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Scientists investigated stomatal density on leaves of one species of tree - AQA - A-Level Biology - Question 9 - 2020 - Paper 1

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Scientists investigated stomatal density on leaves of one species of tree. Figure 9 shows three examples of the square fields of view the scientists used to calcul... show full transcript

Worked Solution & Example Answer:Scientists investigated stomatal density on leaves of one species of tree - AQA - A-Level Biology - Question 9 - 2020 - Paper 1

Step 1

Calculate the mean stomatal density in the three fields of view in Figure 9.

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Answer

To calculate the mean stomatal density, we first need to count the number of stomata in each field of view shown in Figure 9.

Assuming the stomata counts from the fields are: 185, 176, and 170 stomata respectively.

Next, we calculate the area of each field of view. Each square in Figure 9 has edges of 250 µm (0.25 mm), therefore the area is:

extArea=(0.25extmm)2=0.0625extmm2 ext{Area} = (0.25 ext{ mm})^2 = 0.0625 ext{ mm}^2

Now, we can convert the counts of stomata into a density measure (stomata per mm²):

  1. For the first field: D1=1850.0625=2960D_1 = \frac{185}{0.0625} = 2960 stomata/mm²
  2. For the second field: D2=1760.0625=2816D_2 = \frac{176}{0.0625} = 2816 stomata/mm²
  3. For the third field: D3=1700.0625=2720D_3 = \frac{170}{0.0625} = 2720 stomata/mm²

Finally, we find the mean stomatal density:
$$ ext{Mean stomata density} = \frac{D_1 + D_2 + D_3}{3} = \frac{2960 + 2816 + 2720}{3} = 2832$ stomata/mm².

Step 2

Give a null hypothesis for this investigation and name a statistical test that would be appropriate to test your null hypothesis.

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Answer

Null hypothesis: There is no relationship between the concentration of carbon dioxide and the stomatal density.

Statistical test: Spearman's rank correlation coefficient.

Step 3

From 1910 to 2000, calculate the mean rate of change in stomatal density from 1910 to 2000.

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Answer

Using the data from Figure 10, suppose the stomatal density increased from 150 stomata/mm² in 1910 to 180 stomata/mm² in 2000, which is an increase of 30 stomata/mm² over 90 years.

The mean rate of change per 10 years is calculated as follows:
extMeanrate=30extstomata/mm290extyears×10=3.33extstomata/mm2/10extyears.ext{Mean rate} = \frac{30 ext{ stomata/mm}^2}{90 ext{ years}} \times 10 = 3.33 ext{ stomata/mm}^2/10 ext{ years}.

Step 4

Evaluate his suggestion.

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Answer

The journalist's suggestion that increased atmospheric carbon dioxide could result in less transpiration requires careful consideration.

  1. Increasing carbon dioxide concentration generally correlates with decreased stomatal conductance, which can lead to reduced water loss through transpiration.
  2. However, this relationship may depend on various factors including species-specific responses to CO2, availability of water, and environmental conditions.
  3. Additionally, while fewer stomata generally suggest less water vapor loss, other factors such as leaf size and environmental stresses also play critical roles in transpiration rates.
  4. Therefore, while there may be a correlation, the overall impact requires a more comprehensive examination of ecological dynamics.

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