The photograph shows a species of fungus growing on a dead tree in a woodland habitat. Wood contains a high proportion of lignin and cellulose. Fungi can break down molecules in wood by releasing extracellular enzymes. A study compared the biodiversity of two woodland habitats. The index of diversity for habitat one was 4.2. The table shows the data obtained from habitat two. (i) An index of diversity (D) is calculated using the formula: $$\[ \mathrm{D}=\frac{\mathrm{N}(\mathrm{~N}-1)}{\sum \mathrm{n}(\mathrm{n}-1)} \]$$ Calculate the index of diversity for habitat two. Use the table and the formula to help you. (3) Answer ......................................................................................... (ii) State which habitat has the higher biodiversity. Give a reason for your answer. (1) ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ (iii) Species D is an endangered plant species. Explain the processes that could be used by a seed bank to conserve this plant species. (3) ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................

Biology
IGCSE&ALevel
EDEXCEL
Exam No:wbi12-01-que-20240518 Year:2024 Question No:8(c)

Answer:

An answer that includes the following points:
- \(\mathrm{N}(\mathrm{N}-1)\) correctly calculated (1)
- \(\quad \sum n(n-1)\) correctly calculated (1)
- calculation of D for habitat two (1)
- habitat 1 because it has \(\{a\) higher index of diversity / more species\} (1)
An explanation that makes reference to three of the following points:
- collection of seeds from multiple plants of species D (to ensure different alleles) (1)
- \{washing / disinfecting / sterilising\} seeds to remove (decomposing) microbes (1)
- x-ray seeds to check \{viability / presence of embryo\} (1)
- \{freeze the seeds / dry the seeds / store in very low temperatures\} to \{prevent germination / maintain viability / prevent growth of microbes / reduce enzyme activity / keep them dormant\} (1)
- \{germination / growth / pollination\} of genetically different plants to collect new seeds (1)

Knowledge points:

CH4 – Plant Structure and Function, Biodiversity and Conservation

Solution:

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