Classification, biodiversity & conservation: A Level Biology questions with answers
Free Classification, biodiversity & conservation questions for A Level Biology (9700), part of Genetics & evolution. Each question has the answer and the marking points an examiner would look for.
Classification, biodiversity & conservation practice questions
Question 1Structured · 10 marksA* challenge
The numbers of individuals of four species found in a quadrat survey were 25, 10, 5, 4.
(a)
Calculate Simpson’s index of diversity, D = 1 − Σ(n/N)2.
[3]
(b)
In a mark–release–recapture study, 40 beetles were marked; later 45 were caught, of which 5 were marked. Estimate the population size and state two assumptions.
[4]
(c)
Explain why conserving biodiversity is important.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
0.604
N = 44 B1 Σ(n/N)2 calculated M1 0.604 A1
3
(b)
360; no migration/births/deaths, marks not lost, marking does not affect survival, random mixing
c1 × c2 ÷ m M1 360 A1 two valid assumptions B2
4
(c)
ecological stability/food webs; source of drugs/genes; economic (tourism) and ethical reasons
any three valid reasons B3
3
Question 2Structured · 12 marks
A student samples a woodland using quadrats and records five plant species with numbers 10, 10, 6, 2, 2.
(a)
Define biodiversity.
[2]
(b)
Calculate Simpson’s index of diversity, D = 1 − Σ(n/N)2.
[3]
(c)
Explain what the value of D shows about the woodland.
[1]
(d)
Describe how the student should place quadrats to obtain representative data.
[3]
(e)
Explain how the mark–release–recapture method estimates population size, and state one assumption.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
variety of ecosystems, species and genetic variation
Any 2 of: the range of ecosystems / habitats; the number of species and their relative abundance; genetic variation within species B2
2
(b)
0.729
N = 30 M1 Σ(n/N)2 = 0.271 M1 D = 0.729 A1
3
(c)
high diversity
high diversity (closer to 1), more stable community B1
1
(d)
random coordinates from grid; random number generator; many quadrats
lay out a grid using tapes B1 use random numbers to generate coordinates B1 place many quadrats to obtain a representative sample B1
3
(e)
mark, release, recapture; N = n₁n₂/m; assumption: marks not lost, no migration, marked mix randomly
Any 3 of: capture and mark individuals, release them; recapture a second sample and count marked individuals; N = (number first sample × number second sample)/number marked in second; assumption: no births, deaths or migration / marks not lost or harmful B3
3
Question 3Structured · 10 marks
This question is about classification and conservation.
(a)
State the three domains and give one feature that distinguishes each.
[3]
(b)
Explain why viruses are not placed in any of the three domains.
[2]
(c)
Describe the roles of zoos and frozen zoos in conserving endangered mammals.
[3]
(d)
Explain why it can be difficult to reintroduce captive-bred animals into the wild.
captive animals may lack skills such as finding food or avoiding predators B1 their natural habitat may have been lost / they may lack immunity to local diseases B1
2
Question 4Structured · 9 marks
A student investigates whether the number of plant species in a quadrat is related to distance from a footpath.
distance from path / m
2
4
6
8
10
12
number of species
3
5
4
8
9
12
Table 4.1
(a)
State a null hypothesis for this investigation.
[1]
(b)
Calculate Spearman’s rank correlation coefficient for these data.
[3]
(c)
The critical value at p = 0.05 for n = 6 is 0.886. State the conclusion.
[2]
(d)
Describe how the student should sample species along a line transect from the path.
[2]
(e)
Explain why Spearman’s rank was chosen rather than Pearson’s correlation coefficient.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
no correlation between distance and number of species
there is no correlation between distance from the path and the number of species B1
1
(b)
0.943
ranks assigned to both variables M1 Σd2 = 2 M1 rs = 1 − 6Σd2/(n(n2 − 1)) = 0.943 A1
3
(c)
significant positive correlation; reject null
|rs| > 0.886 so the correlation is significant B1 reject the null hypothesis (species number increases with distance from the path) B1
2
(d)
quadrats at regular intervals along a tape from the path; record species in each
lay a tape at right angles to the path B1 place quadrats at regular intervals along it and record the number of species in each B1
2
(e)
data not normally distributed / relationship not linear
the data may not be normally distributed / the relationship is not expected to be linear (ignore: small sample) B1
1
Question 5Structured · 10 marks
A student samples ground beetles in two woodlands using pitfall traps. Table 5.1 shows the numbers of each species caught.
species
P
Q
R
S
T
woodland A
15
9
7
5
4
woodland B
30
4
3
2
1
Table 5.1
(a)
Simpson’s index of diversity is D = 1 − Σ(n/N)². Calculate D for woodland A.
[3]
(b)
D for woodland B is 0.419. Compare the diversity of the two woodlands and explain the difference using the data.
[2]
(c)
Explain why species richness alone would not show this difference.
[1]
(d)
Describe how pitfall traps should be used so that the samples from the two woodlands can be compared.
[3]
(e)
State why the beetles must be identified to species level for this calculation.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
0.752
N = 40 C1 Σ(n/N)2 = 0.2475 C1 D = 0.752 (accept 2 or 3 s.f.) A1
3
(b)
A more diverse
woodland A has the higher index, so it is more diverse B1 B has the same species but one species dominates / A has more even numbers of each species B1
2
(c)
both have 5 species; ignores evenness
Any 1 of: both woodlands have five species; richness ignores the relative abundance / evenness of the species B1
1
(d)
same number of traps; placed randomly; same time; checked regularly
Any 3 of: same number and size of traps in each woodland; placed at random (random coordinates); left for the same length of time; emptied regularly to prevent predation B3
3
(e)
index uses number of individuals of each species
D depends on the number of individuals of each species, so each must be correctly assigned to a species B1
1
Question 6Structured · 5 marks
Biodiversity can be measured in different ways.
(a)
Explain what is meant by species richness and species evenness.
[2]
(b)
Describe how the mark–release–recapture method estimates population size.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
number of species; relative abundance
species richness: number of different species B1 evenness: how evenly the individuals are distributed among species B1
2
(b)
mark, release, recapture, Lincoln index
capture and mark a sample, then release B1 recapture a second sample later B1 population = (n₁ × n₂) ÷ number marked in second sample B1
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in