Solved Examples — Class 12 Biology Notes

Section 12 — Solved Examples

This is the dedicated problem set for Chapter 11 (Organisms and Populations). The 30 worked examples below are arranged in three tiers — concept checks (population attributes, definitions, interaction signs), numerical and application (birth and death rates, computing $r$ , doubling time, reading age pyramids, identifying interactions from scenarios), and analytical and multi-concept (reasoning problems that link growth models, competition, predation and mutualism across the whole chapter).

Unlike a plain glossary, most of these ask you to calculate, identify and justify, because that is exactly how a population question earns marks in a Board or NEET paper. Naming an interaction is rarely enough; you are usually asked why both species benefit, how a density changes, or what the sign combination tells you.

How to use this section

Concept Checks (Q1–Q10): Quick recall of attributes, growth terms and interaction types. If you stumble on more than one or two, revisit Sections 2–11.
Numerical and Application (Q11–Q20): The workhorse questions — computing rates, using the growth equations, reading age pyramids and identifying interactions from real examples.
Analytical and Multi-Concept (Q21–Q30): Multi-step reasoning that ties several topics together, the kind that separates a good answer from a full-mark answer.

Total target time: around an hour for a complete revision sweep.

Note: This section is for practice and revision only — there is no quiz at the end. Treat each answer as a model of how much detail an examiner expects.

Memory Capsule — Facts Worth Locking In

Before working through the problems, fix these ten high-yield facts firmly in mind:

#	Fact	Where it is tested
1	Attributes a population has but an individual does not — birth and death rates, sex ratio, age distribution (age pyramid)	Attribute questions
2	Birth rate = new individuals added per member; death rate = deaths per member. Example: 8 new plants on 20 gives 0.4 offspring per lotus per year	Numerical
3	Density change: $N_{t+1} = N_t + [(B + I) - (D + E)]$ Natality and immigration add; mortality and emigration subtract	Growth logic
4	Exponential growth: $\frac{dN}{dt} = rN$ gives a J-shaped curve; integral form $N_t = N_0 e^{rt}$	Growth models
5	$r$ = intrinsic rate of natural increase $= b - d$ . For the Norway rat $r = 0.015$ ; for the flour beetle $r = 0.12$	Numerical
6	Logistic growth: $\frac{dN}{dt} = rN\left(\frac{K - N}{K}\right)$ gives a sigmoid curve; $K$ = carrying capacity (Verhulst-Pearl)	Growth models
7	Age pyramid shape tells growth status — broad base = growing, straight sides = stable, narrow base = declining	Diagram reading
8	Interaction signs — Mutualism (+ +), Competition (- -), Predation (+ -), Parasitism (+ -), Commensalism (+ 0), Amensalism (- 0)	Identify-the-interaction
9	Classic mutualisms — lichen (fungus + alga), mycorrhiza (fungus + root), fig-wasp, Ophrys orchid-bee	Example matching
10	Competition ideas — Gause's competitive exclusion, competitive release, resource partitioning (MacArthur's warblers)	Reasoning

Pro tip: For a numerical, always write the formula first, then substitute; for an interaction question, assign the plus, minus or zero sign to each partner before you name it. Both habits protect easy marks.

Concept Checks (Q1–Q10)

Q1. List the attributes that a population has but an individual organism does not.

Answer: A population has birth rates and death rates, sex ratio, and age distribution (age structure), along with a measurable population density. An individual only has a single birth and a single death; the idea of a rate, a ratio or a distribution has meaning only for a group.

Q2. Define population density and name the symbol used for it.

Answer: Population density is the size of a population in a given habitat at a given time, denoted by $N$ . It is usually reported as the total number of individuals, but where counting is impractical it may be expressed as per cent cover or biomass.

Q3. Why is per cent cover or biomass sometimes a better measure of density than total number?

Answer: When a single individual plays an outsized role, a head count misleads. A single huge banyan with a vast canopy would be recorded as low density beside many small carrot grass plants, yet its ecological role is enormous. In such cases per cent cover or biomass captures the true importance better than numbers.

Q4. Name the four processes that change population density and state which two increase it.

Answer: The four are natality, mortality, immigration and emigration. Natality (births) and immigration increase density; mortality and emigration decrease it.

Q5. What is the intrinsic rate of natural increase, and how is it obtained from birth and death rates?

Answer: The intrinsic rate of natural increase ( $r$ ) is the inherent capacity of a population to grow. It equals the per-capita birth rate minus the per-capita death rate, $r = b - d$ . It is the key parameter for judging how any biotic or abiotic factor affects growth.

Q6. Name the two population growth models and the shape of the curve each produces.

Answer: Exponential (geometric) growth produces a J-shaped curve under unlimited resources; logistic growth produces an S-shaped (sigmoid) curve when resources become limiting. The logistic model is considered the more realistic one.

Q7. What is carrying capacity, and by what letter is it represented?

Answer: Carrying capacity ( $K$ ) is the maximum population size that a given habitat can support with its available resources. Beyond $K$ no further growth is possible, so in logistic growth the population levels off as an asymptote at $K$ .

Q8. Give the sign pairs for mutualism, commensalism and amensalism.

Answer: Mutualism = (+ +) — both species benefit. Commensalism = (+ 0) — one benefits, the other is unaffected. Amensalism = (- 0) — one is harmed, the other is unaffected.

Q9. Name one example each of mutualism and commensalism.

Answer: Mutualism — a lichen (a fungus living with a photosynthesising alga or cyanobacterium), or a mycorrhiza. Commensalism — an orchid growing as an epiphyte on a mango branch, or the cattle egret foraging beside grazing cattle.

Q10. State Gause's Competitive Exclusion Principle in one line.

Answer: Two closely related species competing for the same limiting resources cannot co-exist indefinitely, and the competitively inferior one is eventually eliminated. This holds when resources are limiting but not always otherwise.

Numerical and Application (Q11–Q20)

Q11. A pond had 20 lotus plants last year; through reproduction 8 new plants were added this year. Calculate the birth rate.

Answer: Birth rate is the number of new individuals added per member of the starting population.

$\text{Birth rate} = \frac{8}{20} = 0.4$

So the birth rate is 0.4 offspring per lotus per year. The current population is $20 + 8 = 28$ plants.

Q12. In a laboratory culture of 40 fruitflies, 4 individuals died in one week. Calculate the death rate.

Answer: Death rate is the number of deaths per member over the period.

$\text{Death rate} = \frac{4}{40} = 0.1$

The death rate is 0.1 individuals per fruitfly per week.

Q13. A population of 100 has, over a year, 30 births, 10 deaths, 15 immigrants and 5 emigrants. Find the population at the end of the year.

Answer: Use the density equation $N_{t+1} = N_t + [(B + I) - (D + E)]$

Substituting: $N_{t+1} = 100 + [(30 + 15) - (10 + 5)] = 100 + [45 - 15] = 100 + 30 = 130$ .

The population grows to 130 individuals, because additions (45) exceed losses (15).

Q14. For a population the per-capita birth rate is 0.45 and the per-capita death rate is 0.15. Calculate $r$ and state what the value means.

Answer: $r = b - d = 0.45 - 0.15 = 0.30$ .

A positive $r$ of 0.30 means the population is growing; the instantaneous growth rate is 30% of the current size (dN/dt = 0.30N) under the exponential model.

Q15. A population growing exponentially doubles in size in 3 years. Calculate its intrinsic rate of increase $r$ .

Answer: Start from $N_t = N_0 e^{rt}$

At doubling, $N_t = 2N_0$ , so $2 = e^{rt}$ . Taking natural logs, $\ln 2 = rt$ , giving $r = \dfrac{\ln 2}{t} = \dfrac{0.693}{3} = 0.231$ .

The intrinsic rate of increase is about 0.231 per year.

Q16. A bacterial population starts at 1000 cells and grows exponentially with $r = 0.5$ per hour. Using $N_t = N_0 e^{rt}$ , estimate the size after 4 hours. (Take $e^{2} = 7.39$ .)

Answer: $N_t = N_0 e^{rt} = 1000 \times e^{0.5 \times 4} = 1000 \times e^{2} = 1000 \times 7.39 = 7390$ .

The population reaches roughly 7390 cells after 4 hours, illustrating how fast unimpeded growth builds up.

Q17. An age pyramid has a very broad base that narrows steadily towards the top. What does its shape indicate?

Answer: A broad base means a large proportion of young, pre-reproductive individuals. Such a pyramid is triangular and indicates a growing (expanding) population, because the many young will soon enter the reproductive ages and add still more births.

Q18. Two age pyramids are compared: one has nearly straight, vertical sides; the other is narrow at the base and wider in the middle. Classify each.

Answer: The pyramid with straight, bell-like sides shows roughly equal age groups — a stable (stationary) population. The pyramid that is narrow at the base has fewer young than adults, so births are falling — a declining population that will shrink over time.

Q19. Identify the interaction in each case: (a) a lichen, (b) an orchid on a mango branch, (c) a tiger eating a deer.

Answer: (a) A lichen is mutualism (+ +) — the fungus and the alga both benefit. (b) The orchid on the mango branch is commensalism (+ 0) — the orchid gains support, the tree is unaffected. (c) The tiger and deer show predation (+ -) — the predator benefits, the prey is killed.

Q20. Identify the interaction: (a) cattle egret foraging beside grazing cattle, (b) Cuscuta growing on a hedge plant, (c) sea anemone and clown fish.

Answer: (a) Cattle egret and cattle — commensalism (+ 0): the egret catches insects stirred up by the cattle, which gain nothing but lose nothing. (b) Cuscuta on a host — parasitism (+ -): this leafless, chlorophyll-less plant draws nutrition from the host, harming it. (c) Sea anemone and clown fish — commensalism (+ 0): the fish shelters among the stinging tentacles while the anemone is unaffected.

Analytical and Multi-Concept (Q21–Q30)

Q21. Explain why the exponential model gives a J-shaped curve while the logistic model gives an S-shaped curve, referring to their equations.

Answer: In the exponential model $\frac{dN}{dt} = rN$ the growth rate rises in direct proportion to $N$ with no ceiling, so the curve keeps steepening — a J shape. In the logistic model $\frac{dN}{dt} = rN\left(\frac{K - N}{K}\right)$ the extra term $(K - N)/K$ shrinks towards zero as $N$ approaches $K$ , so growth slows and levels off at the carrying capacity — a sigmoid (S) shape. The logistic curve is more realistic because real resources are finite.

Q22. A tiger census cannot count every animal directly. Explain how ecologists estimate such populations and why absolute counts are often replaced by indirect measures.

Answer: Where individuals are secretive, huge in number or spread over large areas, direct counting is impossible or too time-consuming. Ecologists then use indirect estimates — the tiger census relies on pug marks and faecal pellets. For other cases relative density works well, for example the number of fish caught per trap as an index of the lake's fish population. These methods report status without seeing or counting every individual.

Q23. The prickly pear cactus spread uncontrollably after being introduced into Australia, but was later controlled. Explain the ecological principle behind both the spread and the control.

Answer: The cactus spread because the new land lacked its natural predators, so nothing checked its numbers — an invasive species released from control. It was brought under control by introducing a cactus-feeding moth from its native range, a predator that regulated the prey population. This is the principle behind biological control of pests: a natural predator holds the target species in check.

Q24. When all the starfish Pisaster were removed from an intertidal zone, more than ten invertebrate species disappeared within a year. Explain why removing a predator reduced diversity.

Answer: Pisaster is a keystone predator that preys on several competing invertebrates and keeps any one of them from monopolising space. Once it was removed, the competitively superior species outgrew the rest, and intense interspecific competition drove more than ten species to local extinction. Predators therefore help maintain species diversity by reducing the intensity of competition among prey.

Q25. Two barnacles, Balanus and Chthamalus, live on Scottish rocky coasts. Connell found Balanus excludes Chthamalus from one zone. Name the phenomenon and the related concept of competitive release.

Answer: The larger, competitively superior Balanus dominates the intertidal zone and excludes the smaller Chthamalus — a demonstration of interspecific competition and competitive exclusion. Competitive release is the reverse: when the superior competitor is experimentally removed, the restricted species expands its range dramatically, confirming that competition, not physical limits, had confined it.

Q26. MacArthur observed five warbler species feeding on the same tree without one eliminating the others. Explain how they co-exist despite apparent competition.

Answer: The warblers avoid direct competition through resource partitioning — behavioural differences in where and how they forage on the tree let each use a different part of the resource. By feeding in different zones or at different times, they reduce overlap, so competition is eased and the species co-exist rather than exclude one another. This shows that competing species can evolve mechanisms promoting co-existence.

Q27. A cuckoo (koel) lays its eggs in a crow's nest, and the crow raises them. Name this interaction, classify it, and explain the co-evolution involved.

Answer: This is brood parasitism, a form of parasitism (+ -): the parasitic koel benefits by offloading parental care, while the host crow is harmed by wasting effort on foreign young. Over evolutionary time the koel's eggs have come to resemble the host's eggs in size and colour, so the crow is less likely to detect and eject them — a clear case of host-parasite co-evolution.

Q28. Explain why natural selection has not produced completely harmless parasites, and list two adaptations parasites show.

Answer: A parasite gains food and shelter at the host's expense, so selection favours parasites that exploit the host efficiently; a totally harmless parasite would extract fewer resources and leave fewer offspring, so it is not favoured. Parasites do, however, evolve to avoid killing the host too quickly. Typical adaptations include loss of unnecessary sense organs and the digestive system, adhesive organs or suckers to cling to the host, host specificity, and very high reproductive capacity (any two).

Q29. The fig tree and its wasp show a tight one-to-one partnership. Explain how each partner benefits and why this is called co-evolution.

Answer: This is mutualism (+ +). The wasp pollinates the fig inflorescence while searching for egg-laying sites, and in return the fig offers some of its developing seeds as food for the wasp larvae. Because a given fig species can be pollinated only by its partner wasp species, the two lineages have evolved in tight step with each other — a textbook case of co-evolution of a mutualistic pair.

Q30. The Mediterranean orchid Ophrys achieves pollination by 'sexual deceit'. Explain the mechanism and what would happen to it if the bee's appearance changed.

Answer: One petal of the Ophrys flower mimics the female of a bee species in size, colour and markings. The male bee, deceived, pseudocopulates with the flower and is dusted with pollen, which it carries to the next flower it visits. This depends on precise resemblance, so if the female bee's colouration changed during evolution, pollination success would fall unless the orchid co-evolved to keep matching the bee — again showing tightly linked co-evolution.

End of Section 12

You have now worked through 30 examples spanning population attributes and density, the four processes and the density equation, exponential and logistic growth, computing $r$ and doubling time, reading age pyramids, and every interaction type from predation and competition to parasitism, commensalism and mutualism. For a numerical, write the formula first; for an interaction, assign the signs first — and you will rarely leave marks on the table.