Lessons 19–20 · 2 lessons · A. G‘afurov, A. Abdukarimov, J. Tolipova, O. Ishankulov, M. Umaraliyeva, I. Abdurahmonova. Biology Grade 11, 1st edition. Sharq Publishing and Printing Joint-Stock Company, Tashkent, 2018
20
Laboratory work 2
Textbook: pp. 106–107
GoalSolve food chain and ecological pyramid problems step by step: build a chain, find the trophic level, calculate mass and area.
New words
building a chain · zanjir tuzishtrophic level · trofik darajatenfold step · 10 baravar o‘tishyield (g per m²) · hosildorlik (g/m²)
Explanation
In the practical session problems are solved with one algorithm. 1) Build the food chain starting from the producer and number the trophic levels. 2) Decide to which level the given mass (or energy) belongs. 3) Divide by 10 for each step up and multiply by 10 for each step down; the mass must be dry mass – if wet mass is given, subtract the water share (for example 70 %) to get dry mass. 4) If an area is asked, divide the required plant mass by the yield per 1 m². 5) Write the answer with units and check that it is sensible: the bottom level is always the largest. Remember that the 10 % rule is an average, so results are approximate. Concluding remark: why large predators need large areas and why protecting them protects the whole ecosystem.
Worked examples
Problem: chain grass → hare → fox. The fox’s dry mass is 6 kg. How much grass is needed? Solution: grass (1) → hare (2) → fox (3): two steps down: 6 · 10 · 10 = 600 kg. If the grass yield is 300 g/m² of dry mass, the area = 600 000 g : 300 g/m² = 2000 m².
Problem: a wolf of 50 kg wet mass (water 60 %) feeds in the chain grass → ungulate → wolf. Solution: dry mass = 50 · (1 − 0.6) = 20 kg. Dry mass of grass = 20 · 10 · 10 = 2000 kg. This much plant production is needed for this wolf, which is why wolves are few and live over large territories.
Class activity
In pairs invent your own problem: a 4-link chain, the mass of the last level, a yield – and have another pair solve it. Check the calculations with a calculator. Do not try to catch animals outdoors.
Practice
1
Chain: alga → small fish → pike. If the pike’s dry mass is 4 kg, how many kg dry mass of alga are needed?
400
2
A wheat field yields 500 g/m² (dry). For 5000 g of product, what is the area in m²?
10
3
What is the dry mass in kg of an animal with a wet mass of 30 kg and 70 % water?
9
4
Why must the area around a predatory bird’s nest be large to support its food chain?
Energy falls tenfold at each step, so feeding one predator takes the plant production of a large area.