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Ecology - Species, communities and ecosystems

Grade 12IBBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Species are groups of organisms that can potentially interbreed to produce fertile offspring. Members of a species may be reproductively isolated in separate populations.

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Autotrophs obtain inorganic nutrients, such as CO2CO_2, H2OH_2O, and nitrates (NO3−NO_3^-), from the abiotic environment and convert them into organic compounds.

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Heterotrophs obtain their organic molecules from other organisms via different feeding mechanisms: Consumers (ingestion of living/recently killed organisms), Detritivores (internal digestion of detritus), and Saprotrophs (external digestion of dead organic matter by secreting enzymes).

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A community is formed by populations of different species living together and interacting with each other. An ecosystem is formed by a community and its interactions with the abiotic environment.

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The supply of inorganic nutrients is finite and is maintained by nutrient cycling. While energy flows through an ecosystem and is eventually lost as heat, nutrients like CC, NN, and PP are recycled.

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Ecosystems have the potential to be sustainable over long periods of time, provided there is a continuous energy source (usually sunlight), nutrient availability, and detoxification of waste products.

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Statistical association between species can be tested using the χ2\chi^2 (Chi-squared) test on data obtained by quadrat sampling. A null hypothesis (H0H_0) assumes no significant association.

📐Formulae

χ2=∑(fo−fe)2fe\chi^2 = \sum \frac{(f_o - f_e)^2}{f_e}

fe=Row Total×Column TotalGrand Totalf_e = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}

df=(r−1)(c−1)df = (r - 1)(c - 1)

💡Examples

Problem 1:

In a study of 100 quadrats, Species A and Species B were both present in 20 quadrats. Species A was present alone in 30 quadrats, and Species B was present alone in 10 quadrats. Calculate the expected frequency (fef_e) for the 'Both Species Present' cell to perform a χ2\chi^2 test.

Solution:

First, construct a 2×22 \times 2 contingency table.

  • Both present: 2020
  • Only Species A: 3030
  • Only Species B: 1010
  • Neither present: 100−(20+30+10)=40100 - (20 + 30 + 10) = 40.

Row 1 Total (Species A present): 20+30=5020 + 30 = 50 Column 1 Total (Species B present): 20+10=3020 + 10 = 30 Grand Total: 100100

fe=50×30100=15f_e = \frac{50 \times 30}{100} = 15

Explanation:

The expected frequency (fef_e) is calculated by multiplying the sum of the row and the sum of the column for that specific cell, then dividing by the total number of observations (n=100n=100). This value represents the frequency expected if the distribution of the two species were completely independent (H0H_0).

Problem 2:

Determine the degrees of freedom (dfdf) for a contingency table used to analyze the association between two species in a quadrat study.

Solution:

df=(2−1)×(2−1)=1×1=1df = (2 - 1) \times (2 - 1) = 1 \times 1 = 1

Explanation:

In a standard species association study, the contingency table is 2×22 \times 2 (Species A Present/Absent vs Species B Present/Absent). Using the formula df=(r−1)(c−1)df = (r-1)(c-1), where r=2r=2 and c=2c=2, the degrees of freedom is always 11 for this type of ecological test.

Species, communities and ecosystems - Revision Notes & Key Diagrams | IB Grade 12 Biology