S5 Biology Paper: Energy Flow And Nutrient Cycling In Ecosystems (100 Marks)

S5 Biology Paper: Energy Flow and Nutrient Cycling in Ecosystems (100 marks)

2026-11-11

S5 Biology – ecosystems and energy flow

S5 Biology paper: six questions, 24 parts, 100 marks, 75 minutes. Covers energy flow, energy pyramid efficiency, biomass, population growth and human impact.

Download　Question paper PDF　|　Answer key PDF

Paper details　S5 Biology practice paper with six questions on energy flow, energy pyramids, biomass, population growth and human impact, with answers and marking points.

Question 1: Ecosystem components and energy flow

In a grassland ecosystem the gross primary production (GPP) is 80 000 kJ m⁻² per year and plant respiration uses 20 000 kJ m⁻². Primary consumers assimilate 6 000 kJ m⁻².

#	Question	Answer and marking notes
1	Calculate the net primary production (NPP) of this ecosystem.	Answer: NPP = GPP – respiration = 80 000 – 20 000 = 60 000 kJ m⁻² per year.
2	Calculate the efficiency of energy transfer from producers to primary consumers.	Answer: 6 000 / 60 000 x 100% = 10.0%, so the transfer efficiency is 10.0%.
3	Secondary consumers assimilate 480 kJ m⁻². Calculate the transfer efficiency from primary to secondary consumers.	Answer: 480 / 6 000 x 100% = 8.0%. This is lower than the 10.0% of the previous step, showing energy lost through respiration and excretion.
4	Assume every later trophic level transfers 10% of its energy. Estimate the maximum energy the tertiary consumers can assimilate.	Answer: 480 x 10% = 48 kJ m⁻² per year.

Question 2: Energy pyramids and trophic efficiency

Following Question 1, producers, primary, secondary and tertiary consumers assimilate 60 000, 6 000, 480 and 48 kJ m⁻² per year respectively.

#	Question	Answer and marking notes
5	Explain why an energy pyramid is always upright and never inverted.	Answer: At each transfer most energy is lost through respiration, excretion and unassimilated material, so only a small part is stored; usable energy therefore falls at every step and each lower level holds more energy than the one above.
6	Calculate the overall transfer efficiency from producers to tertiary consumers.	Answer: 48 / 60 000 x 100% = 0.08%.
7	A quaternary consumer needs to assimilate 5 kJ per day. Using 10% efficiency, find how much energy the tertiary consumers must supply and judge whether the present energy is sufficient.	Answer: 5 / 10% = 50 kJ is needed but only 48 kJ is available, so the level cannot support a quaternary consumer.
8	To let a quaternary consumer assimilate 5 kJ, what is the minimum energy producers must fix? Assume 10% efficiency at every step.	Answer: 5 / (0.10 x 0.10 x 0.10 x 0.10) = 5 / 0.0001 = 50 000 kJ.

Question 3: Biomass and productivity

Researchers harvest plants from a 0.25 m² quadrat in a grassland; the dry mass is 45 g. One year later the dry mass of the same quadrat is 229.5 g.

#	Question	Answer and marking notes
9	Calculate the biomass density of the quadrat in g m⁻².	Answer: 45 / 0.25 = 180 g m⁻².
10	Calculate the net productivity for that year in g m⁻² yr⁻¹.	Answer: 229.5 / 0.25 – 180 = 918 – 180 = 738 g m⁻² per year.
11	If respiration accounts for 20% of GPP, calculate the gross primary production for that year.	Answer: GPP = 738 / 0.80 = 922.5 g m⁻² per year.
12	Convert 738 g m⁻² per year into t ha⁻¹ per year (1 hectare = 10 000 m²).	Answer: 738 x 10 000 = 7 380 000 g = 7 380 kg = 7.38 t, so the value is 7.38 t ha⁻¹ per year.

Question 4: The carbon cycle and decomposition

A forest produces 2.4 kg m⁻² of litter each year and the litter contains 45% carbon. The decomposition constant is k = 0.35 per year and follows first order kinetics, N = N₀e^(−kt).

#	Question	Answer and marking notes
13	Calculate the mass of carbon entering the litter pool each year.	Answer: 2.4 x 45% = 1.08 kg C m⁻² per year.
14	Calculate the proportion of litter decomposed within one year.	Answer: 1 – e^(−0.35) = 1 – 0.7047 = 0.2953, about 29.5%.
15	Calculate the mass of carbon decomposed within one year, in grams.	Answer: 1.08 x 0.2953 = 0.3189 kg, about 319 g C m⁻².
16	Calculate the number of years needed for 90% of the litter to decompose.	Answer: t = −ln(0.10) / 0.35 = 2.3026 / 0.35 = 6.58, about 6.6 years.

Question 5: Population growth and limiting factors

A deer population on an island starts with 200 animals. The intrinsic growth rate is r = 0.15 per year and the carrying capacity is K = 800.

#	Question	Answer and marking notes
17	If the population grows exponentially without limit, calculate its size after 3 years, to the nearest whole animal.	Answer: 200 x e^(0.15 x 3) = 200 x 1.5683 = 313.7, about 314 animals.
18	Under logistic growth, at what population size is the growth rate greatest?	Answer: K / 2 = 800 / 2 = 400 animals, where crowding and growth potential balance.
19	Calculate the growth rate dN/dt when the population is 200.	Answer: dN/dt = 0.15 x 200 x (1 – 200 / 800) = 30 x 0.75 = 22.5 animals per year.
20	Calculate the growth rate when the population is 600 and explain why it equals the answer above.	Answer: dN/dt = 0.15 x 600 x (1 – 600 / 800) = 90 x 0.25 = 22.5 animals per year. Both sizes are equally far from K / 2, so the unused growth fraction is the same.

Question 6: Human impact and biological magnification

A pesticide measures 0.0005 ppm in the water, 0.04 ppm in algae, 2.0 ppm in small fish and 12.0 ppm in an eagle. One eagle of mass 2.0 kg eats 0.25 kg of small fish per day.

#	Question	Answer and marking notes
21	Calculate the magnification factor of the pesticide in the eagle relative to the water.	Answer: 12.0 / 0.0005 = 24 000 times.
22	Calculate the concentration factor of the fish relative to the algae.	Answer: 2.0 / 0.04 = 50 times.
23	Calculate the mass of pesticide the eagle takes in each day, in mg.	Answer: 0.25 kg x 2.0 mg kg⁻¹ = 0.5 mg.
24	If the safe daily intake is 0.1 mg kg⁻¹ of body mass, judge whether the eagle exceeds the limit and find the factor.	Answer: the limit is 2.0 x 0.1 = 0.2 mg; the actual 0.5 mg is above it by 0.5 / 0.2 = 2.5 times, which can cause reproductive failure over time.

Find a tutor — free Tutor Registration →

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