biology : class12 Energy Flow Efficiency : Primary Productivity, Secondary Productivity, Gross Productivity, Net Productivity : Productivity MCQ Questions and Answers
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Energy Flow Efficiency
MCQ questions and answers.
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MCQ Questions
1. A plant community has a net primary productivity of 2400 kJ m−2 year−1. Herbivores ingest 600 kJ m−2 year−1, lose 180 kJ m−2 year−1 in faeces, and use 300 kJ m−2 year−1 in respiration. What is the ecological efficiency of energy transfer from plant production to herbivore production?
A.
3.5%
B.
5.0%
C.
8.0%
D.
20.0%
Correct Answer:
5.0%
Explanation:
Assimilated energy = 600 − 180 = 420 kJ m−2 year−1. Herbivore production = 420 − 300 = 120 kJ m−2 year−1. Ecological efficiency from plant NPP to herbivore production = (120/2400) × 100 = 5.0%.
2. An herbivore ingests 720 kJ of plant energy and excretes 180 kJ in faeces. What is its assimilation efficiency?
3. A consumer assimilates 500 kJ of energy from its food and uses 350 kJ in respiration. What is its production efficiency?
A.
20%
B.
30%
C.
50%
D.
70%
Correct Answer:
30%
Explanation:
Secondary production = assimilated energy − respiratory expenditure = 500 − 350 = 150 kJ. Production efficiency = (150/500) × 100 = 30%.
4. A plant community has a gross primary productivity of 18,000 kJ m−2 year−1 and plant respiration of 7,000 kJ m−2 year−1. Herbivores produce 550 kJ m−2 year−1. What percentage of the plant community's net primary productivity is transferred into herbivore production?
A.
3.06%
B.
4.58%
C.
5.00%
D.
7.86%
Correct Answer:
5.00%
Explanation:
Net primary productivity = GPP − plant respiration = 18,000 − 7,000 = 11,000 kJ m−2 year−1. Transfer efficiency to herbivore production = (550/11,000) × 100 = 5.00%.
5. Energy transfer efficiencies between successive trophic levels are 12%, 10%, and 15%. If the net primary productivity available at the producer level is 10,000 kJ m−2 year−1, how much energy becomes production at the fourth trophic level?
A.
1.8 kJ m−2 year−1
B.
18 kJ m−2 year−1
C.
180 kJ m−2 year−1
D.
1,800 kJ m−2 year−1
Correct Answer:
18 kJ m−2 year−1
Explanation:
The energy reaching the fourth trophic level is 10,000 × 0.12 × 0.10 × 0.15 = 18 kJ m−2 year−1. Multiplicative losses at each transfer make the final energy much smaller than the initial NPP.
6. Two consumers each assimilate 800 kJ of energy. Consumer X respires 560 kJ, whereas consumer Y respires 400 kJ. Which statement correctly compares their production efficiencies?
A.
X has 20% and Y has 30% production efficiency
B.
X has 30% and Y has 50% production efficiency
C.
X has 40% and Y has 60% production efficiency
D.
Both have 50% production efficiency
Correct Answer:
X has 30% and Y has 50% production efficiency
Explanation:
X produces 800 − 560 = 240 kJ, so its production efficiency is 240/800 × 100 = 30%. Y produces 800 − 400 = 400 kJ, so its production efficiency is 400/800 × 100 = 50%.
7. Which expression most appropriately represents ecological efficiency between two successive trophic levels?
A.
Energy respired by the higher trophic level divided by its assimilated energy
B.
Energy stored as production at the higher trophic level divided by production at the lower trophic level
C.
Energy lost in faeces divided by energy ingested by the lower trophic level
D.
Gross primary productivity divided by plant respiration
Correct Answer:
Energy stored as production at the higher trophic level divided by production at the lower trophic level
Explanation:
Ecological efficiency measures the fraction of production at one trophic level that is converted into production at the next level. It is calculated as higher-level production divided by lower-level production, usually expressed as a percentage.
8. A plant community has a gross primary productivity of 25,000 kJ m−2 year−1 and a net primary productivity of 15,000 kJ m−2 year−1. Herbivores generate 300 kJ m−2 year−1 of production. What is the energy-transfer efficiency calculated relative to net primary productivity?
A.
0.8%
B.
1.2%
C.
2.0%
D.
3.0%
Correct Answer:
2.0%
Explanation:
When efficiency is calculated relative to plant production available to consumers, NPP is used. Efficiency = (herbivore production/NPP) × 100 = (300/15,000) × 100 = 2.0%.
9. An herbivore ingests 1,000 kJ of food energy and has an assimilation efficiency of 65%. If its production efficiency is 40%, how much energy is lost in faeces?
A.
250 kJ
B.
350 kJ
C.
400 kJ
D.
600 kJ
Correct Answer:
350 kJ
Explanation:
Assimilated energy = 1,000 × 0.65 = 650 kJ. Faecal energy loss = 1,000 − 650 = 350 kJ. Production efficiency is not needed to calculate faecal loss, but it would determine the fraction of assimilated energy converted into new biomass.
10. A herbivore receives 16,000 kJ m−2 year−1 as available plant production. Its assimilation efficiency is 70% and its production efficiency is 25%. How much energy is converted into herbivore production?
A.
700 kJ m−2 year−1
B.
2,800 kJ m−2 year−1
C.
4,000 kJ m−2 year−1
D.
11,200 kJ m−2 year−1
Correct Answer:
2,800 kJ m−2 year−1
Explanation:
Assimilated energy = 16,000 × 0.70 = 11,200 kJ m−2 year−1. Herbivore production = 11,200 × 0.25 = 2,800 kJ m−2 year−1. The combined efficiency is 0.70 × 0.25 = 17.5%.
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