Stomatal Density vs Stomatal Index: Calculations and Sampling

Plant Physiology I · Practical biology for A-level and introductory university study
Stomatal density asks how many stomata occupy an area. Stomatal index asks what proportion of counted epidermal units are stomata. They use different denominators, so they can tell different stories about the same leaf.

One leaf sample. Two different questions.STOMATAL DENSITYS / areaUnits: stomata per mm²STOMATAL INDEX100 × S / (S + E)Units: percent
Original comparison diagram. S = stomata; E = ordinary epidermal cells in the same counted region.

What are stomata?

A stoma is a pore associated with a pair of guard cells in the epidermis. Changes in guard-cell behaviour regulate pore opening and influence gas exchange and water loss. For these counting exercises, one pair surrounding one pore is counted as one stomatal unit. Do not count the two guard cells as two stomata.

Distinguish anatomy from activity. Counting stomata measures their abundance in a sampled surface. It does not directly measure the aperture of each pore, stomatal conductance or the transpiration rate at the moment of observation.

Density and index: the key difference

FeatureStomatal densityStomatal index
QuestionHow many stomata per area?What fraction of counted epidermal units are stomata?
FormulaS/A100 × S/(S + E)
DenominatorMeasured area AStomata plus ordinary epidermal cells
Typical unitStomata/mm²%
Needs area calibration?YesNot for the ratio itself, but comparable sampling still matters

Here E excludes the guard cells already represented by the counted stomatal units. Some specialised studies distinguish additional epidermal categories or use different counting conventions. State your convention and apply it consistently, especially when comparing your result with a published dataset.

Worked example 1: calculate both values

In an invented image dataset, a 0.20 mm² region contains 40 stomata and 360 ordinary epidermal cells. Density = 40/0.20 = 200 stomata/mm². Index = 100 × 40/(40 + 360) = 10%.

The denominator for index is 400, not 360. A calculation of 100 × 40/360 would give approximately 11.1%, but that is a stomata-to-ordinary-cell ratio rather than the defined stomatal index.

Worked example 2: convert image dimensions correctly

A calibrated rectangular field is 500 micrometres (µm) wide and 400 µm high. Since 1 mm = 1,000 µm, the dimensions are 0.5 mm and 0.4 mm. Area = 0.5 × 0.4 = 0.20 mm². If it contains 36 stomata, density = 36/0.20 = 180 stomata/mm².

You may instead calculate 200,000 µm² first and divide by 1,000,000 to convert to mm². The conversion factor is squared because this is area. Dividing a square-micrometre value by 1,000 is a common error.

For a circular microscope field with diameter 0.40 mm, radius is 0.20 mm. Area = π × 0.20² ≈ 0.1257 mm². A count of 25 gives approximately 199 stomata/mm². Using the diameter as the radius would make area four times too large and density four times too small.

Why density can change while index stays the same

Consider a hypothetical patch with 20 stomata and 180 ordinary epidermal cells. At 0.10 mm², density is 200 stomata/mm² and index is 10%. If that same set of cells expands to occupy 0.20 mm² without changes in cell numbers, density becomes 100 stomata/mm², while index remains 10%.

This thought experiment separates cell abundance from surface expansion. Index can help interpret density differences, but it is not universally constant: changes in epidermal development can alter the proportion of stomata as well. Real leaves may change in several ways at once.

Design a fair leaf comparison

Before counting, define the comparison: upper versus lower epidermis, two species, or two growth conditions. Keep leaf developmental stage, region and surface identity clear. A lower surface is termed abaxial; the upper surface is adaxial. Do not assume every plant has the same distribution between the two.

  1. Calibrate the area. Use a valid scale calibration for the image or microscope setup. Magnification alone is not an area measurement, and changing image size can invalidate an unadjusted scale.
  2. Select fields systematically or randomly. Avoid choosing only the clearest or most densely populated patches. Record any exclusions, such as folds or unreadable regions.
  3. Use a boundary rule. For example, include objects touching the top and left edges and exclude those touching the bottom and right. Apply the same rule throughout.
  4. Count both S and E in the same region. Mixing counts from different areas invalidates the index.
  5. Retain the sampling hierarchy. Record plant, leaf, surface and field. Multiple fields from one leaf are subsamples, not independent plants.

Worked example 3: unequal sampled areas

Suppose one field has 20 stomata in 0.10 mm² and a second has 60 in 0.20 mm². Their densities are 200 and 300 stomata/mm². The pooled density for the combined sampled area is (20 + 60)/(0.10 + 0.20) = 266.7 stomata/mm².

The simple average, 250, weights the two fields equally rather than weighting by area. Neither summary should be selected without defining the question. For an estimate across the combined area, use total count/total area. For biological comparisons, calculate and summarise appropriate leaf or plant estimates instead of treating every field as independent.

Frequently asked questions

Does higher density always mean greater water loss?

No. Aperture, stomatal size, leaf temperature, humidity and other factors matter. A density comparison alone cannot establish the actual rate of transpiration.

Can I compare photographs taken at different magnifications?

Yes, if their areas are correctly calibrated and the sampling and counting methods are comparable. Raw counts alone cannot be compared fairly across different field areas.

Should I average the index from every field?

Decide what you are estimating. A pooled cell-based index uses 100 × total S/(total S + total E). An unweighted mean of field percentages answers a different question. Preserve leaf and plant identities when analysing independent biological replicates.

Interactive practice quiz

Choose an answer for each question, then check your score. Explanations appear below each question; your answers stay in this browser tab.

1. Stomatal density is reported as:
Answer and explanation

Density divides the number of stomata by the measured area.

2. The stomatal index formula is:
Answer and explanation

S is the number of stomata and E the ordinary epidermal cells counted in the same area.

3. 40 stomata in 0.2 mm² give a density of:
Answer and explanation

40/0.2 = 200 stomata per square millimetre.

4. S = 30 and E = 270 give an index of:
Answer and explanation

100 × 30/(30 + 270) = 10%. Dividing by E alone gives the wrong denominator.

5. A 500 µm × 400 µm image has area:
Answer and explanation

500 µm = 0.5 mm and 400 µm = 0.4 mm. Their product is 0.2 mm².

6. One pair of guard cells surrounding one pore counts as:
Answer and explanation

Count one stomatal unit, not two guard cells as two stomata.

7. A circular field of diameter 0.4 mm has radius:
Answer and explanation

Radius is half the diameter. Use it in area = πr².

8. When the same cells spread over twice the area, density:
Answer and explanation

The same count divided by double the area gives half the density.

9. Many fields from one leaf are automatically many independent plants.
Answer and explanation

Fields are subsamples. Independent biological replication depends on the sampling design.

10. Greater density alone proves faster transpiration.
Answer and explanation

Pore opening, stomatal size, environment and other factors affect water loss.

Related lessons and sources

Continue with Plant Physiology I and Laboratory Techniques. For more unit-conversion practice, try serial dilution and CFU/mL calculations.

Original teaching diagrams, simulated numerical examples and quiz by PreachBio. Use your course’s specified counting convention when preparing assessed practical work.

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