Organisms and Populations ISC Class 12 Biology: Notes

What Organisms and Populations is about

Weightage: Organisms and Populations is part of the unit Ecology and Environment, which carries 15 of the 70 marks in the ISC Biology Paper 1 (Theory), shared with 2 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Biology syllabus, examination year 2027, distribution of marks

Key concepts

Population attributes and measuring density

Sex ratio: proportion of females to males (e.g. 60% females, 40% males). Natality (birth rate): number of births per unit population per unit time, e.g. if 8 frogs are born in a pond of 40 frogs in a year, natality = 8/40 = 0.2 per frog per year. Mortality (death rate): deaths per unit population per unit time, e.g. 4 deaths among 80 fruit flies in a week = 0.05 per fly per week. Immigration: individuals of the same species entering from elsewhere; emigration: individuals leaving. Population density: number of individuals per unit area or volume; sometimes biomass or percentage cover is a better measure (one huge banyan tree and 200 Parthenium plants: percentage cover or biomass is more meaningful). Ways to measure density: total count (large animals in a small area); percentage cover (plants); pugmarks and faecal pellets (tigers); catch per trap (fish); mark-recapture (estimate N = (marked in first catch × total in second catch) / marked in second catch). Often only relative density is needed (e.g. fish caught per trap).

Worked example

In a pond, 50 fish are caught, marked and released. Later 40 fish are caught, of which 10 are marked. Estimate the population.

Try it first, then show the solutionSolution

1. Assume the proportion of marked fish in the second catch equals that in the whole population.
2. N = (50 × 40)/10 = 200.

Answer: About 200 fish

Age pyramids

A population contains individuals of different ages; plotting the age distribution (percentage in each age group) gives an age pyramid with three groups: pre-reproductive, reproductive and post-reproductive. For humans the pyramid is drawn separately for males and females. Expanding (triangular, broad base): many young individuals; the population will grow rapidly (e.g. many developing countries). Stable (bell-shaped): pre-reproductive and reproductive groups nearly equal; growth stable. Declining (urn-shaped, narrow base): fewer young than reproductive adults; the population will shrink. The shape reflects the growth status of the population.

Worked example

A country's pyramid has a narrow base and a wide middle. What does this suggest about its future population, and why?

Try it first, then show the solutionSolution

1. Urn-shaped pyramid: pre-reproductive individuals are fewer than reproductive ones.
2. As the present reproductive group ages, fewer people will replace them, so the population is declining.

Answer: Declining population

Population growth equation

The size of a population is dynamic. Over a period t, $N_{t+1}=N_t+[(B+I)-(D+E)]$, where B = births, I = immigrants, D = deaths, E = emigrants. Births and immigration increase density; deaths and emigration decrease it. Under normal conditions births and deaths are the most important factors; immigration and emigration matter when a new habitat is being colonised. Factors affecting growth: food availability, predation pressure, weather, competition, disease.

Worked example

A deer population of 500 has 60 births, 30 deaths, 20 immigrants and 10 emigrants in a year. Find the population after a year.

Try it first, then show the solutionSolution

1. $N_{t+1}=500+[(60+20)-(30+10)]$
2. $=500+80-40=540$

Answer: 540 deer

Exponential growth model

When resources (food, space) are unlimited, every species can realise its full innate potential to grow in number. If b and d are the per capita birth and death rates, the increase per unit time is $\frac{dN}{dt}=(b-d)N=rN$, where r is the intrinsic rate of natural increase (Norway rat r = 0.015; flour beetle 0.12; Indian human population (1981) 0.0205). Integral form: $N_t=N_0{e}^{rt}$, where e is the base of natural logarithms (2.71828). Plotting N against time gives a J-shaped curve. Any species growing exponentially under unlimited conditions can reach enormous densities in a short time (Darwin's illustration with elephants). Examples: bacteria in fresh medium, introduced species with no predators, algal blooms.

Worked example

A population of 100 grows exponentially with r = 0.1 per year. Find its size after 10 years.

Try it first, then show the solutionSolution

1. $N_t=N_0{e}^{rt}=100\times e^{0.1\times 10}=100e^1$
2. $=100\times 2.718=272$

Answer: About 272

Logistic growth model and carrying capacity

No population has unlimited resources; individuals compete, and the 'fittest' survive and reproduce. A given habitat has enough resources to support only a maximum possible number, its carrying capacity (K), beyond which no further growth is possible. A population growing in such a habitat shows an initial lag phase, then phases of acceleration and deceleration, and finally an asymptote when density reaches K. N against time gives a sigmoid (S-shaped) curve: Verhulst-Pearl logistic growth: $\frac{dN}{dt}=rN\left(\frac{K-N}{K}\right)$. The factor (K − N)/K is the environmental resistance: when N is small it is close to 1 (near-exponential growth); as N approaches K it approaches 0 and growth stops. Growth rate is maximum at N = K/2. The logistic model is considered more realistic because resources are finite.

Worked example

A population with r = 0.5 per day and K = 1000 has N = 200. Find dN/dt, and the maximum possible growth rate.

Try it first, then show the solutionSolution

1. $\frac{dN}{dt}=0.5\times 200\times \frac{1000-200}{1000}=0.5\times 200\times 0.8=80$ per day.
2. Maximum at N = K/2 = 500: $0.5\times 500\times 0.5=125$ per day.

Answer: 80 per day; maximum 125 per day at N = 500

Life history variation and reproductive fitness

Populations evolve to maximise their reproductive fitness (Darwinian fitness, a high r value) in the habitat in which they live: fitness is measured by the number of surviving, fertile offspring an individual leaves. Under particular selection pressures, organisms evolve the most efficient reproductive strategy. Examples: some organisms breed only once in their lifetime (Pacific salmon, bamboo) while others breed many times (most birds and mammals); some produce a large number of small-sized offspring (oysters, pelagic fishes) while others produce a small number of large-sized offspring (birds, mammals). These life history traits evolve in relation to the constraints of abiotic and biotic components of the habitat.

Worked example

Explain reproductive fitness and give two contrasting life-history strategies with examples.

Try it first, then show the solutionSolution

1. Reproductive fitness is the ability to leave surviving, fertile offspring; selection maximises it (high r).
2. Breed once (Pacific salmon, bamboo) vs many times (most birds, mammals); many small offspring (oysters) vs few large offspring (mammals).

Answer: As above

Population interactions: mutualism, competition, commensalism, amensalism

Mutualism (+/+): both species benefit: lichens (fungus and alga or cyanobacterium), mycorrhizae (fungi and roots), plant-pollinator partnerships (fig and fig wasp; Ophrys orchid and bee by sexual deceit), plant-animal seed dispersal. Competition (−/−): both are harmed when closely related species compete for limited resources, but totally unrelated species can also compete (flamingoes and resident fish for zooplankton in South American lakes). Interference competition: the feeding efficiency of one species is reduced by the interfering presence of another even if resources are abundant. Competitive release: a species whose distribution is restricted by a superior competitor expands its range dramatically when the competitor is removed (Connell's barnacles: removal of Balanus let the smaller Chthamalus spread on Scottish rocky coasts). Gause's competitive exclusion principle: two closely related species competing for the same resources cannot coexist indefinitely; the competitively inferior one is eventually eliminated. Species may coexist by resource partitioning (MacArthur's warblers feeding in different parts of the same tree). Commensalism (+/0): one benefits, the other is neither harmed nor benefited: orchid growing as an epiphyte on a mango branch, barnacles on a whale, cattle egret and grazing cattle, sea anemone and clownfish. Amensalism (−/0): one is harmed, the other unaffected: Penicillium secreting penicillin that kills bacteria.

Worked example

Barnacles grow attached to the back of a whale. Name and explain this interaction.

Try it first, then show the solutionSolution

1. The barnacles gain a place to attach and are carried to new feeding areas; the whale is neither harmed nor benefited.
2. One partner benefits and the other is unaffected: commensalism (+/0).

Answer: Commensalism

Predation and parasitism

Predation (+/−): a way of transferring energy fixed by plants to higher trophic levels; herbivores are 'predators' of plants. Predators keep prey populations under control (prickly pear cactus in Australia was controlled by introducing a cactus-feeding moth) and help maintain species diversity by reducing competition among prey (Paine's experiment: removing the starfish Pisaster led to extinction of many invertebrate species). Anti-predator adaptations: camouflage (cryptic colouring in insects and frogs); poisonous or distasteful prey (monarch butterfly feeding on a poisonous weed as a caterpillar); plant defences: thorns (Acacia, cactus), chemicals (Calotropis produces cardiac glycosides; nicotine, caffeine, quinine, strychnine, opium). Parasitism (+/−): the parasite lives on or in the host, depending on it for food (and shelter) and harming it. Parasites are often host-specific and co-evolve with the host. Adaptations: loss of unnecessary sense organs, adhesive organs or suckers, loss of digestive system, high reproductive capacity; complex life cycles with intermediate hosts (human liver fluke uses a snail and a fish). Ectoparasites feed on the external surface of the host: lice on humans, ticks on dogs, copepods on marine fish, and Cuscuta (a parasitic plant on hedge plants, lacking chlorophyll). Endoparasites live inside the host's body: tapeworm, liver fluke, Plasmodium; their life cycles are more complex and morphology simpler. Brood parasitism: a bird lays its eggs in the nest of its host and lets the host incubate them: the cuckoo (koel) laying eggs in the crow's nest; the parasite's eggs resemble the host's in size and colour to reduce detection.

Worked example

Why are many parasites host-specific? Give three adaptations of an endoparasite such as a tapeworm.

Try it first, then show the solutionSolution

1. Parasite and host tend to co-evolve: as the host evolves ways to reject the parasite, the parasite evolves ways to counter them, so it becomes suited to one host.
2. Tapeworm adaptations: hooks and suckers to cling to the gut wall; loss of the digestive system (it absorbs digested food through its body surface); very high reproductive capacity.

Answer: Co-evolution with the host; suckers and hooks, no digestive system, high reproductive capacity

Organisms and Populations in real life

Each concept, met in an everyday situation. Read the situation, answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Topics in this chapter

• Population Growth: Exponential and Logistic
• Predation, Parasitism and Mutualism
• Population Attributes and Age Pyramids

Organisms and Populations: key facts

Population change

$N_{t+1}=N_t+[(B+I)-(D+E)]$

B births, I immigrants, D deaths, E emigrants.

Natality and mortality

$\text{natality}=\frac{\text{births}}{N\times t},\text{mortality}=\frac{\text{deaths}}{N\times t}$

Per individual per unit time.

Exponential growth

$\frac{dN}{dt}=rN,N_t=N_0{e}^{rt}$

J-shaped curve; r = b − d.

Logistic growth

$\frac{dN}{dt}=rN\left(\frac{K-N}{K}\right)$

S-shaped; maximum rate at N = K/2.

Mark-recapture estimate

$N=\frac{M\times C}{R}$

M marked first, C caught second, R recaptured marked.

Interaction signs

$\text{mutualism }(+,+);\text{competition }(-,-);\text{predation, parasitism }(+,-);\text{commensalism }(+,0);\text{amensalism }(-,0)$

Classify any interaction by its effect on each species.

Organisms and Populations: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Organisms and Populations

The question patterns to expect, and the method that works for each.

Tricks to remember Organisms and Populations

Short memory aids. Say each one aloud once, then use it on a question the same day.

• BIDE

  Births and Immigration add; Deaths and Emigration subtract.

  Use it when: Population change equation.

• J is for 'Just keeps growing'

  J-curve = exponential (no limits); S-curve = logistic (Stops at K).

  Use it when: Growth curves.

• Commensal = Common table

  Commensalism: one eats at the table, the other loses nothing (+/0). Spell it with 'commen'.

  Use it when: Interaction names.

• Cuckoo cons the crow

  Brood parasitism: koel eggs in crow's nest.

  Use it when: Parasite types.

• Gause: two can't share one bed

  Competitive exclusion: identical niches cannot coexist.

  Use it when: Competition.

Organisms and Populations: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Organisms and Populations

The slips examiners see most often, and what to write instead.

• ✗Mistake: Misspelt or vague interaction names

  CISCE examiners' comments, 2024: for an orchid on a mango tree, candidates wrote 'communalism' instead of commensalism, or vague answers like positive interaction, symbiosis or mutualism.

  Wrong working

  Communalism / symbiosis.

  Right working

  Commensalism (+/0).

  ✓Do this: Decide the effect on each partner, then name it exactly.

  Usually costs: 1 mark.

• ✗Mistake: Natality as a difference, not a rate

  CISCE examiners' comments, 2024: for lily plants increasing from 50 to 65 in a year, candidates only subtracted the numbers instead of expressing natality as an annual rate per individual; in 2025, candidates used an incorrect formula or made arithmetic errors for growth rate.

  Wrong working

  Natality = 65 − 50 = 15.

  Right working

  Natality = 15/50 = 0.3 per plant per year.

  ✓Do this: Divide by the initial population and time.

  Usually costs: 1 mark.

• ✗Mistake: Wrong curve or incomplete logistic equation

  CISCE examiners' comments, 2024 and 2025: candidates swapped the J and S figures, explained reproductive health instead of reproductive fitness, wrote dN/dt = r[(K − N)/K] omitting N, drew a J-shaped curve or a species-area curve when asked for the logistic curve, left axes unlabelled, and did not explain the phases.

  Wrong working

  dN/dt = r(K − N)/K

  Right working

  dN/dt = rN(K − N)/K

  ✓Do this: Check that N appears outside the bracket.

  Usually costs: 1-2 marks.

• ✗Mistake: Wrong examples of parasitism

  CISCE examiners' comments, 2025: candidates explained types of parasitism without examples or a proper definition, gave orchids on mango, fungi or bacteria as examples, and listed commensalism and amensalism as types of parasitism.

  Wrong working

  Orchid on mango is an ectoparasite.

  Right working

  Ectoparasite: lice on humans, Cuscuta on hedge plants; endoparasite: tapeworm; brood parasite: koel in crow's nest.

  ✓Do this: Learn two named examples for each type.

  Usually costs: 1-2 marks.

Practice questions on Organisms and Populations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   The S-shaped growth curve is described by:
   (a) dN/dt = rN
   (b) dN/dt = rN(K − N)/K
   (c) N = K
   (d) dN/dt = K/N

   Hint

   Logistic.

   Answer

   (b)

2. Q2 · 1 mark · MCQBoard level

   Lichen is an example of:
   (a) commensalism
   (b) parasitism
   (c) mutualism
   (d) amensalism

   Hint

   Both benefit.

   Answer

   (c)

3. Q3 · 1 mark · Very short answerBoard level

   Define carrying capacity.

   Hint

   Maximum number.

   Answer

   The maximum population size that the resources of a habitat can support indefinitely.

4. Q4 · 1 mark · Very short answerBoard level

   In a pond of 120 fish, 18 die in a year. Calculate the mortality rate.

   Hint

   Deaths per individual.

   Answer

   18/120 = 0.15 per fish per year.

5. Q5 · 1 mark · Very short answerHarder board level

   A pond has 30 water hyacinth plants; by the end of the year 15 new plants have been produced (no deaths). Calculate natality.

   Hint

   15 new plants.

   Answer

   15/30 = 0.5 per plant per year.

6. Q6 · 2 marks · Short answerHarder board level

   Why is the logistic model considered more realistic than the exponential model?

   Hint

   Finite resources.

   Answer

   No habitat has unlimited resources; competition increases as numbers rise, so growth slows and stops at the carrying capacity, which the logistic model includes through (K − N)/K.

7. Q7 · 2 marks · Short answerHarder board level

   A population of 200 grows exponentially at r = 0.05 per year. Estimate its size after 20 years (e = 2.718).

   Hint

   N₀e^{rt}.

   Answer

   $N=200e^{0.05\times 20}=200e^1=543.6$, about 544.

8. Q8 · 2 marks · Short answerBoard level

   Distinguish between ectoparasites and endoparasites with two examples each.

   Hint

   Location.

   Answer

   Ectoparasites feed on the surface of the host: lice on humans, ticks on dogs (Cuscuta on plants). Endoparasites live inside the host: tapeworm, liver fluke (Plasmodium).

9. Q9 · 2 marks · Short answerHarder board level

   Explain with an example how plants defend themselves against herbivores.

   Hint

   Morphological and chemical.

   Answer

   Morphological: thorns of Acacia and cactus. Chemical: Calotropis produces highly poisonous cardiac glycosides, so cattle avoid it; other plants make nicotine, caffeine, quinine.

10. Q10 · 2 marks · Short answerBoard level

    What is brood parasitism? How has the parasite adapted?

    Hint

    Koel and crow.

    Answer

    A parasitic bird lays its eggs in the host's nest and lets the host incubate and rear them (koel in crow's nest). The parasite's eggs resemble the host's in size and colour, reducing the chance of the host ejecting them.

11. Q11 · 5 marks · Long answerHarder board level

    (a) Draw the three types of age pyramids and state what each indicates. [3]
    (b) What is reproductive fitness? Give two examples of life history variations. [2]

    Hint

    Triangular, bell, urn.

    Answer

    (a) Triangular (broad base): expanding population; bell-shaped: stable; urn-shaped (narrow base): declining. Bars for pre-reproductive, reproductive and post-reproductive groups, males and females. (b) The ability to leave the maximum number of surviving, fertile offspring (high r). Pacific salmon and bamboo breed once; birds breed many times; oysters produce many small offspring, mammals few large ones.

12. Q12 · 2 marks · Short answerHOTS / competency

    Flamingoes and resident fish in South American lakes compete although unrelated. Explain, and say what Gause's principle predicts if resources are very limited.

    Hint

    Shared food.

    Answer

    Both feed on the same zooplankton, so their common resource is limited and each reduces the other's food (competition, −/−). If the resource is very limited and they cannot partition it, the inferior competitor would eventually be excluded from the lakes.

13. Q13 · 2 marks · Short answerHOTS / competency

    A population at its carrying capacity of 600 has r = 0.4. Calculate dN/dt at N = 300 and N = 600, and explain the difference.

    Hint

    Logistic.

    Answer

    At 300: 0.4 × 300 × (300/600) = 60 per unit time (the maximum, since N = K/2). At 600: 0.4 × 600 × 0 = 0. At K, births balance deaths because environmental resistance is maximal.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Define population.Show answer

   A group of individuals of the same species in a defined area, sharing resources and interbreeding.

2. 2.What is r in dN/dt = rN?Show answer

   The intrinsic rate of natural increase (b − d).

3. 3.Shape of an expanding age pyramid?Show answer

   Triangular with a broad base.

4. 4.What is competitive release?Show answer

   Expansion of a species' range when a superior competitor is removed.

5. 5.Example of amensalism?Show answer

   Penicillium inhibiting bacteria by secreting penicillin.

6. 6.Example of an ectoparasitic plant?Show answer

   Cuscuta.

7. 7.Who stated the competitive exclusion principle?Show answer

   Gause.

8. 8.What is percentage cover used for?Show answer

   Measuring plant population density where counting individuals is not meaningful.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you calculate natality and mortality as rates?
• ☐Can you use N(t+1) = N(t) + (B + I) − (D + E)?
• ☐Can you draw and label J and S curves with equations?
• ☐Can you explain why growth is fastest at K/2?
• ☐Can you name the three age pyramid shapes?
• ☐Can you define reproductive fitness with examples?
• ☐Can you classify interactions by +/−/0 with examples?
• ☐Can you name ecto-, endo- and brood parasites?

Organisms and Populations: frequently asked questions