Grade by Grade
F. Population Dynamics · Population Dynamics

Estimating Population Size: Quadrats, Transects and Mark-Recapture

F3.2 F2.1 A1.2 A1.6 A1.8 A1.13About 70 minutes
Learning goals
  • I can choose a suitable method (total count, quadrats, transects, mark-recapture or a newer technology) to estimate the size of a given population.
  • I can estimate density and population size from quadrat data, including correct edge rules and unit conversions.
  • I can calculate population size with the Lincoln-Petersen and Chapman mark-recapture formulas.
  • I can identify sources of error and bias in population estimates and explain how to reduce them.

Words to know

census (total count)
Counting every individual in a population. Possible only for small, visible or easily photographed populations.
quadrat
A square (or circular) frame of known area used to sample plants or slow-moving organisms.
percent cover
The percentage of a quadrat's area covered by a species, used when individuals are hard to tell apart (grasses, mosses).
transect
A line across a habitat along which organisms are recorded at intervals, used to show change along a gradient.
mark-recapture
A method for mobile animals: mark a sample, release it, recapture a second sample and use the fraction marked to estimate N.
Chapman estimator
A corrected mark-recapture formula, N = (M + 1)(C + 1) ÷ (R + 1) − 1, that is less biased when R is small.
environmental DNA (eDNA)
Traces of DNA that organisms shed into water or soil. Testing a water sample for a species' DNA shows whether it is present.
sampling bias
A systematic error caused by the way samples are chosen, so the estimate is pushed consistently too high or too low.

Why estimate?

To calculate a growth rate, set a fishing quota or decide whether a species is at risk, you first need N. Counting every individual (a census) is sometimes possible: biologists photograph seabird colonies from above and count every nest, and caribou herds are photographed from aircraft when they gather after calving. For most populations, though, a census is impossible, so ecologists sample and then scale up.

Remember from Grade 11: in Diversity of Living Things you used quadrats and the Lincoln-Petersen mark-recapture formula N = (M × C) ÷ R. This lesson sharpens those tools and adds ways to judge how trustworthy an estimate is.

Quadrats, done properly

  1. Choose a quadrat size that usually contains several individuals but is quick to count: about 0.25 m² for small plants, 1 m² or larger for shrubs.
  2. Place quadrats randomly (use random numbers to pick grid coordinates) or systematically (every 5 m along a line). Never pick spots that "look good"; that is sampling bias.
  3. Use an edge rule. Plants that touch the border are a problem: count them on two sides only (for example, top and left) and ignore them on the other two. On average this counts half the edge plants, which is correct.
  4. Record counts (for individuals you can tell apart) or percent cover (for grasses and mosses).
  5. Scale up: mean count per quadrat ÷ quadrat area = density; density × total area = population estimate.
one 0.25 m² quadratinside: count (7)on top or left edge: count (2)on bottom or right edge: do not count (2)Count = 7 + 2 = 9 plantsDensity = 9 ÷ 0.25 m² = 36 plants/m²
The edge rule: plants touching the top or left side are counted, plants touching the bottom or right side are not. Here the count is 9 in a 0.25 m² quadrat, so the density is 36 plants/m².
Worked example

Let's estimate a population from quadrats

Eight 0.25 m² quadrats are placed randomly in a 1.2 hectare (12 000 m²) meadow. Counts of common milkweed stems: 2, 0, 3, 1, 4, 2, 0, 4. Estimate the density and the total population.

  1. Mean per quadrat: (2 + 0 + 3 + 1 + 4 + 2 + 0 + 4) ÷ 8 = 16 ÷ 8 = 2.0 stems per quadrat.
  2. Density: 2.0 ÷ 0.25 m² = 8.0 stems/m².
  3. Convert hectares: 1 ha = 10 000 m², so 1.2 ha = 12 000 m².
  4. Population: 8.0 stems/m² × 12 000 m² = 96 000 stems.
  5. Report sensibly: eight small quadrats in a big meadow give a rough estimate, so write "about 1 × 10⁵ stems" or "roughly 96 000", not 96 000 exactly.

Density ≈ 8.0 stems/m²; population ≈ 96 000 stems (about 1 × 10⁵).

Transects

When conditions change across a habitat (from a lakeshore up into forest, or from a road edge into a woodlot), random quadrats can miss the pattern. A line transect is a tape stretched across the gradient; you record every organism touching the line at fixed intervals. A belt transect places quadrats every few metres along the tape, so you get density at each distance. Transects are also used for animals: a biologist walks a fixed route and counts every bird heard or seen, which is the basis of many breeding bird surveys.

Mark-recapture, sharpened

For mobile animals, use mark-recapture:

  • Capture and mark M animals, then release them.
  • Later, capture C animals and count how many are marked (R).
  • If marked animals have mixed evenly, R ÷ C ≈ M ÷ N, so the Lincoln-Petersen estimate is N = (M × C) ÷ R.

When R is small (less than about 10), the Lincoln-Petersen formula tends to overestimate N, and if R = 0 it cannot be used at all. Ecologists then use the Chapman estimator:

N = (M + 1)(C + 1) ÷ (R + 1) − 1

Assumptions (check every one when you evaluate a study):

  1. The population is closed between samples: no births, deaths, immigration or emigration.
  2. Marks are not lost and do not change behaviour or survival (a bright paint mark might attract predators).
  3. Marked animals mix randomly back into the population.
  4. Every animal has the same chance of capture. "Trap-happy" animals that return for bait make R too high, so N is underestimated; "trap-shy" animals make R too low, so N is overestimated.
1. Catch and mark M2. Release and mix3. Catch C, count R markednet holds M = 6; mark them24 fish, 6 marked, mixedC = 8 caught, R = 2 markedN ≈ (M × C) ÷ R = (6 × 8) ÷ 2 = 24 fish (the pond really has 24)
Mark-recapture in three steps. Because 2 of the 8 fish in the second catch are marked, about one quarter of the population is marked, so the 6 marked fish are about one quarter of N.
Worked example

Let's compare the Lincoln-Petersen and Chapman estimates

A biologist nets and fin-clips 60 brook trout in a small headwater lake. Two weeks later she nets 45 brook trout, and 9 of them have clipped fins. Estimate N both ways.

  1. Identify: M = 60, C = 45, R = 9.
  2. Lincoln-Petersen: N = (60 × 45) ÷ 9 = 2700 ÷ 9 = 300 trout.
  3. Chapman: N = (61 × 46) ÷ 10 − 1 = 2806 ÷ 10 − 1 = 280.6 − 1 = 279.6, about 280 trout.
  4. Why they differ: R = 9 is small, so one extra or one fewer recapture changes the answer a lot. The Chapman value corrects the slight upward bias of Lincoln-Petersen.
  5. Check an assumption: two weeks is short for trout, so births and deaths are probably small, but some marked fish may have died from handling; if so, R is too low and both estimates are too high.

Lincoln-Petersen: 300 trout; Chapman: about 280 trout. Both suggest roughly 300, and the Chapman value is the better estimate for this small R.

Newer tools

  • Camera traps photograph animals that pass a sensor. Individuals with unique markings (the spots of a lynx or the stripes of a bobcat) can be "recaptured" in photos without ever being handled.
  • Environmental DNA (eDNA): a litre of lake water can contain DNA shed by fish. Testing for a species' DNA shows whether it is present (for example, early detection of invasive carp), though turning eDNA amounts into exact population sizes is still an area of research.
  • Genetic mark-recapture: hair snags (barbed wire around a scent lure) collect bear hair, and DNA identifies each individual bear.
  • Aerial surveys from aircraft or drones count moose, caribou and waterfowl over large areas, and acoustic recorders identify birds, frogs and bats by their calls.
  • Community science: volunteers in the Christmas Bird Count (organized in Canada by Birds Canada) add decades of data on bird populations.

How good is the estimate?

  • Precision is how close repeated estimates are to each other. More quadrats or larger samples give more precise estimates.
  • Accuracy is how close an estimate is to the true value. Bias (non-random quadrats, trap-happy animals, lost marks) makes an estimate inaccurate even if it is precise.
  • Report a range when you can, for example the lowest and highest of several trials, and use significant figures that match your data (A1.13).
  • Repeat: running mark-recapture several times and averaging reduces random error.
Just for fun

Field biologists have a saying: the population was estimated at 280 trout, give or take a few trout who refused to fill out the survey.

Hands-on activity

Estimate a "population" two ways and test the result

You need: a large bowl, about 3 cups of dried pasta (one shape), a marker, a measuring cup, a 30 cm × 30 cm baking sheet or tray, a ruler, a calculator.

Part A: quadrat estimate on a tray

  1. Spread about 1 cup of pasta evenly over the tray. Draw (or imagine) the tray as nine 10 cm × 10 cm squares.
  2. Pick three squares with random numbers (roll a die twice: row and column, rerolling 4 to 6 or using 1-2 = row 1, 3-4 = row 2, 5-6 = row 3). Count the pasta in each chosen square, using the edge rule.
  3. Mean per square × 9 = estimate. Then count every piece on the tray and calculate the percent error.

Part B: mark-recapture in the bowl

  1. Put the remaining pasta in the bowl. Scoop out about 30 pieces, mark each with the marker (M), let the ink dry and return them.
  2. Mix well. Scoop a second sample (C) and count the marked pieces (R).
  3. Calculate both the Lincoln-Petersen and the Chapman estimates. Return the scoop and repeat four more times.
  4. Count every piece. Which formula's mean was closer? Which trials were furthest off, and why?

What to look for: the quadrat estimate is usually within about 10 to 25% of the true count; more squares would improve it. In mark-recapture, single trials vary a lot when R is small, and the Chapman estimates are usually a little lower than the Lincoln-Petersen ones. Keep pasta away from small children.

Your turn

1

Which method fits best: (a) yellow lady's-slipper orchids in a fen, (b) white suckers in a river, (c) change in plants from a beach into a forest, (d) black bears in a large park?

2

Six 1 m² quadrats contain 5, 3, 6, 2, 4, 4 plants. The field is 500 m². Estimate the population.

3

Explain the edge rule for quadrats and why it gives a fair count.

4

40 snapping turtles are marked. Later 30 are caught and 5 are marked. Find N by Lincoln-Petersen.

5

Use the Chapman estimator for the turtle data in the previous question.

6

A mark-recapture study uses bait, and marked mice keep coming back to the traps. Is N overestimated or underestimated? Explain.

7

What is the difference between precision and accuracy in a population estimate?

More practice: warm-up

1

What does the letter R stand for in mark-recapture?

2

How many square metres are in 1 hectare?

3

When is percent cover used instead of counting individuals?

4

What is eDNA?

5

Why should quadrats be placed randomly?

More practice: on your own

1

Ten 0.5 m² quadrats in a 2.0 ha old field contain a total of 35 goldenrod plants. Estimate the population.

2

In a study, 25 northern leopard frogs are toe-tagged. Later 20 frogs are caught; 0 are tagged. Why can't you use Lincoln-Petersen, and what does Chapman give?

3

A paint mark on beetles makes them easier for birds to see. How does this affect the estimate?

4

Explain why a mark-recapture study of a fish population should not span the spawning season.

5

A student's five mark-recapture trials on the same bean population give 210, 260, 190, 240, 300. Find the mean and the range, and comment on precision.

Challenge

1

A study marks 120 painted turtles. In the second sample of 80, 16 are marked. (a) Find N with Lincoln-Petersen and Chapman. (b) In fact, 10 marked turtles lost their marks before the second sample. If those 10 had kept their marks and been caught at the same rate as other marked turtles, about how many more marked turtles would have been recaptured? (c) What does this tell you about the effect of lost marks?

2

Design a study to estimate the white-tailed deer population of a 50 km² conservation area in winter. Include the method, how you would avoid bias, and one assumption you would need.

3

Two methods estimate a pond's frog population: Method A gives 410, 395, 402 (true value 520); Method B gives 470, 590, 540. Which is more precise, which is more accurate, and what could cause Method A's pattern?

Check your understanding

Quiz

1.

In a mark-recapture study, M = 50, C = 40 and R = 8. What is the Lincoln-Petersen estimate?

2.

The Chapman estimator is preferred over Lincoln-Petersen when:

3.

Which method best shows how vegetation changes from a pond edge up a hillside?

4.

Marked animals become trap-shy and avoid the traps. The estimate of N will be:

5.

A quadrat count gives 12 plants in a 0.5 m² quadrat. The density is:

6.

Which is an assumption of mark-recapture?

For parents and teachers

Big idea (F3.2, F2.1, A1.2, A1.8, A1.13): most populations are estimated by sampling. Students choose a method, use quadrats with an edge rule and unit conversions, calculate Lincoln-Petersen and Chapman mark-recapture estimates, and judge bias, precision and accuracy.

How to help: the pasta activity needs only kitchen items. Ask your teen to explain one assumption of mark-recapture and what happens to N if it is broken; this is a favourite test question.

Common difficulties:

  • Forgetting to convert hectares to m² (1 ha = 10 000 m²) or count per quadrat to count per m².
  • Direction of bias: a larger R gives a smaller N, and a smaller R gives a larger N.
  • Chapman arithmetic: add 1 to M, C and R, then subtract 1 at the end.

Sources: Lincoln-Petersen and Chapman estimators and their assumptions from standard ecology and wildlife-management texts; Birds Canada for the Christmas Bird Count.