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Ecology

227 questions

All types227Multiple choice105True / False13Numeric8Matching59Ordering42
Question 21Question →

In ecological studies, while pyramids of numbers and biomass can sometimes be inverted in specific ecosystems, a pyramid of energy is always upright. Which of the following reasons explains why a pyramid of energy can never be inverted?

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Answer: Energy is continuously lost as heat at each successive trophic level during metabolic processes.

Answer

Energy is continuously lost as heat at each successive trophic level during metabolic processes.
The correct answer emphasizes that energy is progressively dissipated as heat due to cellular respiration and metabolic work at every trophic step. As a result of this unidirectional and inefficient energy transfer, higher trophic levels inevitably store less usable energy than preceding ones, keeping energy pyramids strictly upright.

Step-by-Step Solution

1
Identify the primary source and movement of energy in an ecosystem.
Solar energy is captured by green plants (producers) and converted into chemical energy.
Primary producers form the base of the energy pyramid and hold the maximum total energy in the ecosystem.
2
Apply thermodynamic principles to trophic energy transfers.
At each consumer level, organism respiration and heat dissipation reduce available energy by roughly 90%.
Because energy transfer is unidirectional and continually diminished, every successive trophic level contains less energy than the level below it.

Key Concept

Unidirectional energy flow and heat dissipation across trophic levels
Question 22Question →

In an aquatic ecosystem consisting of microscopic algae, zooplankton, small fish, and predatory birds, ecological measurements recorded over a seasonal cycle revealed that the standing crop biomass of zooplankton periodically exceeded that of the algae. Despite this biomass inversion, the pyramid of energy for this ecosystem remained strictly upright. Which of the following statements best accounts for why a pyramid of energy can never be inverted in a functional ecosystem?

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Answer: Energy is continuously dissipated as metabolic heat and lost to entropy at each successive trophic transfer, ensuring unidirectional flow.

Answer

Energy is continuously dissipated as metabolic heat and lost to entropy at each successive trophic transfer, ensuring a strictly unidirectional flow.
The correct answer emphasizes that energy transfer across trophic levels obeys the laws of thermodynamics. Because organisms expend energy on cellular respiration and metabolic processes, energy is lost as unrecoverable heat at every transfer step. Consequently, the energy available to successive trophic levels always decreases, keeping energy pyramids strictly upright regardless of seasonal biomass fluctuations.

Step-by-Step Solution

1
Distinguish between standing crop biomass and energy flow per unit time.
Recognize that biomass represents a static measurement at a single instant, while energy flow measures total productivity rates over time.
Phytoplankton have a very high turnover rate and rapid reproduction, allowing a small standing biomass to support a larger biomass of zooplankton.
2
Apply thermodynamic principles to energy transfer across trophic levels.
Determine that only approximately 10% of total energy at one trophic level is incorporated into organic tissue at the next level, while ~90% is dissipated via cellular respiration, excretion, and metabolic heat.
The Second Law of Thermodynamics dictates that energy transformations are inefficient, increasing environmental entropy.
3
Evaluate the structural behavior of ecological pyramids of energy.
Conclude that because energy flow is strictly unidirectional and experiences inevitable net loss at each step, lower trophic levels must always contain more total energy rate than higher levels.
Pyramids of energy reflect rates of production over time, making an inverted energy pyramid physically impossible in a stable natural ecosystem.

Key Concept

Unidirectional Energy Dissipation and Invariance of Upright Energy Pyramids
Estimated Time:1m 30s
Question 23Question →

Match each ecological measuring instrument with the specific abiotic factor it is designed to measure.

Click a left item, then click its matching right item

Items

Rain gauge
Six's thermometer
Barometer
Wind vane

Matches

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Answer

Rain gauge matches Amount of precipitation; Six's thermometer matches Diurnal temperature extremes; Barometer matches Atmospheric pressure; Wind vane matches Direction of air currents.
Each ecological measuring instrument is matched to its corresponding physical parameter: the rain gauge measures precipitation depth, Six's thermometer records the daily highest and lowest temperatures, the barometer detects atmospheric pressure, and the wind vane points to the direction of air currents.

Step-by-Step Solution

1
Identify the primary function of a rain gauge.
The rain gauge quantifies rainfall volume.
Rainfall accumulates in a funnel and graduated container to measure total precipitation depth.
2
Determine the instrument designed to capture temperature range across a 24-hour cycle.
Six's maximum and minimum thermometer tracks diurnal temperature extremes.
Six's thermometer uses dual indicators moved by expanding liquid to retain markers at maximum and minimum temperature levels.
3
Associate atmospheric pressure with its specific field instrument.
The barometer measures atmospheric pressure.
Barometers measure changes in air pressure exerted by atmospheric gases.
4
Distinguish between instruments measuring wind motion properties.
The wind vane determines wind direction.
A wind vane aligns with wind flow to indicate direction, while an anemometer measures wind speed.

Key Concept

Measurement of Abiotic Ecological Factors
Estimated Time:1m 0s
Question 24Question →

In an ecological field investigation assessing microclimatic, edaphic, and atmospheric variables across diverse habitats, match each ecological measurement requirement on the left with its corresponding measuring instrument on the right.

Click a left item, then click its matching right item

Items

Quantifying microclimatic atmospheric moisture by evaluating temperature differentials caused by evaporative cooling.
Determining light intensity and photosynthetically active radiation at different vertical canopy strata.
Measuring hydrogen ion concentration (pHpH) directly in an edaphic soil solution sample.
Evaluating ambient barometric pressure variations along an altitudinal gradient in a montane ecosystem.

Matches

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Answer

Relative humidity measured via evaporative cooling pairs with the wet and dry bulb psychrometer. Light intensity measurement across canopy layers pairs with the photometer (lux meter). Edaphic soil solution hydrogen ion concentration pairs with the soil pH meter. Atmospheric pressure evaluation along altitudinal gradients pairs with the aneroid barometer.
Each ecological parameter matches its designated measurement instrument based on standard field measurement principles in ecology: relative humidity is quantified using a psychrometer via temperature depression caused by evaporation; solar light intensity is quantified using a photometer; soil hydrogen ion concentration (pHpH) is measured using a pH meter probe; and atmospheric pressure at varying altitudes is quantified using an aneroid barometer.

Step-by-Step Solution

1
Identify the physical or chemical ecological factor described in each item on the left.
Item 1 refers to relative humidity/evaporation; Item 2 refers to light intensity; Item 3 refers to soil pH; Item 4 refers to atmospheric pressure.
Correct matching requires linking environmental variables to their specific underlying physical/chemical parameters.
2
Correlate each identified parameter with its specialized measuring tool and operational mechanism.
Relative humidity correlates with the psychrometer; light intensity correlates with the photometer; soil hydrogen ion concentration correlates with the soil pH meter; barometric pressure correlates with the aneroid barometer.
Each abiotic factor requires a specific physical sensor or transducer calibrated to detect and measure that parameter accurately.

Key Concept

Ecological Measuring Instruments and Abiotic Factor Quantitation
Question 25Question →

Consider the following organisms residing in a West African savanna ecosystem: Agama lizards, Star grass, Martial eagles, and Grasshoppers. Arrange these organisms in sequence from the trophic level containing the HIGHEST available energy to the trophic level containing the LOWEST available energy.

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Answer

Star grass → Grasshoppers → Agama lizards → Martial eagles
In accordance with the second law of thermodynamics, radiant energy fixed by primary producers (Star grass) is progressively lost as metabolic heat and waste as it flows through primary consumers (Grasshoppers), secondary consumers (Agama lizards), and tertiary consumers (Martial eagles). Consequently, available energy is always highest at the base of the food chain (trophic level 1) and lowest at the apex (trophic level 4).

Step-by-Step Solution

1
Identify the trophic role and position of each organism in the savanna food chain.
Star grass is a primary producer (trophic level 1); Grasshopper is a primary consumer/herbivore (trophic level 2); Agama lizard is a secondary consumer/carnivore (trophic level 3); Martial eagle is a tertiary consumer/apex predator (trophic level 4).
Trophic position dictates the direction of nutrient flow and relative energy content within an ecological community.
2
Apply Lindeman's efficiency rule (10% law) regarding energy transfer across trophic levels.
Energy decreases progressively from lower to higher trophic levels because approximately 90% of transferred energy is lost as heat via cellular respiration and unconsumed biomass at each link.
The second law of thermodynamics requires energy pyramids to remain upright, with energy concentration greatest at the base and lowest at the top.
3
Order the organisms from highest available energy to lowest available energy.
The correct sequence is Star grass (Producer, Level 1) → Grasshoppers (Primary Consumer, Level 2) → Agama lizards (Secondary Consumer, Level 3) → Martial eagles (Tertiary Consumer, Level 4).
Energy attenuation along a food chain mandates that producers hold the highest energy content while top predators hold the lowest.

Key Concept

Trophic energy attenuation and ecological pyramid hierarchy
Estimated Time:1m 30s
Question 26Question →

An ecologist conducting a field study along a coastal cliff needs to measure both the angle of slope of the terrain and the atmospheric pressure at different elevations. Which pair of ecological instruments should the ecologist select for these measurements?

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Answer: Clinometer and barometer

Answer

The ecologist should select a clinometer to measure the angle of slope and a barometer to measure atmospheric pressure.
The correct answer correctly pairs the clinometer, which measures slope gradient and terrain inclination, with the barometer, which quantifies atmospheric pressure.

Step-by-Step Solution

1
Identify the ecological factors specified in the study
The parameters are terrain slope angle (a topographic factor) and atmospheric pressure (a climatic factor).
Matching abiotic parameters to their appropriate measurement instruments requires categorizing each parameter correctly.
2
Determine the correct instrument for measuring slope angle
A clinometer (or abney level) is specifically designed to measure gradient or angle of inclination.
Topographic measurements of slope steepness require an instrument that quantifies angles relative to the horizontal.
3
Determine the correct instrument for measuring atmospheric pressure
A barometer measures atmospheric pressure in units such as mmHg or hPa.
Climatic variations with altitude involve changes in air pressure, which are measured using a barometer.

Key Concept

Measurement of topographic and climatic abiotic factors using ecological instruments
Question 27Question →

A farmer notices a rapid loss of soil nitrogen in a waterlogged maize field that had been treated with ammonium-based fertilizer. Soil analysis confirms that ammonium ions (NH4+\text{NH}_4^+) were first oxidized to nitrites (NO2−\text{NO}_2^-), then further oxidized to nitrates (NO3−\text{NO}_3^-), which were subsequently reduced to gaseous nitrogen (N2\text{N}_2) under anaerobic conditions. Which sequence of bacteria is sequentially responsible for these three specific biochemical transformations?

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Answer: Nitrosomonas, Nitrobacter, and Pseudomonas

Answer

The sequence Nitrosomonas, Nitrobacter, and Pseudomonas correctly identifies the organisms performing ammonium oxidation, nitrite oxidation, and denitrification respectively.
The first step of nitrification (oxidation of ammonium NH4+\text{NH}_4^+ to nitrite NO2−\text{NO}_2^-) is performed by Nitrosomonas. The second step (oxidation of nitrite NO2−\text{NO}_2^- to nitrate NO3−\text{NO}_3^-) is performed by Nitrobacter. Under waterlogged, oxygen-depleted soil conditions, anaerobic denitrifying bacteria such as Pseudomonas convert soil nitrates back into gaseous elemental nitrogen (N2\text{N}_2), causing a loss of soil fertility.

Step-by-Step Solution

1
Identify the bacterium responsible for converting ammonium ions (NH4+\text{NH}_4^+) to nitrite ions (NO2−\text{NO}_2^-).
Nitrosomonas is the nitrifying bacterium that carries out the first step of nitrification.
Ammonium oxidation is an aerobic process mediated by chemoautotrophic bacteria like Nitrosomonas.
2
Identify the bacterium responsible for converting nitrite ions (NO2−\text{NO}_2^-) to nitrate ions (NO3−\text{NO}_3^-).
Nitrobacter completes nitrification by oxidizing nitrite to nitrate.
Nitrate is the primary form of nitrogen absorbed by plants, generated through nitrite oxidation by Nitrobacter.
3
Identify the bacterium responsible for converting nitrates (NO3−\text{NO}_3^-) into gaseous nitrogen (N2\text{N}_2) in waterlogged, anaerobic soil.
Pseudomonas (or Thiobacillus denitrificans) conducts denitrification, returning nitrogen gas to the atmosphere.
Waterlogged soils lack molecular oxygen, forcing facultative anaerobes like Pseudomonas to use nitrate as a terminal electron acceptor.

Key Concept

Nitrification and Denitrification Pathways in the Nitrogen Cycle
Estimated Time:2m 0s
Question 28Question →

In a savanna ecosystem, primary producers have a gross primary productivity (GPP) of 20000 kcal m−2 yr−120{}000\text{ kcal m}^{-2}\text{ yr}^{-1}, but consume 50%50\% of this energy through autotrophic cellular respiration (RAR_A). Assuming a constant ecological efficiency of 10%10\% for energy transfer between consecutive trophic levels, what is the total amount of energy per square meter per year available to tertiary consumers?

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Answer: 10 kcal m−2 yr−110\text{ kcal m}^{-2}\text{ yr}^{-1}

Answer

The total energy available to tertiary consumers is 10 kcal m−2 yr−110\text{ kcal m}^{-2}\text{ yr}^{-1}.
To find the energy available to tertiary consumers, first calculate Net Primary Productivity: NPP=GPP−RA=20000−10000=10000 kcal m−2 yr−1\text{NPP} = \text{GPP} - R_A = 20{}000 - 10{}000 = 10{}000\text{ kcal m}^{-2}\text{ yr}^{-1}. Then apply the 10%10\% transfer efficiency across three consumer steps: primary consumers receive 1000 kcal m−2 yr−11{}000\text{ kcal m}^{-2}\text{ yr}^{-1}, secondary consumers receive 100 kcal m−2 yr−1100\text{ kcal m}^{-2}\text{ yr}^{-1}, and tertiary consumers receive 10 kcal m−2 yr−110\text{ kcal m}^{-2}\text{ yr}^{-1}.

Step-by-Step Solution

1
Calculate Net Primary Productivity (NPP) of the primary producers
NPP=GPP−RA=20000−(0.50×20000)=10000 kcal m−2 yr−1\text{NPP} = \text{GPP} - R_A = 20{}000 - (0.50 \times 20{}000) = 10{}000\text{ kcal m}^{-2}\text{ yr}^{-1}
Only energy stored as organic biomass after metabolic respiration is available to herbivores.
2
Calculate energy transferred to primary consumers (trophic level 2)
Energy2=10% of 10000=1000 kcal m−2 yr−1\text{Energy}_2 = 10\% \text{ of } 10{}000 = 1{}000\text{ kcal m}^{-2}\text{ yr}^{-1}
According to the 10%10\% law of energy transfer, only one-tenth of available energy passes to primary consumers.
3
Calculate energy transferred to secondary consumers (trophic level 3)
Energy3=10% of 1000=100 kcal m−2 yr−1\text{Energy}_3 = 10\% \text{ of } 1{}000 = 100\text{ kcal m}^{-2}\text{ yr}^{-1}
Apply the 10%10\% transfer efficiency from primary consumers to secondary consumers.
4
Calculate energy transferred to tertiary consumers (trophic level 4)
Energy4=10% of 100=10 kcal m−2 yr−1\text{Energy}_4 = 10\% \text{ of } 100 = 10\text{ kcal m}^{-2}\text{ yr}^{-1}
Apply the 10%10\% transfer efficiency from secondary consumers to tertiary consumers.

Key Concept

Calculation of net primary productivity and progressive thermodynamic energy attenuation across trophic levels.
Question 29Question →

Match each ecological concept or trophic entity on the left with its corresponding characteristic regarding energy flow and pyramid structure on the right.

Click a left item, then click its matching right item

Items

Primary Producers
Primary Consumers
Pyramid of Energy
Pyramid of Numbers (Parasitic)

Matches

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Answer

Primary Producers match with converting light energy into chemical energy at the foundation level; Primary Consumers match with occupying the second trophic level; Pyramid of Energy matches with being strictly upright due to thermodynamic heat loss; Pyramid of Numbers (Parasitic) matches with an inverted shape where one host supports many parasites.
Each item correctly aligns with fundamental ecological principles: autotrophs fix solar energy into biomass; herbivores occupy the second trophic level; energy pyramids remain exclusively upright due to thermodynamic dissipation at each level; and parasitic pyramids of numbers invert because many organisms feed on a single larger host.

Step-by-Step Solution

1
Identify the biological role of Primary Producers
They fix solar energy into organic matter at the base trophic level.
Autotrophs are the entry point of energy into ecosystems.
2
Identify the position and role of Primary Consumers
They occupy the second trophic level (herbivores).
They obtain energy by consuming primary producers.
3
Analyze the thermodynamic constraint on the Pyramid of Energy
It must always be upright.
Energy transfer between trophic levels is never 100% efficient due to metabolic heat loss.
4
Analyze structural exceptions in Pyramids of Numbers
Parasitic chains yield inverted pyramids.
A single tree or animal host supports many smaller parasites.

Key Concept

Energy Flow, Food Chains, and Ecological Pyramids
Question 30Question →

Match each ecological succession stage or concept in List I with its corresponding description in List II.

Click a left item, then click its matching right item

Items

Pioneer species
Seral community
Climax community
Secondary succession

Matches

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Answer

Pioneer species matches the first organisms to colonize bare habitats; Seral community matches the intermediate transitional stage; Climax community matches the final stable, self-sustaining community; Secondary succession matches succession in disturbed areas with pre-existing soil.
Each ecological succession stage is accurately matched to its definition: Pioneer species are the initial colonizers, seral communities are intermediate stages, the climax community is the stable endpoint, and secondary succession takes place where soil already exists after a disturbance.

Step-by-Step Solution

1
Identify pioneer organisms
Pioneer species are the earliest colonizers of new or bare land.
They initiate soil formation and modify harsh abiotic conditions.
2
Identify seral stages
Seral communities are the intermediate stages between pioneer colonization and climax equilibrium.
Species replace one another as soil depth and nutrients increase.
3
Identify climax community characteristics
Climax community represents the final mature stage.
It maintains equilibrium with climate and environmental conditions.
4
Distinguish secondary succession
Secondary succession occurs post-disturbance where organic soil already exists.
Soil presence speeds up colonization compared to primary succession on bare rock.

Key Concept

Stages and Types of Ecological Succession
Question 31Question →

In ecological succession, pioneer species and environmental conditions vary depending on substrate characteristics and whether the process represents primary or secondary succession. Which of the following correctly matches each ecological habitat scenario on the left with its appropriate pioneer community or successional characteristic on the right?

Click a left item, then click its matching right item

Items

Bare lava flow on a newly formed volcanic island
Deforested farmland abandoned after intensive cultivation
Freshly exposed aquatic substrate in a newly formed oxbow lake
Unvegetated maritime sand dune deposit behind a shoreline

Matches

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Answer

Bare lava flow matches with crustose lichens and mosses capable of soil formation through organic acid secretion. Deforested farmland abandoned after cultivation matches with herbaceous weeds and grasses arising from a pre-existing soil seed bank. Freshly exposed aquatic substrate in an oxbow lake matches with microscopic phytoplankton and submerged aquatic macrophytes trapping organic silt. Unvegetated maritime sand dune deposit matches with halophytic and drought-resistant sand-binding perennial grasses.
Each habitat matches its specific pioneer assemblage: primary xerosere on bare lava requires crustose lichens for biological rock weathering; secondary succession on abandoned farmland utilizes residual topsoil and dormant seed reserves for rapid weed growth; hydrarch succession in oxbow lakes begins with microscopic algae and submerged plants accumulating silt; and psammosere succession on mobile sand dunes requires rhizomatous grasses to stabilize windblown sand.

Step-by-Step Solution

1
Differentiate between primary and secondary succession conditions across the given habitats.
Identified bare lava flow, oxbow lake, and sand dunes as primary succession seres (devoid of soil), and abandoned farmland as a secondary succession sere (containing pre-existing fertile soil).
Secondary succession begins on intact soil containing viable seeds and organic nutrients, whereas primary succession requires substrate modification and initial soil formation.
2
Match pioneer species adaptations to substrate physical and chemical requirements.
Paired crustose lichens with bare rock, phytoplankton with aquatic substrate, sand-binding grasses with dunes, and weed seed bank colonizers with farmland soil.
Pioneer adaptations directly address limiting factors such as lack of substrate anchorage, lack of organic soil, water depth, or substrate mobility.

Key Concept

Substrate characteristics and pioneer organism adaptations across primary and secondary ecological seres
Question 32Question →

In an agricultural ecosystem, grasshoppers and palm-weevils feed directly on oil palm fronds, while praying mantises prey exclusively on grasshoppers. Agama lizards consume both grasshoppers and praying mantises. If a target pest control measure drastically reduces the grasshopper population, which of the following changes will occur in the energy flow of this food web?

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Answer: Energy transfer to praying mantises will decline, causing Agama lizards to obtain a larger proportion of their energy from alternative prey pathways.

Answer

Energy transfer to praying mantises will decline, causing Agama lizards to obtain a larger proportion of their energy from alternative prey pathways.
Because energy flows unidirectionally through food chains, removing a key primary consumer (grasshoppers) reduces the energy available to its direct predator (praying mantises). Flexible higher-level predators (Agama lizards) must adapt by utilizing alternative trophic pathways to satisfy their metabolic energy requirements.

Step-by-Step Solution

1
Analyze the trophic positions in the food web.
Oil palm fronds are primary producers, grasshoppers and palm-weevils are primary consumers, praying mantises are secondary consumers, and Agama lizards act as both secondary and tertiary consumers.
Identifying trophic connections clarifies how energy moves through the different feeding links.
2
Determine the impact of reducing the grasshopper population on energy flow.
A reduction in grasshopper numbers decreases the energy transferred to praying mantises, which feed solely on grasshoppers.
Energy flow between trophic levels depends on biomass consumption.
3
Evaluate the response of higher trophic levels.
Agama lizards, having alternative prey (such as palm-weevils or other insects), will shift their predation effort to compensate for reduced energy input from the mantis/grasshopper line.
Food webs provide alternative pathways for energy flow when specific populations fluctuate.

Key Concept

Food Web Dynamics and Energy Redistribution
Question 33Question →

An ecologist conducting an edaphic investigation needs to quantify the percentage of organic content (humus) in a soil sample without interference from soil moisture. Arrange the following procedural steps in the correct chronological sequence to complete this measurement:

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Answer

The correct chronological sequence is: (1) Heat the fresh soil sample in a drying oven at 105°C to constant mass; (2) Weigh the crucible containing the thoroughly dried soil (m1m_1); (3) Strongly heat the dried soil sample over a Bunsen burner until all organic components combust; (4) Transfer the hot crucible into a desiccator to cool; (5) Reweigh the cooled crucible containing the remaining inorganic mineral residue (m2m_2).
The procedure relies on isolating water loss from organic mass loss. First, heating at 105°C drives off all soil moisture without burning organic material. Weighing the dried soil establishes the initial dry mass. High-temperature ignition burns off the organic matter completely. Placing the hot residue in a desiccator prevents atmospheric water absorption during cooling. Finally, reweighing the cooled residue provides the mass of mineral matter remaining, allowing calculation of humus content.

Step-by-Step Solution

1
Separate soil water measurement from organic matter combustion.
Oven drying at 105∘C105^\circ\text{C} evaporates capillary and hygroscopic water without scorching organic matter.
If soil is ignited directly without prior drying at 105∘C105^\circ\text{C}, the combined mass loss of water and organic matter will skew the humus calculation.
2
Measure baseline dry soil mass (m1m_1).
Establishes a precise starting mass consisting strictly of dry mineral matter + organic matter.
Accurate calculation of percentage loss requires knowing the exact initial dry mass of the soil sample.
3
Perform high-temperature ignition.
Organic matter (humus) is completely oxidized to carbon dioxide and water vapor, leaving inorganic ash.
Organic compounds break down completely only when subjected to direct strong heating with a Bunsen burner.
4
Cool the sample under dry conditions.
The crucible and mineral residue reach thermal equilibrium without re-absorbing atmospheric water vapor.
Weighing hot apparatus creates convection currents that alter balance readings, and exposed cooled ash rapidly absorbs atmospheric moisture.
5
Obtain final mass (m2m_2) and determine humus percentage.
Percentage of humus is calculated using m1−m2m1×100%\frac{m_1 - m_2}{m_1} \times 100\%.
The difference between initial dry mass and final burnt residue mass equals the total organic content present.

Key Concept

Determination of Edaphic Factors: Soil Organic Matter Content via Loss on Ignition
Question 34Question →

An ecology student investigated the population of *Tridax procumbens* in a cassava farmland in Ogun State using a 0.5 m×0.5 m0.5\text{ m} \times 0.5\text{ m} quadrat frame. Across 10 random quadrat throws, the total number of *Tridax procumbens* plants counted was 80. What is the estimated population density of *Tridax procumbens* per square metre (m−2\text{m}^{-2}) in the farmland?

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Answer: 32 plants m−232\text{ plants m}^{-2}

Answer

The population density of *Tridax procumbens* is 32 plants m−232\text{ plants m}^{-2}.
The area of one quadrat frame is 0.5 m×0.5 m=0.25 m20.5\text{ m} \times 0.5\text{ m} = 0.25\text{ m}^2. Across 10 throws, the total area sampled is 10×0.25 m2=2.5 m210 \times 0.25\text{ m}^2 = 2.5\text{ m}^2. Dividing the total number of plants counted (80) by the total sampled area (2.5 m22.5\text{ m}^2) yields an accurate population density of 32 plants m−232\text{ plants m}^{-2}.

Step-by-Step Solution

1
Calculate the area of a single quadrat frame.
Area of 1 quadrat=0.5 m×0.5 m=0.25 m2\text{Area of 1 quadrat} = 0.5\text{ m} \times 0.5\text{ m} = 0.25\text{ m}^2
Determines the spatial coverage of one quadrat throw.
2
Calculate the total area sampled across all quadrat throws.
Total sampled area=10×0.25 m2=2.5 m2\text{Total sampled area} = 10 \times 0.25\text{ m}^2 = 2.5\text{ m}^2
Accounting for all 10 sampling units used during field sampling.
3
Divide the total count of organisms by the total sampled area.
Population Density=80 plants2.5 m2=32 plants m−2\text{Population Density} = \frac{80\text{ plants}}{2.5\text{ m}^2} = 32\text{ plants m}^{-2}
Population density is defined as the number of individuals of a species per unit area.

Key Concept

Quadrat Sampling and Population Density Calculation
Question 35Question →

In a freshwater pond ecosystem, energy flows sequentially through distinct trophic levels following the principle of ecological energy transfer. Consider the following aquatic organisms: Freshwater pike, Water fleas (Daphnia), Microscopic green algae (Chlorella), and Minnows.

Arrange these organisms in order from the HIGHEST available energy to the LOWEST available energy.

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Answer

The correct order from highest available energy to lowest available energy is: Microscopic green algae (Chlorella) → Water fleas (Daphnia) → Minnows → Freshwater pike.
In any ecosystem, primary producers (green algae) fix solar energy into biochemical energy and possess the greatest amount of available energy. As energy transfers to primary consumers (water fleas), secondary consumers (minnows), and tertiary consumers (freshwater pike), roughly 90% of the energy at each level is dissipated as metabolic heat. Consequently, available energy decreases continuously from producers up to apex carnivores.

Step-by-Step Solution

1
Identify the trophic role of each organism in the pond ecosystem.
Microscopic green algae are primary producers (TL1), Water fleas are primary consumers (TL2), Minnows are secondary consumers (TL3), and Freshwater pike are tertiary consumers (TL4).
Determining trophic positions is necessary to trace the direction of energy transfer along the food chain.
2
Apply the 10% law of energy transfer (Lindeman's efficiency principle).
Energy decreases by approximately 90% at each successive trophic level due to metabolic respiration, movement, excretion, and heat dissipation.
The second law of thermodynamics requires that energy available to subsequent trophic levels decreases steadily from producers to top carnivores.
3
Arrange the organisms from maximum available energy (Trophic Level 1) to minimum available energy (Trophic Level 4).
Microscopic green algae (Chlorella) [TL1] → Water fleas (Daphnia) [TL2] → Minnows [TL3] → Freshwater pike [TL4].
Energy pyramids are strictly upright, meaning energy content is highest at the base and lowest at the apex.

Key Concept

Trophic Energy Transfer and the 10% Law
Estimated Time:1m 15s
Question 36Question →

In an aquatic ecosystem, a biologist observes that the pyramid of biomass is inverted, with the standing crop of zooplankton exceeding that of the phytoplankton at any given time. However, the pyramid of energy for the same ecosystem remains upright. Which of the following best explains why a pyramid of energy can never be inverted in any natural ecosystem?

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Answer: Energy transfer between trophic levels is accompanied by continuous loss of heat energy due to metabolic activities.

Answer

Energy transfer between trophic levels is accompanied by continuous loss of heat energy due to metabolic activities.
The pyramid of energy reflects the rate of energy flow and productivity per unit area over a given period. As energy moves from one trophic level to the next, a large portion (around 90%) is lost as heat through respiration and metabolic processes. Consequently, the energy available at a higher trophic level is strictly less than at the preceding level, ensuring the pyramid of energy is always upright.

Step-by-Step Solution

1
Analyze the nature of energy flow in ecological systems.
Energy enters ecosystems primarily as solar radiation and is converted to chemical energy by primary producers, flowing unidirectionally through trophic levels.
Understanding the source and direction of energy flow sets the foundation for evaluating pyramid structures.
2
Apply the second law of thermodynamics and the 10% energy transfer rule.
At each trophic transition, approximately 90% of energy is lost through cellular respiration, movement, excretion, and heat dissipation, leaving only about 10% for the next trophic level.
Because energy decreases progressively at every higher trophic level over time, the total energy content at a lower level must always exceed that of a higher level.
3
Distinguish between standing biomass and energy productivity rate.
While standing biomass at a single point in time can be inverted due to rapid turnover of phytoplankton, the total energy fixed and transferred per unit time always yields an upright energy pyramid.
Pyramids of energy represent productivity over time, making an inverted energy pyramid physically impossible in a self-sustaining ecosystem.

Key Concept

Unidirectional Energy Flow and Thermodynamic Constraints on Energy Pyramids
Estimated Time:1m 0s
Question 37Question →

The northernmost ecological zone in Nigeria is characterized by sparse vegetation, low rainfall, and prolonged dry seasons. Which biome does this region belong to?

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Answer: Sahel savanna

Answer

Sahel savanna
The Sahel savanna is Nigeria's northernmost terrestrial biome. It borders the semi-arid regions, receiving minimal annual rainfall and supporting drought-resistant vegetation like acacia and short grasses.

Step-by-Step Solution

1
Identify the geographical orientation and abiotic profile described
The region is located at the northernmost boundary of Nigeria with low rainfall and long dry seasons
Vegetation zones in Nigeria follow a precipitation gradient from south to north
2
Match the environmental profile to the correct Nigerian biome
The Sahel savanna forms the northernmost vegetation band adjacent to the Sahara desert edge
Sahel savanna receives the least rainfall (300-500 mm annually) among Nigerian terrestrial biomes

Key Concept

Distribution and Abiotic Profiles of Nigerian Local Biomes
Estimated Time:45s
Question 38Question →

Arrange the following biological transformations of nitrogen in the correct sequence, starting from organic waste breakdown and ending with the release of free nitrogen gas into the atmosphere.

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Answer

The correct sequence starts with ammonification of organic waste into ammonium ions, followed by Nitrosomonas oxidation to nitrites, Nitrobacter oxidation to nitrates, and finally denitrification to nitrogen gas by Pseudomonas.
The biological nitrogen cycle begins with ammonification (converting organic matter to ammonium), followed by a two-stage nitrification process where Nitrosomonas oxidizes ammonium to nitrite, and Nitrobacter oxidizes nitrite to nitrate. Finally, anaerobic denitrifying bacteria such as Pseudomonas reduce nitrate to atmospheric nitrogen gas.

Step-by-Step Solution

1
Identify the starting compound and initial biochemical step.
Decomposers perform ammonification, breaking organic proteins down into ammonium ions (NH4+NH_4^+).
Organic waste must first be converted into inorganic nitrogenous forms before nitrification can occur.
2
Determine the first oxidation stage of nitrification.
Nitrosomonas converts ammonium ions (NH4+NH_4^+) into nitrite ions (NO2−NO_2^-).
Nitrification proceeds in two distinct bacterial steps, starting with ammonium oxidation.
3
Determine the second oxidation stage of nitrification.
Nitrobacter converts nitrite ions (NO2−NO_2^-) into nitrate ions (NO3−NO_3^-).
Nitrate is the primary oxidized form utilized by plants and susceptible to denitrification.
4
Identify the final step returning nitrogen to the atmosphere.
Denitrifying bacteria like Pseudomonas reduce nitrates (NO3−NO_3^-) back into gaseous nitrogen (N2N_2).
Denitrification completes the biogeochemical cycle by converting fixed nitrogen back into gaseous form.

Key Concept

Nitrogen Cycle Bacterial Transformations
Estimated Time:1m 30s
Question 39Question →

In a deep freshwater lake ecosystem (lentic habitat), an ecological survey measured light penetration and metabolic activity across different depth zones. Zone A is the shallow shore region with rooted vegetation. Zone B is the sunlit open-water layer dominated by phytoplankton. Zone C is the deep layer situated below the light compensation level, where respiratory oxygen consumption exceeds photosynthetic oxygen production. Which of the following correctly identifies Zone C and describes the primary metabolic role of its resident biological community?

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Answer: Profundal zone; populated mainly by heterotrophic decomposers and detritivores dependent on organic fallout from upper layers

Answer

Zone C is the profundal zone, populated mainly by heterotrophic decomposers and detritivores dependent on organic fallout from upper layers.
In a lentic (freshwater lake) ecosystem, the region extending below the light compensation level where light intensity drops below 1%1\% of surface illumination is termed the profundal zone. Because photosynthesis cannot take place in this aphotic zone, the biological community consists exclusively of heterotrophic organisms, such as benthic detritivores, fungi, and bacteria, which feed on decaying organic detritus sinking from the sunlit littoral and limnetic zones above.

Step-by-Step Solution

1
Analyze the environmental parameters described for Zone C in the freshwater lake ecosystem
Zone C is described as being deep and below the light compensation level, meaning cellular respiration exceeds photosynthesis (R>PR > P).
The light compensation depth marks the boundary where photosynthetic rate equals respiratory rate; below it lies the aphotic profundal zone.
2
Match the depth profile and light conditions to standard freshwater lake zonation
Shallow shore = Littoral (Zone A); sunlit open surface water = Limnetic (Zone B); deep dark open water = Profundal (Zone C).
Lake zonation classifies regions based on proximity to shore and depth of light penetration.
3
Deduce the metabolic role of organisms living in an aphotic environment
Without sunlight, primary production via photosynthesis is absent, so resident organisms must be heterotrophs, scavengers, detritivores, and decomposers.
Energy input in aphotic benthic/profundal zones relies entirely on organic matter (detritus) sinking from the euphotic zone above.

Key Concept

Freshwater Lake Zonation and Trophic Organization
Estimated Time:1m 30s
Question 40Question →

During extended periods of waterlogging in agricultural soils, anaerobic conditions develop rapidly. Which ecological process is enhanced under these oxygen-deficient conditions, leading to a direct depletion of soil nitrogen usable by plants?

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Answer: Conversion of nitrates into atmospheric nitrogen gas

Answer

Conversion of nitrates into atmospheric nitrogen gas
Under oxygen-depleted (anaerobic) conditions such as flooded or waterlogged soils, denitrifying bacteria reduce nitrates into gaseous nitrogen. This process, termed denitrification, releases nitrogen gas into the atmosphere and causes a loss of available soil nutrients.

Step-by-Step Solution

1
Identify environmental condition
Waterlogging leads to oxygen depletion (anaerobic soil conditions).
Water fills soil pore spaces, restricting oxygen diffusion needed for aerobic respiration.
2
Determine metabolic process favored by anaerobic conditions
Denitrifying bacteria reduce soil nitrates to nitrogen gas.
Anaerobic organisms such as PseudomonasPseudomonas species utilize nitrate (NO3−\text{NO}_3^-) as an electron acceptor when oxygen is scarce.
3
Assess ecological impact on soil nitrogen
Plant-usable nitrogen escapes from the soil into the atmosphere as gaseous dinitrogen (N2\text{N}_2).
Gaseous nitrogen cannot be directly absorbed by crops, leading to depleted soil fertility.

Key Concept

Denitrification under anaerobic conditions
Estimated Time:1m 0s
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