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Question #46083
The Question
A large, continuous tropical rainforest is heavily deforested, leaving behind only a few small, isolated patches preserved as 'sacred groves'. Ecological assessment of these fragmented patches, compared to the original continuous forest, will most likely reveal which of the following dynamics?
Options
- A.
A decreased perimeter-to-area ratio, leading to an expanded core habitat and a Z-value (slope of species-area relationship) stabilizing between 0.1 and 0.2.
- B.
An increased perimeter-to-area ratio, intensifying the edge effect and exposing interior species to ecotone conditions, resulting in a steeper Z-value (e.g., 0.6 to 1.2).
- C.
An increased perimeter-to-area ratio with reduced edge effects, allowing the Z-value to remain constant at 0.1 to 0.2 regardless of the habitat's isolation.
- D.
A decreased perimeter-to-area ratio, transforming the ecotone into a core habitat, thereby decreasing the Z-value below 0.1 due to high species density.
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Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Regulatory_Failure
Understand how NOT, EXCEPT, and logical constraints are used to manipulate exam results.
Concepts Behind This Question
Nutrient_Cycling_Reservoirs|Nutrient_Immobility|Biosphere_Reserves_Architecture
Why This Question Is Tricky
Common Mistakes
Conceptual Pivot: Students often memorize the NCERT line that Z-values of 0.6-1.2 apply to 'entire continents' but fail to synthesize that isolated fragmented habitats (like small sacred groves) function ecologically as 'islands'.
In island biogeography, isolation and high edge effects cause the species-area slope to steepen dramatically.
The trap is failing to link the geometric concept of perimeter-to-area ratio with the ecological concepts of edge effect and the steepening of the Z-value.
Quick Facts About This Question
Question Information
| Subject | Biology |
| Chapter | Biodiversity and Conservation - Class 12 |
| Concept | Biosphere Reserves Architecture |
| Difficulty | 8 |
| Cognitive Load | 5 |
| Target Solve Time | 120 sec |
| Access | Free |