BiologyISC Class 12 · Biodiversity and its Conservation

Patterns of Biodiversity: Latitudinal Gradient and Species-Area Relationship

Updated 15 October 2026 · Fact-checked

Species diversity generally decreases from the equator towards the poles (latitudinal gradient) because the tropics have had long undisturbed evolutionary time, a less seasonal environment and more solar energy. Within a region, species richness rises with area up to a limit (Humboldt): log S = log C + Z log A, with Z about 0.1-0.2, but 0.6-1.2 for whole continents.

Understand Patterns of Biodiversity

1. Latitudinal gradient. Diversity of plants and animals is uneven. As a rule, the farther a place lies from the equator, the fewer species it has. Barring a few exceptions, the tropics (23.5° N to 23.5° S) hold more species than temperate or polar regions.

Place	Latitude	Bird species
Colombia	near the equator	about 1400
New York	41° N	105
Greenland	71° N	56
India (mostly tropical)		more than 1200

A tropical forest such as Ecuador's has up to 10 times as many vascular plant species as an equal temperate area (Midwest USA). The Amazon rain forest has the greatest biodiversity on earth: more than 40 000 plant species, 3000 fishes, 1300 birds, 427 mammals, 427 amphibians, 378 reptiles and more than 1 25 000 invertebrates.

Why are the tropics richer? Three hypotheses.
1. Time: speciation is a function of time. Temperate regions suffered frequent glaciations; tropical latitudes stayed relatively undisturbed for millions of years, giving a long evolutionary time for diversification.
2. Constancy: tropical environments are less seasonal, more constant and predictable, which promotes niche specialisation and greater species diversity.
3. Energy: more solar energy in the tropics gives higher productivity, which may indirectly support greater diversity.

2. Species-area relationship (Alexander von Humboldt). During his travels in South American forests, Alexander von Humboldt, a German naturalist and geographer, noticed that within a region the number of species found rises as a larger area is explored, but only up to a limit.

On ordinary axes the curve is a rectangular hyperbola.
On logarithmic axes it is a straight line: log S = log C + Z log A, equivalent to S = C A^Z.
S: species richness | A: area | Z: slope (regression coefficient) | C: Y-intercept.
Z lies between 0.1 and 0.2 regardless of taxon or region (plants in Britain, birds in California, molluscs in New York state).
For very large areas such as whole continents, the slope is much steeper: Z = 0.6 to 1.2; for frugivorous birds and mammals in tropical forests of different continents, Z = 1.15. A steeper slope means species richness rises much faster with area, because separate continents have largely different sets of species (high turnover between regions).

Using the equation.

To find S: compute A^Z (or use logs), then multiply by C.
To compare two regions: compute S for each; the larger S is more diverse.
Effect of changing area: S₂/S₁ = (A₂/A₁)^Z, so halving or quartering a habitat removes a predictable share of species.

Diagram checklist. Graph 1: species richness (S) on the y-axis against area (A) on the x-axis: a curve rising steeply then flattening (rectangular hyperbola). Graph 2: log S against log A: a straight line; label slope Z and intercept log C. Label both axes on both graphs.

Patterns of Biodiversity: key facts

Species-area (log form): $lo g S = lo g C + Z lo g A$; Straight line on log-log axes.
Species-area (power form): $S = C A^{Z}$; Same relationship without logs.
Typical Z: $Z \approx 0.1 - 0.2 (regions); 0.6 - 1.2 (continents)$; 1.15 for frugivorous birds and mammals across continents.
Change of area: $\frac{S _{2}}{S _{1}} = (\frac{A _{2}}{A _{1}})^{Z}$; Predicts loss of species when habitat shrinks.

How to answer Patterns of Biodiversity questions

ISC asks the latitudinal gradient with reasons, Humboldt's species-area relationship with graph and equation, and numericals finding S for given C, A and Z.

1Write S = C A^Z (or log S = log C + Z log A) before substituting.
2Multiply C by A^Z; never divide A by C.
3If Z = 1, S = C × A directly.
4Compare regions using the computed S and name the more diverse one.
5For reasons tropics are richer, give the three hypotheses: time (no glaciation), constancy (niche specialisation), energy (productivity).
6Draw both graphs (hyperbola; straight line on log scale) with labelled axes.

Patterns of Biodiversity in real life

The idea, met in an everyday situation. Answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Shrinking a reserve

Suppose a state proposes cutting a wildlife reserve that holds 120 bird species to half its area for a highway project, and conservationists use Z = 0.2 for this region.

Estimate how many bird species the smaller reserve can support in the long run.

Try it first, then show the solutionSolution

S₂/S₁ = (A₂/A₁)^Z = (0.5)^0.2.
(0.5)^0.2 ≈ 0.871.
S₂ ≈ 120 × 0.871 ≈ 104.5, so about 104-105 species.
About 15 to 16 species (120 − 104.5) may eventually be lost, even though half the area remains.

Answer: About 104 species; roughly 15 to 16 species could disappear.

Memory hook: Halve the land, lose a slice of life.

Tricks to remember Patterns of Biodiversity

Short memory aids. Say each one aloud once, then use it on a question the same day.

T-C-E for tropics
Time (no ice ages), Constancy (less seasonal), Energy (more sunlight).
Use it when: Three reasons for tropical richness.
Z small at home, big abroad
Z is 0.1-0.2 within a region but 0.6-1.2 across continents.
Use it when: Z-value questions.
Cats Adore Zebras
S = C × A to the power Z.
Use it when: Writing the equation in order.

Patterns of Biodiversity: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Latitudinal gradient · 3 marks · Short answerBoard level

What is the latitudinal gradient in biodiversity? Give two hypotheses that explain it.

Try it first, then show the solutionSolution

Species diversity generally decreases from the equator towards the poles; tropics (23.5° N-23.5° S) have the most species (1 mark).
Tropics remained relatively undisturbed for millions of years (no frequent glaciations), giving long evolutionary time for speciation (1 mark).
Tropical environments are less seasonal, constant and predictable, promoting niche specialisation (or: more solar energy gives higher productivity) (1 mark).

Answer: Diversity falls towards the poles; time, constancy and energy explain the tropical excess.

Example 2: Species-area graph · 3 marks · Short answerBoard level

Who described the species-area relationship? Write its equation and name each term. What does the graph look like on ordinary and logarithmic scales?

Try it first, then show the solutionSolution

Alexander von Humboldt; species richness increases with area up to a limit (1 mark).
log S = log C + Z log A; S = species richness, A = area, Z = slope (regression coefficient), C = Y-intercept (1 mark).
Ordinary scale: rectangular hyperbola; logarithmic scale: straight line (1 mark).

Answer: Humboldt; log S = log C + Z log A; hyperbola, straight line on log scale.

Example 3: Calculating S · 2 marks · Short answerHarder board level

For a region, C = 5 and Z = 0.2. Calculate the species richness expected in an area of 59 049 km² (note that 59 049 = 3¹⁰).

Try it first, then show the solutionSolution

S = C A^Z = 5 × (3¹⁰)^0.2 = 5 × 3² (1 mark).
S = 5 × 9 = 45 species (1 mark).

Answer: 45 species.

Example 4: Comparing two islands with Z = 1 · 3 marks · Short answerHarder board level

Island X has A = 45 000 km² and C = 8; island Y has A = 30 000 km² and C = 15. Taking Z = 1, find S for each and state which is richer.

Try it first, then show the solutionSolution

With Z = 1, S = C × A (1 mark).
X: 8 × 45 000 = 360 000; Y: 15 × 30 000 = 450 000 (1 mark).
Y is richer despite its smaller area, because its intercept C is higher (1 mark).

Answer: X 360 000, Y 450 000; Y is more diverse.

Example 5: Meaning of steep slopes · 3 marks · Short answerHarder board level

If an area is doubled, by what factor does species richness increase when Z = 0.15 and when Z = 1.15? What does the steeper slope tell us?

Try it first, then show the solutionSolution

Z = 0.15: factor 2^0.15 ≈ 1.11 (about 11% more species) (1 mark).
Z = 1.15: factor 2^1.15 ≈ 2.22 (more than double) (1 mark).
A steep slope (as across continents) means each new area adds many new species, because different continents hold largely different species (1 mark).

Answer: About 1.11 versus 2.22; steep Z means high turnover of species between large regions.

Example 6: Habitat loss forecast · 4 marks · Short answerHOTS / competency

Suppose a forest region is reduced to one quarter of its area. (a) Using Z = 0.15, what fraction of species is expected to survive? (b) What fraction would survive if the region behaved like a continent with Z = 1? (c) Why are conservationists alarmed by fragmentation even when 'only' a part of a forest is cleared?

Try it first, then show the solutionSolution

(a) (1/4)^0.15 ≈ 0.81, so about 81% survive (about 19% lost) (1½ marks).
(b) (1/4)^1 = 0.25: only 25% survive (1 mark).
(c) Species richness falls with area, so even partial clearing causes predictable losses; small fragments also cannot support animals needing large territories, so actual losses can be greater (1½ marks).

Answer: (a) About 81%. (b) 25%. (c) Losses scale with area and fragments fail wide-ranging species.

Where students lose marks in Patterns of Biodiversity

The slips examiners see most often, and what to write instead.

Mistake: Dividing instead of multiplying
In 2024 examiners reported that candidates divided instead of multiplying C and A, or wrote the formula without calculating.
Do this: S = C A^Z; with Z = 1, S = C × A.
Usually costs: 1 to 2 marks.
Mistake: Naming the wrong region
Examiners (2024) found Africa named instead of Europe as the more diverse, after wrong arithmetic.
Do this: Compare the computed S values, not the areas alone.
Usually costs: 1 mark.
Mistake: Wrong factors for high biodiversity
Examiners (2024) saw conservation methods or the species-area relationship described when factors increasing biodiversity were asked.
Do this: Give the tropical factors: undisturbed history, constant climate, more solar energy and productivity.
Usually costs: 1 mark.
Mistake: Unlabelled species-area graph
Graphs without axes or without the log-scale version lose marks.
Do this: Label S and A (or log S and log A); show slope Z.
Usually costs: 1 mark.

Exam tips

Write Humboldt's full name: Alexander von Humboldt.
Quote the Z ranges with their contexts (regions versus continents).
Give the tropics' latitude range (23.5° N to 23.5° S).
Show each step of A^Z; round only at the end.
Link high Z to the large differences in species between continents.

Practice questions on Patterns of Biodiversity

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
For birds within California, the slope of the log-log species-area line is closest to:
(a) 0.01 (b) 0.15 (c) 1.15 (d) 2.5
Hint
Within one region, not across continents.
Answer
(b) 0.15 (Z within a region lies between 0.1 and 0.2).
Q2 · 2 marks · Short answerBoard level
Give the number of bird species in Colombia, New York and Greenland, and state the pattern shown.
Hint
Equator to 71° N.
Answer
Colombia about 1400, New York (41° N) 105, Greenland (71° N) 56. Species diversity decreases from the equator towards the poles (latitudinal gradient).
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Tropical regions have greater species diversity than temperate ones.
Reason (R): Tropical environments are more seasonal and less predictable.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Constancy.
Answer
(c). Tropical environments are less seasonal and more constant.
Q4 · 2 marks · Short answerHarder board level
A survey finds 20 species in 100 m² and 40 species in 3200 m² of the same habitat. Calculate Z.
Hint
Z = log(S₂/S₁)/log(A₂/A₁).
Answer
Z = log(40/20)/log(3200/100) = log 2/log 32 = log 2/(5 log 2) = 0.2.
Q5 · 2 marks · Short answerHarder board level
Using the log form, find S when log C = 0.5, Z = 0.2 and A = 10⁵ km².
Hint
log A = 5.
Answer
log S = 0.5 + 0.2 × 5 = 1.5, so S = 10^1.5 ≈ 31.6, i.e. about 32 species.
Q6 · 2 marks · Short answerHarder board level
Why is the Amazon rain forest called the region with the greatest biodiversity on earth? Give four figures.
Hint
Plants, fishes, birds, invertebrates.
Answer
It is home to more than 40 000 species of plants, 3000 of fishes, 1300 of birds, 427 of mammals, 427 of amphibians, 378 of reptiles and more than 1 25 000 invertebrates (any four), with perhaps two million insect species still undescribed.
Q7 · 3 marks · Short answerHOTS / competency
Suppose a polar island and an equatorial island have the same area and the same C. Which would you expect to have more species, and why might the species-area equation alone fail to show it?
Hint
C carries the latitude effect.
Answer
The equatorial island, because of the latitudinal gradient (long undisturbed time, constant climate, more solar energy and productivity). If C and A are the same, the equation gives the same S; in reality C would be much larger for the tropical island, so the latitude effect appears through C, not through A or Z.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose two regions have Z = 1: region P with A = 48 000 km² and C = 12, and region Q with A = 52 000 km² and C = 9.
(a) Calculate S for each. (1)
(b) Which is more diverse? (1)
(c) Give one factor, other than area, that could raise biodiversity in a region. (1)
Hint
Multiply, do not divide.
Answer
(a) P: 12 × 48 000 = 576 000; Q: 9 × 52 000 = 468 000. (b) P, although it is smaller. (c) Tropical location with an undisturbed history, less seasonal climate, more solar energy and productivity (any one).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Who proposed the species-area relationship?Show answer
Alexander von Humboldt.
2.Equation of the species-area relationship?Show answer
log S = log C + Z log A.
3.Z for frugivorous birds and mammals across continents?Show answer
1.15.
4.Number of bird species in Greenland?Show answer
56.
5.Shape of the species-area curve on an ordinary scale?Show answer
Rectangular hyperbola.
6.With Z = 1, C = 12, A = 48 000 km², S = ?Show answer
576 000.

Patterns of Biodiversity: frequently asked questions

What is the latitudinal gradient of biodiversity?

The general decrease in species diversity from the equator towards the poles.

Why do tropics have more species?

They were undisturbed by glaciations for millions of years, have a less seasonal, more constant climate that favours niche specialisation, and receive more solar energy, raising productivity.

What is the species-area relationship?

Species richness increases with area up to a limit; on a log scale log S = log C + Z log A.

What does Z mean?

The slope of the log-log line (regression coefficient); about 0.1-0.2 within regions and 0.6-1.2 across continents.

How do you calculate species richness from C, A and Z?

Use S = C A^Z: raise the area to the power Z and multiply by C.