Ecology & Plant Physiology

12 methods · 35 questions

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Keep the two abundance measures apart and scale each correctly. Frequency of occurrence is the fraction of quadrats in which a species appears (its numbers are irrelevant); density is the mean count per quadrat scaled to the whole area by (total area divided by quadrat area). Any estimate is valid only from a random, representative sample.

Trigger: quadrat, transect or kite-diagram data with options that mix fractions, per-square-metre values and totals.

Instances:
(i) frequency equals quadrats containing the species divided by quadrats sampled; a distractor equal to a mean or a total count is a density, not a frequency;
(ii) density in the field equals the mean per quadrat times (area divided by quadrat area); convert the quadrat side to metres first (a $0.5 m$ side gives $0.25 m^{2}$ );
(iii) reject any scaling from a biased sample (all sweeps at the edge) or any 'always' claim about quadrat size, since random quadrats vary and a small quadrat is only usually, not always, less likely to contain the species;
(iv) read a kite diagram by the width of its shading (wider means more individuals) and count overlapping kites to find the richest sampling point.

Linked questions (3)

NSAA 2017 Section 1, Part D (Biology questions) · Q57

Ten randomly placed quadrats were used to estimate the frequency of occurrence of three plant species in a field.

The table gives the number of individuals of species X, Y and Z in each quadrat.

quadrat number 12345678910 species X 411302171467 species Y 10509822403 species Z 512139426020

What was the correct lowest frequency of occurrence calculated from X or Y or Z?

A0
B0.2
C0.7
D1.0
E1.4
F5.2
G52.0

Worked solution

Answer:C

The frequency of occurrence of a species is the fraction of the quadrats in which it is present, whatever its numbers.

Species X is absent only from quadrat 4, so its frequency is

\frac{9}{10} = 0.9

. Species Y is absent from quadrats 2, 4 and 9:

\frac{7}{10} = 0.7

. Species Z is absent from quadrats 8 and 10:

\frac{8}{10} = 0.8

.

The lowest frequency of occurrence is 0.7. (5.2 is the mean number of individuals of Y per quadrat and 52 their total: these are densities, not frequencies.)

The answer is C.

NSAA 2018 Section 1, Part D (Biology questions) · Q58

A student investigated the tadpole population in a large pond.

A net with a rectangular opening measuring

0.1 m \times 0.2 m

was swept through the water for a fixed distance of

1 m

. This was repeated 10 times.

All the sweeps were made at the edge of the pond as the student had no waders or boat.

The number of tadpoles in each sweep was recorded in the table.

sweep number number of tadpoles 12021233240546874889121020

The student made the following statements.

1. Each sweep sampled

0.02 m^{3}

of water.
2. The frequency of occurrence of the tadpoles was 90%.
3. An accurate estimate of the population size of tadpoles in the pond could be calculated using this data if the total volume of water was known.

Which of the statements about the investigation is/are correct?

Anone of them
B1 only
C2 only
D3 only
E1 and 2 only
F1 and 3 only
G2 and 3 only
H1, 2 and 3

Worked solution

Answer:E

Each sweep samples a volume equal to the area of the net opening multiplied by the distance swept:

0.1 \times 0.2 \times 1 = 0.02 m^{3}

, so statement 1 is correct.

Tadpoles were caught in 9 of the 10 sweeps (none in sweep 4), so their frequency of occurrence is

\frac{9}{10} = 90

per cent, and statement 2 is correct.

Statement 3 is wrong: every sweep was taken at the edge of the pond, so the sample is not random and does not represent the whole pond. Scaling it up to the total volume of water would not give an accurate estimate of the population.

The answer is E.

NSAA 2019 Section 1, Part D (Biology questions) · Q59

An ecologist used a

50 cm \times 50 cm

square quadrat to estimate the number of meadow buttercups present in a field with an area of

50 m^{2}

. The quadrat was distributed randomly on ten occasions in the field and the number of buttercups counted in each quadrat.

quadrat 12345678910 number of buttercups 310042119611153

How many buttercups were there estimated to be in the

50 m^{2}

field?

A368
B460
C920
D1840
E4600
F18 400
G45 000

Worked solution

Answer:D

The total counted in the ten quadrats is

3 + 10 + 0 + 4 + 21 + 19 + 6 + 11 + 15 + 3 = 92

, a mean of

9.2

buttercups per quadrat.

Each quadrat is

0.5 m \times 0.5 m = 0.25 m^{2}

, so the field of

50 m^{2}

holds

50 \div 0.25 = 200

quadrats.

Estimated number

= 9.2 \times 200 = 1840

.

The answer is D.