A-LevelBiologyBiodiversity and conservation9700/42 · May/June 2025 · Q99 Marks

The biodiversity of an area can be assessed using a variety of sampling methods.

(a)Outline how a frame quadrat could be used to assess the biodiversity of plants in a field.[4]
(b)(i)A student investigated whether the height of the soft rush plant, Juncus effusus, decreases with an increase in altitude on a hillside in the United Kingdom. • 12 sites were chosen at increasing altitudes. • The mean height of 10 plants was calculated at each altitude. Spearman's rank correlation was used to assess the relationship between the height of the plants and altitude. The equation for Spearman's rank correlation (rS) is: rS = 1 - (6 x ΣD²) / (n³ - n) Key to symbols: D = difference in rank between each pair of measurements n = number of pairs of items in the sample ΣD² was calculated to be 550. Calculate the Spearman's rank correlation for these data. Give your answer to three decimal places. answer[3]
(b)(ii)The null hypothesis for this investigation is: there is no correlation between the altitude and the height of the soft rush plants. Table 9.1 shows the critical values for Spearman's rank correlation. [Table 9.1] Use your value of Spearman's rank correlation and Table 9.1 to state and explain if the null hypothesis is correct.[2]

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Mark scheme

Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.

[4] (a) any four from: 1 ref. to quadrat of, known / same / standard, area / size ; 2 ref. to setting out a grid ; 3 random number generator (to produce coordinates) for quadrat placement; 4 use a key; 5 count number of species / numbers of individuals of each species / abundance / percentage cover / ACFOR / Braun- Blanquet; 6 ref. to large sample size / repeats; 7 (use results to calculate) Simpson's Index of Biodiversity / species density;
[3] (b)(i) -0.923 ;;; allow 2 marks if no minus allow 2 marks if not three decimal places allow 2 marks if incorrect value of n used if n is 10 answer = – 2.333 if n is 120 answer = 0.998
[2] (b)(ii) rs > critical value or 0.923 > 0.504 ; null hypothesis rejected as there is a, significant / strong / negative, correlation or null hypothesis rejected as correlation not due to chance / probability that correlation due to chance is less than 5%; ecf from 9(b)(i)

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