Fieldwork Investigations

Question

Step-by-step solution

1. Step 1

   Why sampling is necessary: Counting every individual plant in the meadow (400 m²) would be impractical and extremely time-consuming — the meadow may contain thousands or tens of thousands of individual plants. Sampling provides a representative estimate of the whole population from a manageable subset. Provided samples are taken randomly and in sufficient number, the results can be used to make valid inferences about the entire meadow. Random sampling is essential to avoid bias — if the student chose where to place quadrats based on personal preference, she might unconsciously pick areas with more interesting species, giving an unrepresentative result.

2. Step 2

   Random sampling method: (1) Lay two measuring tapes along adjacent edges of the meadow (one along the 20 m north edge, one along the 20 m west edge) to create a coordinate grid. (2) Use a random number generator (or random number table) to generate pairs of coordinates — e.g. (7, 13), (3, 18), (15, 5). (3) Place the corner of the quadrat at each generated coordinate. (4) Record all species present and their abundance/coverage within the quadrat. Repeat for a minimum of 10 quadrats (more is better — the more quadrats, the more representative the estimate). This ensures every part of the meadow has an equal chance of being sampled.

3. Step 3

   Percentage cover calculation for buttercup:

   • Area covered by buttercup = 8 plants × 0.02 m² = 0.16 m²
   • Quadrat area = 0.5 × 0.5 = 0.25 m²
   • Percentage cover = (0.16 ÷ 0.25) × 100 = 64%

   Note: It is possible for total percentage covers in a quadrat to exceed 100% if plants are growing on top of each other (e.g. clover growing beneath taller grass) — percentage cover records the proportion of ground surface covered when viewed from directly above.

4. Step 4

   Comparison of abundance measures: Percentage cover (used for grasses and ground-cover plants that form mats) is often more meaningful than counting individuals (used for discrete, well-defined plants like buttercups or daisies). For some species (e.g. grass, moss, algae), counting individuals is impossible — percentage cover is the only practical measure. Examiners may ask you to explain which measure is more appropriate for a given species.

Question

Step-by-step solution

1. Step 1

   Line transect vs. belt transect:

   • Line transect: a tape measure is stretched across the study area (here, from shoreline to 40 m inland); the species touching or directly beneath the tape at specified intervals (e.g. every 5 m) are recorded. Quick but records only presence/absence at a point — no abundance data.
   • Belt transect: a strip of defined width (e.g. 0.5 m or 1 m either side of the tape) is sampled at each interval using quadrats placed within the strip. Records both presence AND abundance (% cover, frequency, or density) across the defined strip. The belt transect gives much more quantitative data on species distribution across the gradient than a line transect.

2. Step 2

   What a kite diagram shows: A kite diagram is a graphical method for displaying the distribution and relative abundance of multiple species along a transect. The width of the 'kite' (a diamond shape) at each transect point represents the percentage cover (or frequency) of that species at that point. The wider the kite, the more abundant the species at that location. Kite diagrams allow the distribution of multiple species to be compared visually on the same graph, clearly showing zonation — the bands of different species at different distances along the transect.

   Pattern for marram grass: The kite for marram grass is narrow at 10 m, widens to its maximum at 20–25 m (where % cover peaks at 40%), then narrows again to 5% at 40 m. The kite is absent (zero width) at 0–5 m. This shows marram grass is a mid-dune specialist peaking in abundance at 20–25 m from the shoreline.

3. Step 3

   Why marram grass dominates at 10–30 m but not the shoreline or far inland:

   Absent at shoreline (0–5 m): The shoreline is the most physically extreme zone — regularly inundated by seawater, high salt concentrations, mobile sand, and wave disturbance. Marram grass cannot tolerate regular saltwater inundation. Only highly salt-tolerant pioneer species (sea rocket, sea couch) can survive here.

   Peaks at 10–25 m (yellow dunes / foredune): This is the optimal zone for marram grass. It is specially adapted to deep, mobile sand: (1) extensive rhizome (underground stem) network that spreads rapidly through loose sand; (2) leaves that roll inward in dry conditions to reduce transpiration; (3) can be buried by sand accumulation and continue growing upward — the plant actually traps and accumulates sand, building the dune. Marram thrives in the harsh but not waterlogged conditions of the young mobile dune.

   Declines at 30–40 m (older dunes): As dunes age and stabilise (fixed dunes), soil develops, moisture increases, and conditions become less extreme. Marram grass is outcompeted by less specialised but faster-growing species (grasses, herbs, shrubs) in these more hospitable conditions. Marram grass is a pioneer species — it creates conditions that eventually lead to its own replacement through succession.

Question

Step-by-step solution

1. Step 1

   BMWP Score for Site A (upstream): The BMWP (Biological Monitoring Working Party) score is calculated by summing the scores of each FAMILY of organism found, regardless of how many individuals of that family are present.

   • Stonefly nymphs: score 10
   • Mayfly nymphs: score 10
   • Freshwater shrimp: score 6
   • Water louse: score 3
   • Total BMWP score = 10 + 10 + 6 + 3 = 29

   Note: BMWP scores each family only once — if 12 stonefly nymphs and 8 mayfly nymphs are found, they contribute 10 + 10 = 20 (not 12×10 + 8×10). Individual counts do not affect the score.

2. Step 2

   BMWP Score for Site B (downstream):

   • Bloodworms (midge larvae, Chironomidae): score 2
   • Rat-tailed maggots (drone fly larvae, Syrphidae): score 1
   • Sludge worms (Tubificidae): score 1
   • Total BMWP score = 2 + 1 + 1 = 4

   Water quality assessment:

   • Site A: BMWP = 29 — presence of pollution-sensitive stonefly and mayfly nymphs (score 10 species) indicates good to excellent water quality. High dissolved oxygen, low BOD, minimal pollutant load. This is typical clean, fast-flowing river water.
   • Site B: BMWP = 4 — presence only of highly pollution-tolerant bloodworms, rat-tailed maggots and sludge worms (low score species) indicates severely polluted water. Low dissolved oxygen, high BOD, high nutrient load — consistent with agricultural runoff from the farm between sites A and B.

3. Step 3

   Why indicator species are more valid than chemical testing over time: (1) Integrate pollution over time: Chemical water tests give a snapshot of conditions at the exact moment of sampling — a river could appear clean on a chemical test if sampled hours after a pollution event has passed downstream. Macroinvertebrates live in the river for weeks to months; their presence or absence reflects average conditions over the entire period since they colonised. A single pollution event that killed all stonefly nymphs would show up in a biological survey weeks later, even if water chemistry has returned to normal. (2) Respond to cumulative effects: Chemical tests measure individual pollutants in isolation. Biological surveys reflect the integrated effect of all pollutants plus temperature, flow rate, substrate, and other environmental variables — the overall ecological condition of the river. (3) Cost-effective: Biological surveys require trained ecologists but not expensive laboratory chemical analysis equipment. They can be carried out in the field by trained volunteers. (4) Limitations of indicator species: Cannot identify the specific pollutant causing water quality decline; requires skilled taxonomists for identification; some species respond to physical habitat quality as well as water chemistry — a reach may have poor BMWP score due to channel modification rather than chemical pollution.

Question

Step-by-step solution

1. Step 1

   Site 1 calculations (ancient woodland):

   First, find N (total individuals): N = 18 + 12 + 8 + 5 + 7 = 50

   Now calculate n(n−1) for each species:

   • Oak: 18 × 17 = 306
   • Ash: 12 × 11 = 132
   • Hazel: 8 × 7 = 56
   • Holly: 5 × 4 = 20
   • Field maple: 7 × 6 = 42

   Σn(n−1) = 306 + 132 + 56 + 20 + 42 = 556

   N(N−1) = 50 × 49 = 2450

   D = 1 − (556 ÷ 2450) = 1 − 0.227 = D = 0.77

2. Step 2

   Site 2 calculations (plantation):

   N = 48 + 2 = 50

   n(n−1) for each species:

   • Sitka spruce: 48 × 47 = 2256
   • Douglas fir: 2 × 1 = 2

   Σn(n−1) = 2256 + 2 = 2258

   N(N−1) = 50 × 49 = 2450

   D = 1 − (2258 ÷ 2450) = 1 − 0.922 = D = 0.078

3. Step 3

   Interpretation of results:

   • Site 1 (D = 0.77): A value close to 1 indicates high species diversity. The ancient woodland has 5 tree species in relatively even proportions — no single species completely dominates. Simpson's D of 0.77 means there is a 77% probability that two individuals chosen at random from this site will be different species — a useful way to explain the index.
   • Site 2 (D = 0.078): A value close to 0 indicates very low species diversity. The plantation is almost entirely Sitka spruce — a near-monoculture. Despite having the same total number of trees (50), the extreme dominance of one species makes it very low in diversity.
   • This demonstrates why species evenness (not just richness) is critical to biodiversity — both sites have similar numbers of trees but vastly different Simpson's D values.

4. Step 4

   Conservation value implications:

   • The ancient woodland (Site 1) has far greater conservation value: multiple tree species support a diverse range of insects, birds, fungi, and ground flora. Oak alone supports over 280 species of insect in the UK; ash supports specialist communities. Ancient woodland is irreplaceable — it has taken centuries to develop its ecological complexity and cannot be created in a human lifetime.
   • The plantation (Site 2) has very low conservation value as a habitat: monoculture Sitka spruce shades out all ground flora, provides food and habitat for very few native species, and creates uniform structure with few ecological niches. It does provide timber and some carbon storage, but it cannot substitute for the biodiversity services of ancient woodland.
   • Management implication: conservationists should prioritise protecting and expanding Site 1. Site 2 could potentially be improved by diversifying tree species — adding native broadleaves at margins and in clearings — to increase D toward 0.5–0.7.

Question

Step-by-step solution

1. Step 1

   Measurements and equipment:

   (1) Light intensity: use a lux meter (light sensor/data logger) at ground level within each quadrat. Record in lux (lx). Measurements should all be taken at the same time of day (e.g. 12:00–13:00) to control for the effect of sun angle. Repeat measurements on at least 3 different days to account for cloud cover variation.

   (2) Plant species diversity on woodland floor: place 0.5 m × 0.5 m quadrats at each sampling point; record all plant species present and estimate their percentage cover using DAFOR scale (Dominant, Abundant, Frequent, Occasional, Rare) or direct percentage estimation. Calculate Simpson's Diversity Index (D) for each quadrat using species counts.

   (3) Canopy cover (as a proxy for shading): use a densiometer (convex mirror with a grid) or photograph the canopy through a hemispherical (fisheye) lens to estimate the percentage of sky obscured by leaves — the canopy closure percentage. This confirms the gradient from open to shaded sites.

   (4) Optional: soil pH measured using a calibrated pH meter (or pH indicator paper) in a soil-water suspension. Take soil samples from the top 5 cm at each quadrat location.

2. Step 2

   Sampling strategy:

   Use a systematic sampling approach along a belt transect from the woodland edge (open, high light) to the woodland interior (deep shade). Place the transect along a straight line from the open edge inward, measuring 20–40 m depending on woodland size.

   At 5 m intervals along the transect, place a 0.5 m × 0.5 m quadrat and record: (a) light intensity (lux meter), (b) canopy closure %, (c) all plant species present with % cover, (d) soil pH. This systematic approach ensures the full gradient from open to shaded is sampled.

   Repeat with 3 parallel transects spaced 5 m apart across the woodland to account for spatial variation not related to shading. Use at least 5 quadrat positions along each transect (minimum n = 15 quadrats total).

3. Step 3

   Data analysis and presentation:

   (1) Calculate D (Simpson's Diversity Index) for each quadrat.

   (2) Plot a scatter graph of light intensity (x-axis, lux) against Simpson's D (y-axis) — each quadrat is one data point. Add a line of best fit. If the hypothesis is correct, a positive correlation should be visible: higher light → higher D.

   (3) Plot a kite diagram along the transect showing: (a) canopy closure% and (b) percentage cover of each plant species — this visualises how species composition changes along the light gradient.

   (4) Plot a bar chart comparing mean D values at low-light quadrats (<500 lux), medium-light (500–2000 lux), and high-light (>2000 lux) sites, with error bars showing standard deviation — this quantifies the difference between light zones.

   (5) Calculate mean light intensity and mean D for each transect position to smooth random variation.

4. Step 4

   Major source of error and minimisation:

   Error — Confounding variables (soil type and moisture): Plant species diversity may differ between woodland edge and interior not only because of light differences, but also because soil moisture, soil texture, and soil nutrient content change along the transect. For example, the woodland interior may have deeper, more acidic soils with more leaf litter — which independently affects which plant species grow there, regardless of light. This means any observed change in D could be caused by soil differences rather than light differences, threatening the internal validity of the investigation.

   Minimisation: (1) Measure soil pH and soil moisture (using a soil moisture meter) at each quadrat, then statistically control for these variables in analysis (e.g. using multiple regression or by only comparing quadrats with similar soil conditions). (2) Choose a woodland site with relatively uniform underlying geology and soil type (e.g. all on the same geological formation) to reduce baseline variation. (3) Take soil samples from the same depth (5 cm) at all locations using a standardised soil corer.