculate the change in net carbon balance from the 1980s to the 2000s. [1] (c) Calculate the percentage increase in area burned. [1] (d) Explain how warming can change a boreal forest from a net carbon sink to a net source. [3]

5. (a) Explain why early breakout of landfast ice threatens emperor penguin breeding. [2] (b) Explain how the loss of Arctic sea ice affects walruses. [2]

6. In an upwelling zone, the annual fish catch falls from 180 to 126 thousand tonnes after several warm years.

(a) Calculate the percentage change in catch. [2] (b) Explain how warmer surface water could cause this fall. [3]

7. The mean elevation of a mountain bird species rises from 1,480 m to 1,620 m over 40 years. Temperature falls by 6.5 °C per 1,000 m (supplied).

(a) Calculate the upslope shift and the rate in metres per year. [2] (b) Calculate the temperature change this shift is equivalent to. [1] (c) Suggest two reasons why a species living near a summit is at greater risk than one on the lower slopes. [2]

8. Surface ocean pH falls from 8.20 to 8.10. Coral calcification falls from 2.4 to 1.7 mg CaCO₃ cm⁻² day⁻¹.

(a) Calculate by what percentage the hydrogen ion concentration increases. [2] (b) Calculate the percentage change in calcification. [1] (c) Explain how raised water temperature causes coral bleaching. [2] (d) Explain why loss of corals can lead to collapse of a reef ecosystem. [2]

9. Supplied carbon sequestration rates (t C ha⁻¹ year⁻¹): non-native plantation 3.0, native forest regeneration 1.9, restored peatland 0.6. Each is applied to 120 ha for 25 years.

(a) Calculate the carbon sequestered by native regeneration. [1] (b) Calculate how much more carbon the plantation sequesters than native regeneration. [1] (c) Evaluate the use of plantations compared with native regeneration and peatland restoration. [4]

10. (HL only) In a woodland, the peak of caterpillar biomass occurred on day 142 and peak demand by nestlings on day 146 in 1990. In 2020 the caterpillar peak was on day 127 and nestling demand peaked on day 134.

(a) Calculate the gap between the peaks in 2020. [1] (b) Calculate the mean shift per decade of the caterpillar peak. [1] (c) Explain how different cues could cause the two peaks to move by different amounts. [2] (d) Explain the consequence for the bird population. [2] (e) Suggest how you could test whether photoperiod acts as the cue for nest building. [2]

11. (HL only) A spruce bark beetle population produces 25 surviving daughters per female per generation in a simplified model.

(a) Calculate the increase in numbers from the start to the end of the year with one generation and with two generations. [2] (b) Explain why warming can allow a second generation in a year. [2] (c) Outline one consequence for spruce forests. [2]

12. (HL only) Winter survival of tawny owls in a mild winter is 0.62 for the grey variant and 0.80 for the brown variant.

(a) Calculate the relative fitness of the grey variant, with brown as 1.00. [2] (b) Brown owls are 25% of a population. Using the fitness values and assuming only survival differs, calculate the proportion that are brown after one generation. [2] (c) Explain how reduced snow cover can lead to evolution in this species. [3]

Answers

1. Positive correlation [1]. A third factor could drive both, or temperature could be driving carbon dioxide [1]; other evidence, such as laboratory absorption of infrared radiation, is needed to show cause [1]. [3] Examiner insight: Answers that name the correlation as positive but only repeat “correlation is not causation” without a reason usually miss the explanation point.

2. (a) Change = 418 − 379 = 39 ppm [1]; 39 ÷ 379 × 100 = 10.3% [1] (b) 39 ÷ 17 = 2.29 ppm per year [1] (c) Methane: 136 ÷ 1,775 × 100 = 7.7%, which is less than 10.3% [1] Examiner insight: The base for a percentage change is the starting reading, and ppm and ppb cannot be compared as raw numbers.

3. Warming thaws permafrost [1]; frozen organic matter becomes available to decomposers [1]; in waterlogged anaerobic conditions methanogens release methane (aerobic decay releases carbon dioxide) [1]; methane is a greenhouse gas that increases heat retention, so warming rises [1]; more warming melts more permafrost, so the process repeats and amplifies itself [1] [5] Examiner insight: A loop that stops at “methane is released” shows the cause but not the feedback; state that the result increases the original change.

4. (a) A threshold beyond which a system changes state and is hard to reverse [1] (b) −8 − 52 = −60 g C m⁻² year⁻¹ [1] (c) (2.0 − 0.8) ÷ 0.8 × 100 = 150% [1] (d) Warmer, drier conditions with less winter snow cause drought and lower primary production [1]; fires become more frequent and intense and burn legacy carbon in litter and soil [1]; carbon release then exceeds uptake, so the forest becomes a net source [1] [6] Examiner insight: Quote the sign change from positive to negative when asked about sink to source; a size of change alone does not show the switch.

5. (a) Emperor penguins breed on landfast ice [1]; early breakout before chicks are ready causes chicks to be lost, lowering breeding success [1] (b) Walruses use sea ice to rest and nurse young [1]; with less ice they crowd on land or swim farther to feed, raising energy cost and lowering survival [1] [4] Examiner insight: Each habitat answer needs the use of the ice and the consequence; naming the animal and “habitat loss” scores only once.

6. (a) (126 − 180) ÷ 180 [1] × 100 = −30% [1] (b) Warm surface water is less dense and stratifies [1]; cold nutrient-rich deep water cannot rise, so surface nutrients are used up and phytoplankton production falls [1]; less energy passes to zooplankton and fish, so catch falls [1] [5] Examiner insight: A chain of cause and effect from temperature to catch is needed; “less food” alone is too vague.

7. (a) 1,620 − 1,480 = 140 m [1]; 140 ÷ 40 = 3.5 m per year [1] (b) 140 ÷ 1,000 × 6.5 = 0.91 °C [1] (c) No higher habitat to move to [1]; smaller area near the summit means smaller population, and competition from species moving up [1] [5] Examiner insight: State the unit with each value; convert metres to kilometres in part (b) before applying the temperature gradient.

8. (a) 10^0.10 = 1.26 [1], so an increase of about 26% [1] (b) (1.7 − 2.4) ÷ 2.4 × 100 = −29.2% [1] (c) Heat stress causes the coral to expel its symbiotic algae (zooxanthellae) [1]; the coral loses photosynthetic products and its colour, and may die [1] (d) Corals build the reef structure and provide food and shelter [1]; their loss removes habitat for many species, so the community collapses [1] [7] Examiner insight: Because pH is logarithmic, a 0.10 change is not a 10% change; answers that treat it linearly lose both marks.

9. (a) 1.9 × 120 × 25 = 5,700 t C [1] (b) Plantation 3.0 × 120 × 25 = 9,000, so difference = 3,300 t C [1] (c) Plantation gives the fastest uptake [1]; but it may support fewer species and be vulnerable to pests or fire [1]; native regeneration is slower but supports biodiversity [1]; peat stores 1,800 t C here, but it protects existing carbon, so a judgement should depend on aim and site [1] [6] Examiner insight: An evaluation needs a judgement with a reason; listing pros and cons without choosing does not complete it.

10. (a) 134 − 127 = 7 days [1] (b) (142 − 127) ÷ 3 = 5.0 days per decade earlier [1] (c) One population may respond to temperature, which has changed [1]; the other may use photoperiod, which has not, so it shifts less [1] (d) Nestlings are fed after the food peak [1]; fewer fledge, so recruitment falls and the population may decline [1] (e) Keep birds in controlled temperature with different day lengths [1]; record nest-building date for each group, with replicates [1] [8] Examiner insight: A suitable test changes one variable only; keep temperature the same when the variable is day length.

11. (a) One generation: ×25 [1]; two generations: 25 × 25 = ×625 [1] (b) Beetles are ectotherms so development speeds up with temperature [1]; the warm season lasts long enough to complete another life cycle [1] (c) Larger populations attack more trees [1]; stressed spruce are killed, reducing the carbon store [1] [6] Examiner insight: Compare the two totals explicitly; the key is that the second generation multiplies the first, not adds to it.

12. (a) 0.62 ÷ 0.80 [1] = 0.78 (0.775) [1] (b) Brown contribution = 0.25 × 1.00 = 0.25; grey = 0.75 × 0.775 = 0.581 [1]; brown proportion = 0.25 ÷ 0.831 = 0.30 [1] (c) Variation exists in colour [1]; with less snow, brown owls survive and reproduce more [1]; the brown allele becomes more frequent over generations [1] [7] Examiner insight: Describe change in a population across generations; “the owls change colour” is a common wrong claim.

Where marks are usually lost

• Using the end value as the base when finding percentage change.
• Treating pH as a linear scale.
• Describing a feedback without showing that it amplifies the original change.
• Naming ocean acidification and bleaching but mixing up their causes.
• Giving range shift causes but no evidence or direction.
• Not linking a climate variable to a biological effect and then a population-level result.
• Leaving out units, or mixing ppm and ppb.
• HL only: describing individuals adapting rather than populations evolving.

Next steps

• Review with the revision notes and the study guide.
• Tick off outcomes on the IB DP Biology checklist and visit the course hub.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

IB Diploma Programme Biology guide, first assessment 2025, International Baccalaureate Organization (section D4.3).