Invasive species and testing for competition

Invasive species and testing for competitionSpec C4.1.13, C4.1.14, C4.1.15

Endemic and invasive species

An endemic species is native to an area. An introduced species becomes invasive when it has a competitive advantage in resource acquisition over endemic species and spreads at their expense. Often its natural predators, pathogens and competitors were left behind.

Local example (UAE): mesquite, Prosopis juliflora, is native to the Americas and was planted in the UAE from the 1970s to combat desertification. It grows fast, sets abundant seed that livestock spread, and can switch between shallow lateral roots and a deep taproot, so it takes up groundwater more effectively than native plants. It forms dense thickets, and the native ghaf tree, Prosopis cineraria, is the native plant most affected where they grow together. Another well-studied example: the grey squirrel, introduced to Britain from North America, outcompetes the endemic red squirrel for seeds such as acorns, which it digests more efficiently.

Testing for interspecific competition

• Laboratory experiments: grow each species alone and then together under controlled conditions. In Gause's experiments, Paramecium aurelia and P. caudatum both thrived alone, but together P. aurelia outcompeted P. caudatum.
• Field observations by random sampling: record where each species is found. If they rarely occur together, competition may be keeping them apart.
• Field manipulation by removal: remove one species from plots and compare with control plots. On rocky shores, when the barnacle Semibalanus (Balanus) was removed, the barnacle Chthamalus survived lower down the shore than before.

If one species is more successful in the absence of the other, competition is indicated but not proven, because other factors could explain the result. Hypotheses can be tested by both experiments (the investigator manipulates a variable and controls others) and observations (natural patterns are recorded without intervention); experiments give stronger evidence of cause.

Chi-squared test for association

Record the presence or absence of two species in many random quadrats and arrange the counts in a 2 × 2 contingency table. The null hypothesis (H₀) is that there is no association: the two species are distributed independently.

1. Expected frequency for each cell = (row total × column total) ÷ grand total.
2. For each cell calculate (O − E)² ÷ E, where O is observed and E expected.
3. Add them: χ² = Σ (O − E)² ÷ E.
4. Degrees of freedom = (rows − 1) × (columns − 1) = 1 for a 2 × 2 table. Critical value at p = 0.05 is 3.84.
5. If χ² is greater than 3.84, reject H₀: there is a significant association. Compare O and E to see whether it is positive (found together more than expected) or negative (less than expected).