Biodiversity and Conservation is a core Class 12 Biology chapter because it links ecological principles (variety of life, population genetics, species richness) with real-world conservation decisions. Board exams and competitive tests often ask you to apply quantitative biodiversity indices (Shannon/Simpson), use species–area relationships, and understand genetic consequences of small populations and habitat fragmentation. Therefore, mastering these concepts helps you score well both in theory and problem-based MCQs.
15
Minutes
10
Questions
1 / -0
Marking
Q1. In a 1 m^2 grassland quadrat the counts of four plant species were 10, 30, 30 and 30 individuals. Using the Shannon diversity index where is the proportion of species , what is the approximate value of (round to two decimal places)?
0.97
1.20
1.31
1.45
Q2. A conservation planner must choose between one large reserve of area and four equal smaller reserves whose total area equals . Using the species–area relationship and assuming species in different small reserves are completely distinct (no overlap), the ratio of total species in the four small reserves to species in the single large reserve equals . For , which of the following values is closest to that ratio?
1.14
2.64
0.38
4.00
Q3. A population has effective population size and initial heterozygosity . Using the formula , what is the expected heterozygosity after generations (round to two decimal places)?
0.80
0.72
0.54
0.65
Q4. Assertion (A): For critically small wild populations (census size < 20), transferring individuals into a well-managed captive breeding program is often more effective at maintaining genetic heterozygosity over several decades than leaving them in situ when habitat continues to degrade.
Reason (R): In captivity, controlled breeding and husbandry can increase the effective population size and reduce environmental and demographic stochasticity, thereby slowing the per‑generation loss of heterozygosity described by .
Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.
Q5. Assertion (A): Establishing a wildlife corridor between two isolated habitat patches generally reduces extinction probability for species with limited dispersal by facilitating recolonization and gene flow.
Reason (R): Corridors reduce extinction risk primarily by increasing the edge‑to‑area ratio of habitat patches, which enhances habitat heterogeneity at patch boundaries.
Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.
Q6. Using the species–area relationship with , a forest fragment loses half its area while originally supporting 100 species. According to the model, approximately how many species would remain after the area is halved?
about 50 species
about 92 species
about 84 species
about 75 species
Q7. Two communities each have total individuals . Community I has species counts ; Community II has species counts . Using Simpson's index of diversity
, which statement is correct?
Community I has higher Simpson's diversity index than Community II
Community II has higher Simpson's diversity index than Community I
Both communities have equal Simpson's diversity index
Cannot determine without knowing species identities
Q8. Two isolated populations have effective population sizes and . Using the approximate rate of loss of heterozygosity per generation , by what factor does the smaller population lose heterozygosity faster than the larger one?
2 times faster
5 times faster
4 times faster
10 times faster
Q9. Assertion (A): According to the species–area relationship with a typical exponent , fragmenting a continuous habitat of area into several smaller patches (keeping total area constant) will predict a larger combined species richness (by summing for each fragment) than the richness predicted for the unfragmented area.
Reason (R): For the function is concave, and for positive parts with one has , so .
A is true but R is false
Both A and R are true and R correctly explains A
Both A and R are true but R does not explain A
Both A and R are false
Q10. A captive-breeding programme maintains 24 adults. Compare two strategies: X with , ; Y with , . Using , which statement is correct about effective population size and its implication for retaining genetic diversity?
Strategy Y gives , which is times the under X (); Y will retain genetic variation longer
Strategy X gives , which is times the under Y; X will retain genetic variation longer
Both strategies give the same because total adults are equal
depends only on total adults, so neither strategy changes genetic-retention potential