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Testing for Distribution & Abundance (CIE A Level Biology)

Revision Note

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Testing for Distribution & Abundance

  • The distribution of a species describes how it is spread throughout the ecosystem
  • The abundance of a species is the number of individuals of that species
  • The distribution and abundance of a species in an area can be assessed using different practical methods:
    • Frame quadrats
    • Line and belt transects
    • Mark - release - recapture

Frame quadrats

  • Some ecosystems are very complex with large numbers of different species of different sizes
  • For the sake of logistics, sampling is often used to estimate the distribution and abundance of species
  • When carrying out sampling, square frames called quadrats can be used to mark off the area being sampled
  • Quadrats of different sizes can be used depending on what is being measured and what is most suitable in the space the samples are being made in
    • These are usually made of wood or metal and measure 25 cm × 25 cm, 50 cm × 50 cm or 1 m × 1 m
  • Quadrats must be laid randomly in the area to avoid sampling bias
    • This random sampling can be done by converting the sampling area into a grid format and labelling each square on the grid with a number
    • Then a random number generator is used to pick the sample points

  • Once the quadrat has been laid on the chosen sample point the abundance of all the different species present can be recorded

Results from quadrats

  • The results from the quadrats can be used to calculate the predicted frequency and density of a species within an area
  • Species frequency is the probability that the species will be found within any quadrat in the sample area
    • The number of quadrats that the species was present in is divided by the total number of quadrats and then multiplied by 100
    • For example, if bluebells were found in 18 out of 50 quadrats the species frequency would be (18÷50) x 100 = 36%

  • Species density indicates how many individuals of that species there are per unit area
    • The number of individuals counted across all quadrats is divided by the total area of all the quadrats
    • For example, if 107 bluebells were found across 50 quadrats that are 1m2 each the species density would be 107/50 = 2.14 individuals per m2

  • It can sometimes be difficult to count individual plants or organisms. When this is the case percentage cover of the species within the quadrat can be estimated instead
    • The quadrat is divided into 100 smaller squares
    • The number of squares the species is found in is equivalent to its percentage cover in that quadrat
    • For example, if grass is found in 89 out of 100 squares in the quadrat then it has a percentage cover of 89%

Percentage Cover using Quadrats Diagram

Percentage Cover in Quadrats

Percentage cover of grass in a quadrat - There may be some squares lacking any species and other squares with multiple species overlying one another. Therefore the total percentage cover of a single quadrat can sometimes be over or under 100%

Line & Belt transects

  • Throughout some areas, there can be changes in the physical conditions
    • For example, there may be changes in altitude, soil pH or light intensity

  • When investigating the species distribution in these kinds of areas, systematic sampling is more appropriate
  • Methods using transects can help show how species distribution changes with the different physical conditions in the area
    • A transect is a line represented by a measuring tape, along which sample are taken

  • For a line transect:
    • Lay out a measuring tape in a straight line across the sample area
    • At equal distances along the tape record the identity of the organisms that touch the line. For example, every 2m
    • This produces qualitative data

  • For a belt transect:
    • Place quadrats at regular intervals along the tape and record the abundance of each species within each quadrat
    • This produces quantitative data

Line and Belt Transects Diagram

Line and Belt Transects, downloadable AS & A Level Biology revision notes

A line transect and belt transect is carried out in the field

Mark - release - recapture

  • The methods above are only useful for stationary organisms
  • Different methods are required for estimating the number of individuals in a population of mobile animals
  • The mark-release-recapture method is used in conjunction with the Lincoln Index
  • For a single species in the area:
    • The first large sample is taken. As many individuals as possible are captured / collected, counted and marked in a way that won’t affect their survival
    • They are returned to their habitat and allowed to mix randomly with the rest of the population
    • When a sufficient amount of time has passed another large sample is captured
    • The numbers of marked and unmarked individuals within the sample are counted
    • The proportion of marked to unmarked individuals is used to calculate an estimate of the population size
    • The formula for the calculation is:

 straight N equals fraction numerator straight n subscript 1 cross times straight n subscript 2 over denominator straight m subscript 2 end fraction

    • Where:
      • N = population estimate
      • n1 = number of marked individuals released
      • n2 = number of individuals in the second sample (marked and unmarked)
      • m2 = number of marked individuals in the second sample

Worked example

Lincoln Index with mark-release-recapture

Scientists wanted to investigate the abundance of leafhoppers (Cicadellidae species) in a small grassy meadow. They used sweep nets to catch a large sample of leafhoppers from the meadow.

sweep net leafhopper in a sweep net

A sweep net used to collect insects

Kieren, CC BY 3.0, via Wikimedia Commons

A leafhopper after capture in a sweep net

Donald Hobern from Copenhagen, Denmark, CC BY 2.0, via Wikimedia Commons

Each insect was marked on its underside with non-toxic, waterproof paint and then released back into the meadow. The following day, another large sample was captured from the same area using sweep nets.

  • No. caught and marked in first sample  (n1) = 236
  • No. caught in second sample (n2) = 244
  • No. of marked individuals in the second sample (m2)  = 71

Using the equation

  straight N equals fraction numerator straight n subscript 1 cross times straight n subscript 2 over denominator straight m subscript 2 end fraction

N equals fraction numerator 236 cross times 244 over denominator 71 end fraction equals fraction numerator 57 space 584 over denominator 71 end fraction equals 811

N (estimated population size) = 811 leafhoppers

Examiner Tip

You will be provided with the formula for Lincoln’s index in the exam. You need to be able to carry out the calculation to estimate population size from mark-release-recapture data, as you could be asked to do this in the exam.

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Phil

Author: Phil

Expertise: Biology

Phil has a BSc in Biochemistry from the University of Birmingham, followed by an MBA from Manchester Business School. He has 15 years of teaching and tutoring experience, teaching Biology in schools before becoming director of a growing tuition agency. He has also examined Biology for one of the leading UK exam boards. Phil has a particular passion for empowering students to overcome their fear of numbers in a scientific context.