Biodiversity — patterns and global trends
NEET regularly asks about the species-area relationship, latitudinal gradients, and quantitative values associated with biodiversity patterns. The high-frequency confusion point: mixing up the Z-values for small regions versus entire continents in Alexander von Humboldt's and Robert MacArthur & E.O. Wilson's species-area work.
What the species-area relationship says. Arrhenius described a power-law relationship: within a biogeographic region, the number of species (S) increases with area (A) as log S = log C + Z log A, where C is a y-intercept (constant varying by taxon and region) and Z is the regression slope (NCERT Class 12 Biology Chapter 13, page 220).
The Z-value trap. For smaller areas within a continent — say, counting plant species across English counties — Z typically falls in the range 0.1–0.2. When the analysis scales up to entire continents (comparing, say, tropical South America with temperate Europe), the slope steepens to 0.6–1.2. NEET distractors routinely swap these ranges.
Latitudinal gradient. Species diversity increases from the poles toward the equator. Three commonly tested explanations: (1) tropics had more evolutionary time (no glaciation disruption), (2) tropical environments are relatively stable and predictable, (3) more solar energy in the tropics supports higher productivity and more niche specialisation.
Rivet-popper hypothesis. Paul Ehrlich's analogy: losing species from an ecosystem is like popping rivets from an airplane wing — a few losses may not matter, but beyond a threshold the system fails. NEET uses this to test whether students can distinguish it from other biodiversity hypotheses.
Watch-out. When a question gives you a log S vs. log A graph and asks for the "expected Z for a small island," the answer is 0.1–0.2 — not 0.6–1.2. Read whether the question specifies a continent-scale comparison or a within-region comparison before selecting.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
In the species-area relationship described by Alexander von Humboldt, the equation log S = log C + Z log A, what does 'Z' represent?
Answer: C. In the Arrhenius equation log S = log C + Z log A, Z is the slope (regression coefficient) of the log-log plot of species number against area (NCERT Class 12 Biology Chapter 13, page 220).
Why A is wrong: A is wrong because S (not Z) represents the number of species. Z is a dimensionless slope value.
Why B is wrong: B is wrong because the y-intercept is log C, not Z. C varies by taxon and region.
Why D is wrong: D is wrong because area is represented by A, not Z. Z is the slope of the species-area curve.
What is the typical range of the slope (Z) in the species-area relationship when the analysis is confined to a small region within a continent?
Answer: A. For species-area curves within a small region of a continent (e.g., English counties), Z values range from 0.1 to 0.2. Values of 0.6–1.2 apply when comparing across entire continents (NCERT Class 12 Biology Chapter 13, page 220).
Why B is wrong: B is wrong because 0.3–0.5 falls between the two documented ranges and is not a standard Z-value cited in NCERT.
Why C is wrong: C is wrong because Z = 0.6–1.2 applies to continent-scale comparisons, not within-region analyses. This is a high-frequency NEET swap.
Why D is wrong: D is wrong because Z values above 1.2 are not documented for any standard species-area analysis in NCERT.
Which of the following is NOT a commonly cited explanation for the latitudinal gradient in species diversity?
Answer: B. The three standard NCERT explanations for higher tropical diversity are: (1) longer evolutionary time, (2) environmental stability, (3) more solar energy supporting greater productivity. Higher speciation rates in temperate zones due to seasonal pressure is not one of them (NCERT Class 12 Biology Chapter 13, page 220).
Why A is wrong: A is wrong to choose as the answer because evolutionary time IS one of the three standard explanations — tropics remained undisturbed by glaciation.
Why C is wrong: C is wrong to choose as the answer because greater solar energy driving higher productivity IS the third standard explanation.
Why D is wrong: D is wrong to choose as the answer because environmental stability and predictability IS one of the three standard explanations for tropical diversity.
A researcher studying bird species across islands within the Caribbean finds log S = 1.2 + 0.15 log A. If the study were expanded to compare bird diversity across entire continents, how would you expect the slope to change?
Answer: B. Within a region, Z falls in 0.1–0.2. When the analysis scales to whole continents, Z steepens to 0.6–1.2 because continent-to-continent differences in evolutionary history and climate amplify species turnover (NCERT Class 12 Biology Chapter 13, page 220).
Why A is wrong: A is wrong because scaling UP from regional to continental comparisons increases Z, not decreases it.
Why C is wrong: C is wrong because the slope does not remain constant across spatial scales; continent-level Z is markedly steeper than within-region Z.
Why D is wrong: D is wrong because the species-area relationship always has a positive slope — more area supports more species, never fewer.
In the species-area equation log S = log C + Z log A, a habitat patch has Z = 0.2 and currently supports 100 species. If the area is reduced to 10% of its original size (i.e., log A decreases by 1), what is the approximate new species count?
Answer: A. log S_new = log S_old + Z × Δ(log A) = log 100 + 0.2 × (−1) = 2 − 0.2 = 1.8. So S_new = 10^1.8 ≈ 63 species. A 90% area reduction with Z = 0.2 leads to roughly a 37% species loss, not a 90% loss (NCERT Class 12 Biology Chapter 13, page 222).
Why B is wrong: B is wrong because 80 would correspond to only a 20% species loss, which underestimates the impact. The correct calculation gives 10^1.8 ≈ 63.
Why C is wrong: C is wrong because 50 species would require log S = 1.7, meaning Z × Δlog A = −0.3, not −0.2.
Why D is wrong: D is wrong because assuming species drop proportionally to area (90% area loss = 90% species loss) ignores the power-law nature of the relationship. The actual loss is ~37%, not 90%.
Paul Ehrlich's rivet-popper hypothesis compares species loss in an ecosystem to:
Answer: C. Ehrlich used the airplane rivet analogy: losing a few rivets (species) may not affect flight (ecosystem function), but continued loss crosses a critical threshold and causes collapse (NCERT Class 12 Biology, Chapter 13).
Why A is wrong: A is wrong because the rivet-popper hypothesis specifically emphasises a threshold effect, not equal criticality of every component.
Why B is wrong: B is wrong because the hypothesis does not predict a sequential order of loss starting with top predators; it is about cumulative loss crossing a threshold.
Why D is wrong: D is wrong because the hypothesis describes a non-linear, threshold-based collapse, not a continuous linear decline.
A student claims: "Since Colombia has about 1,400 bird species and India has about 1,200, both countries have essentially the same biodiversity." What is the primary flaw in this reasoning?
Answer: D. Raw species counts without accounting for area differences are misleading. India's land area (~3.3 million km²) is far larger than Colombia's (~1.1 million km²), so India's species density per unit area is substantially lower. The species-area relationship (log S = log C + Z log A) shows that larger areas inherently support more species, so comparing raw totals without area normalisation is flawed (NCERT Class 12 Biology Chapter 13, page 219).
Why A is wrong: A is wrong because the latitudinal gradient explains a general trend, not identical counts. Both countries span multiple latitudes and have different biogeographic histories.
Why B is wrong: B is wrong because bird species inventories for both countries are well-documented and comparable. The flaw is in not normalising for area, not in data unreliability.
Why C is wrong: C is wrong because biodiversity is measured at multiple levels — genetic, species, and ecosystem. Species-level comparison is valid; the problem here is ignoring area.
Region X has 500 species in 10,000 km² and Region Y has 200 species in 1,000 km². Both follow log S = log C + Z log A with the same C. Which region has a higher Z value?
Answer: B. B is correct, and the point of the question is to notice what the data does NOT fix. Rearranging log S = log C + Z log A gives Z = (log S − log C)/log A. With log 500 = 2.699 over log 10000 = 4, Z_X = (2.699 − log C)/4. With log 200 = 2.301 over log 1000 = 3, Z_Y = (2.301 − log C)/3. Subtracting: Z_Y − Z_X = [4(2.301 − log C) − 3(2.699 − log C)]/12 = (1.107 − log C)/12. So the sign of the difference depends entirely on log C. If log C < 1.107, meaning C < 12.8, then Z_Y is larger; if C > 12.8, Z_X is larger; and at C = 12.8 exactly they are equal. Knowing only that the two regions SHARE a value of C does not tell you what that value is, so no comparison can be made. Being told two quantities are equal is not the same as being told what they equal.
Why A is wrong: A is wrong not because Region X definitely has the smaller Z, but because it cannot be established either way. Region X does have the higher Z whenever C > 12.8, which the data neither rules in nor out.
Why C is wrong: C is wrong because the two regions have equal Z only in the single special case C = 12.8. For every other value of C the two Z values differ, so equality is not the general answer.
Why D is wrong: D is wrong for the same reason as A: Region Y has the higher Z only when C < 12.8. That happens to cover the small values of C typical of real species-area data, but the question supplies no value of C, so it remains an assumption rather than a result.
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Given
A tropical island group has 1,000 plant species across a total area of 50,000 km². Z = 0.18 for this region. A conservation plan proposes to protect only 5,000 km² of the island group.
Required
Estimate the number of plant species expected in the protected 5,000 km².
Concept
The species-area relationship (Arrhenius equation): log S = log C + Z log A. We can use the existing data to find C, then apply it to the reduced area.
Formula
log S = log C + Z log A
Substitution
From the full region: log 1000 = log C + 0.18 × log 50000
3 = log C + 0.18 × 4.699
3 = log C + 0.846
log C = 2.154
For the protected area: log S_new = 2.154 + 0.18 × log 5000
log S_new = 2.154 + 0.18 × 3.699
log S_new = 2.154 + 0.666
Calculation
log S_new = 2.820
S_new = 10^2.820 ≈ 661 species
Note: 1,000, 50,000, and 5,000 are treated as exact (problem-defined values), so they do not constrain significant figures. The Z-value (0.18) limits precision to 2 significant figures.
Final answer
Approximately 660 species would be expected in the protected 5,000 km² area.
This means protecting 10% of the area would retain roughly 66% of the species — not 10%. This non-linear relationship between area and species loss is the conceptual trap NEET exploits.
Common trap
Assuming species loss is proportional to area loss (i.e., 90% area loss = 90% species loss, leaving only 100 species). The power-law relationship means species loss is always less severe than area loss, which is why the Z-value matters.
Similar NEET-style question
"An ecologist records 800 species in a 20,000 km² forest (Z = 0.15). If deforestation reduces the forest to 2,000 km², how many species would remain?" (Apply the same method: find C from the original data, then recalculate S for the reduced area.)
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Three levels: genetic, species, ecosystem. Hotspots: 35 globally; 4 in India (Western Ghats + Sri Lanka, Eastern Himalaya, Indo-Burma, Sundaland). Latitudinal: tropics > temperate > polar. Sp-area: log S = log C + Z log A (Z ~ 0.1-0.2).
-- NCERT Class 12 Biology, Chapter 13, p. 219Number of species S in area A increases as a power law. Z (slope) = 0.1-0.2 for small areas, 0.6-1.2 for continents.
| Symbol | Quantity | SI Unit |
|---|---|---|
| S | species number | - |
| A | area | - |
| Z | slope | - |
| C | intercept | - |
More in Ecology and Environment: 3 exam traps and mistakes · 2 formulas · 1 question pattern from its other lessons.
5 questions from NEET 2020, 2024, 2026. Answers verified against NTA official keys.
According to Robert May, the global species diversity is about :
Which of the following regions of the globe exhibits highest species diversity ?
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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