NCERT Class 12 Biology Ch 11 Organisms and Populations Solutions

Memory peg. Whenever a property is a ratio, rate, or distribution across many organisms, it belongs to the population, not the individual. That single rule generates the whole list.

The contrast becomes clean when stated as pairs.

• Individual: born or dies. Population: birth rate, death rate (per capita, per unit time).
• Individual: male or female. Population: sex ratio.
• Individual: has one age. Population: age distribution / age pyramid (Fig. 11.1: expanding, stable, declining).
• Individual: one organism. Population: density N (numbers, biomass, or per cent cover).
• Individual: no notion of growth in number. Population: growth rate dN/dt and the intrinsic rate of natural increase r = b - d.

1. Write the four-process balance equation on the answer sheet to anchor the rate idea: N_t+1 = N_t + (B + I)_gain - (D + E)_loss. B = births, D = deaths, I = immigration, E = emigration during the time interval t → t+1.
2. Cite one NCERT-style numeric example to lock the marks: a laboratory fruit fly culture of 40 individuals with 4 deaths in a week gives a death rate 4/40 = 0.1 per fruit fly per week. The ``per fly per week'' phrasing only makes sense for a population.
3. Close with the age-pyramid distinction: the shape (broad-based expanding, columnar stable, narrow-based declining) is a graphical attribute only the population can have, since a single individual is just a horizontal slice at one age.

Why this matters. Population ecology is the bridge between ecology and evolution: natural selection acts on populations (because selection needs differential birth/death rates), which is exactly why these rate attributes are so important.

Population-only attributes: birth rate, death rate, sex ratio, age distribution, population density, and population growth rate (with the intrinsic rate of natural increase r).

Strategic angle. The whole question reduces to ``invert N_t = N₀ e^rt for r when the doubling time is given''. The doubling-time formula that drops out is T_d = ln 2r r = ln 2T_d, which is worth memorising for board and competitive exams (NEET, specifically, has asked questions in 2018 and 2021 that hinge on it).

1. State the exponential growth law from NCERT Sec. 11.1.2 explicitly: dNdt = rN N_t = N₀ e^rt. Define each symbol: r is the per capita difference of birth and death rates (r = b - d), N₀ the initial density, N_t the density after time t.
2. Identify ``doubling time'' as the time T_d at which N_t / N₀ = 2. Substitute into the integrated form: 2 = e^{r T_d} ln(2) = r T_d. Hence r = ln 2T_d.
3. For this question T_d = 3 years, so r = 0.69313 = 0.23103 yr^-1. Numerator ln 2 = 0.6931; division 0.6931 ÷ 3 = 0.23103 (keep four significant figures).
4. Compare with NCERT benchmarks. The text lists r values: humans (India, 1981) = 0.0205 yr^-1 (doubling time ≈ 34 yr), Norway rat = 0.015 yr^-1 (T_d ≈ 46 yr), flour beetle = 0.12 yr^-1 (T_d ≈ 5.8 yr). At r = 0.231 yr^-1 the population in this question is growing roughly twice as fast as flour beetles in laboratory culture – a very high but not biologically impossible value.

Why this matters. The doubling-time relationship is the bridge between the differential equation dN/dt = rN and observable population data; it lets ecologists infer r from a single census-pair without calculus.

r = ln 23 ≈ 0.231 yr^-1.

Structural angle. Plants face a unique problem – they cannot run. Every defensive trait they have evolved either raises the cost of an attack (mechanical barriers) or lowers the value of the meal (chemical deterrents). Group the answer into these two evolutionary classes the NCERT chapter explicitly names, and supply a textbook example for each.

1. Morphological line – physical barriers. Lead with the most familiar examples – thorns of Acacia and Cactus, which the textbook names side-by-side. Add spines (modified leaves on cacti), prickles (outgrowths of the stem epidermis as in roses), trichomes / stinging hairs (e.g. on the nettle Urtica dioica), and tough silica-rich leaf margins of grasses. Each device works by raising the energetic or physical cost of feeding – torn mouths, slower mastication, wear on tooth enamel.
2. Chemical line – secondary metabolites. Cite Calotropis producing cardiac glycosides as the flagship NCERT example. The chapter explicitly says this is why ``you never see any cattle or goats browsing on this plant''. Then list the named alkaloids the textbook calls out: nicotine (tobacco), caffeine (coffee/tea), quinine (Cinchona), strychnine (Strychnos), opium (poppy). Add tannins and terpenoids for completeness.
3. Mechanisms by which chemicals act on the herbivore.
   • Make the herbivore sick on eating (cardiac glycosides slow the heart).
   • Inhibit feeding (bitter tannins).
   • Disrupt digestion (tannins bind dietary protein).
   • Disrupt insect reproduction or development.
   • Kill the herbivore at high doses (strychnine, certain alkaloids).
4. Evolutionary context – the extra-mark close. About 25% of all insect species are phytophagous and plants cannot flee, so the evolutionary pressure has been intense. This has generated one of the most diverse classes of secondary metabolites known. Many compounds (caffeine, quinine, opium, morphine) are commercially extracted by humans for unrelated uses, but evolved originally as anti-herbivore weapons.

Why this matters. The chemicals plants evolved as anti-herbivore defences are the raw material of pharmacology: morphine, atropine, artemisinin, digitalis and many anti-cancer agents all began life as plant defences. Crop breeders also actively select against such chemicals (to make the crop edible) and for morphological defences (silica leaf-edges in rice, glands in cotton).

Morphological: thorns (Acacia, cactus), spines, prickles, trichomes, silica edges. Chemical: cardiac glycosides (Calotropis), alkaloids (nicotine, caffeine, quinine, strychnine, opium), tannins, terpenoids.

Sign-table angle. Treat every species-interaction question as a two-step process: (1) assign a +, -, or 0 to each partner, then (2) look up the row in the interaction matrix (NCERT Table 11.1). The algorithm cannot go wrong if you do these two steps in order.

1. Partner A – the orchid. An orchid is an epiphyte: a plant that grows attached to another plant, purely for support, with no nutritional dependency. Perched on the mango branch, it gains (i) canopy-level sunlight (vital for photosynthesis), (ii) good air circulation around its aerial roots (orchids absorb atmospheric moisture through specialised velamen tissue), and (iii) refuge from terrestrial herbivores, ground floods and leaf-litter rot. Without the mango it would have to grow at ground level with much less light. Hence the orchid is benefitted: assign orchid = +.
2. Partner B – the mango. The orchid's roots cling to the bark but do not penetrate the cambium, xylem or phloem. It does not extract sap, nutrients or water from the mango. Therefore the mango neither gains (no service rendered) nor loses (no resource stolen). Assign mango = 0.
3. Look up the interaction matrix. The pair (+, 0) appears in NCERT Table 11.1 against exactly one row: commensalism. No ambiguity.
4. Cite parallels from the NCERT chapter to lock the application mark. Two other (+, 0) pairs the textbook lists: (i) the cattle egret foraging beside grazing cattle (egret = +, cattle = 0 – the cattle stir up insects the egret eats, but lose nothing), and (ii) barnacles on the back of a whale (barnacles = +, whale = 0 – barnacles get a free ride through nutrient-rich waters, the whale is unaffected). Naming any one parallel is a reliable way to earn the application mark.
5. Contrast with parasitism. If the orchid were a parasite (e.g. Cuscuta, the dodder, which the NCERT chapter explicitly cites), it would pierce the host's vascular tissue with haustoria to extract sap; the host would weaken; the sign pair would be (+, -) and the name would be parasitism. The fact that an orchid does not have haustoria is the diagnostic distinction.

Why this matters. Distinguishing commensalism from parasitism hinges on asking one focused question: is the host losing anything? If yes, it is parasitism. If no, it is commensalism. The orchid–mango case is the standard textbook benchmark for the ``no, the host loses nothing'' branch of this decision.

Commensalism – the orchid (epiphyte) is benefitted, the mango tree is unaffected. Sign pair (+, 0), NCERT Table 11.1.

Quick reading. Biocontrol = the predation principle applied to agriculture. The pest is the prey, the released agent is the predator (or parasitoid or pathogen), and the goal is to keep the pest below the economic injury threshold – not zero, but low enough that crop loss is acceptable.

1. State the principle. In nature, predators control prey densities through density-dependent feeding: more prey ⇒ more predation, fewer prey ⇒ less predation. This negative-feedback loop is what stabilises prey numbers. Biocontrol deliberately reproduces this loop in a crop field by importing or releasing a host-specific natural enemy of the pest.
2. Anchor with the NCERT example. Opuntia (prickly pear cactus) was introduced into Australia in the early 1920s. With no natural predator present, it spread across millions of hectares of rangeland – an exponential, J-shaped invasion. The moth Cactoblastis cactorum, imported from Opuntia's native habitat in Argentina, lays eggs on the cactus pads; the caterpillars bore into and consume the cactus tissue. Within a few years cactus density was driven back to manageable levels. This is the canonical illustration of predation-based control.
3. Add a second, modern example to widen the answer. The soil bacterium Bacillus thuringiensis (Bt) is sprayed on cotton and brinjal to kill lepidopteran larvae through its crystal toxin (Cry protein). The same predation principle applies, with the ``predator'' replaced by a pathogenic bacterium. Modern transgenic Bt cotton has the Cry gene engineered directly into the plant.
4. Caveat the chapter explicitly teaches. A predator that is too efficient drives the prey extinct and then collapses for lack of food. NCERT states ``predators in nature are prudent''. Biocontrol agents are therefore screened in advance for two properties: (a) host-specificity (they will not attack non-pest native species), and (b) prudence (they regulate, not eradicate). This is why exotic biocontrol introductions are preceded by years of host-range testing in quarantine.
5. Add the diversity-protecting role for the extra mark. NCERT mentions Paine's classic Pacific intertidal experiment: removing the starfish Pisaster caused more than 10 invertebrate species to go extinct within a year due to interspecific competition. Predators thus also protect biodiversity, not just suppress one prey species.

Why this matters. Biocontrol is the foundation of modern integrated pest management (IPM), which combines biological, cultural and minimal-chemical methods to keep pests below the economic injury threshold while protecting non-target organisms, pollinators and soil microbiota.

Predators regulate prey populations. Biological pest control releases a host-specific natural enemy (predator, parasitoid or pathogen) of the pest, exploiting this regulatory predation pressure to suppress the pest below the damage threshold. NCERT example: the Cactoblastis moth controlling Opuntia in Australia.

Definition-pair angle. Write the two definitions as a deliberate contrast: emphasise ``same species, one area'' for population and ``many species, one area'' for community. Examiners explicitly look for both phrases.

1. Population – full definition. A population is a group of individuals of one species living together in a well-defined geographical area, sharing or competing for similar resources, and capable of interbreeding (sexual or, by extension in ecological usage, asexual). It is the basic unit of intraspecific (within-species) ecology and the unit on which natural selection acts (different birth/death rates among sub-groups within the population).
2. NCERT-listed population examples. Cormorants in a wetland; rats in an abandoned dwelling; teakwood trees in a forest tract; bacteria in a culture plate; lotus plants in a pond. Pick any one for the example mark.
3. Community – full definition. A community (or biological/biotic community) is the assemblage of all populations of different species that occupy the same area at the same time and interact through feeding, competition, mutualism, parasitism, etc. It is the unit of interspecific (between-species) ecology.
4. Community-level example. A pond community contains lotus plants (Population A), fishes (Population B), algae, zooplankton, frogs and aquatic insects (further populations) – all interacting through predation (fish eat zooplankton), competition (algae compete with lotus for light) and other relationships.
5. Key distinction – one-line memory peg. Population is intraspecific (one species, many individuals); community is interspecific (many species sharing one habitat).
6. Where each fits in Chapter 11. The population-level section (11.1.1–11.1.3) covers attributes, growth and life history. The community-level section (11.1.4 onwards) covers the six interspecific interactions. Together they set up Chapter 12 (Ecosystem), which adds the abiotic compartment.

Why this matters. Diagnosing the level (population vs community) is a recurring exam skill: ``the orchid–mango interaction'' is a population-pair question (commensalism), while ``the forest floor biodiversity of the Western Ghats'' is a community-level question (interspecific interaction web). Read the question once and identify the level before writing.

Population: group of individuals of one species in a defined geographical area, sharing resources and capable of interbreeding.
Community: assemblage of all populations of different species in the same area, interacting through ecological processes.

Sign-table angle. For each interaction-type part, the marker expects two things: (i) a one-line definition that names the +/-/0 sign pair, and (ii) one named NCERT example. Stick to the textbook examples – they are the safest under board marking schemes.

1. (a) Commensalism (+, 0). One species benefits, the other is neither benefitted nor harmed. Orchid on a mango branch – the orchid (epiphyte) gains a canopy perch and sunlight; the mango is unaffected because the orchid does not pierce its vascular tissue. Backup textbook examples: cattle egret beside grazing cattle, barnacles on a whale, clownfish among sea-anemone tentacles.
2. (b) Parasitism (+, -). The parasite gains nourishment and shelter; the host is harmed (reduced survival, growth, reproduction; may become more vulnerable to predation). The human liver fluke, a trematode that depends on two intermediate hosts (a snail and a fish) to complete its life cycle, is the cleanest NCERT example. Backup examples: lice on humans, ticks on dogs, Cuscuta on hedge plants, Plasmodium (malarial parasite) in human RBCs, and brood parasitism by the cuckoo on the crow.
3. (c) Camouflage. Note: camouflage is a prey-defence trait, not a sign-pair interaction; it does not appear in Table 11.1. It is the cryptic colouration, pattern or shape that hides prey from predators. Use NCERT's phrase: ``some species of insects and frogs are cryptically-coloured (camouflaged) to avoid being detected easily by the predator''. Backup examples: leaf insect (Phyllium), stick insect (Phasmida), the polar bear's white coat.
4. (d) Mutualism (+, +). Both species benefit. Three flagship NCERT examples, any one of which suffices: lichens (fungus provides anchorage and water; alga or cyanobacterium provides organic carbon from photosynthesis); mycorrhizae (fungi help plant roots absorb nutrients; plant supplies sugars); fig–wasp (one-to-one species-specific relationship, Fig. 11.4 – wasp pollinates fig; fig provides developing seeds as food for wasp larvae).
5. (e) Interspecific competition (-, -). Two species compete for the same limited resource (food, space, mates, light); each species' fitness, measured by its intrinsic rate of increase r, is reduced in the presence of the other. The flamingo–fish case in shallow South American lakes – both compete for the zooplankton – is short, neat and directly NCERT-quoted. Backup examples: Balanus excluding Chathamalus (Connell's barnacle experiment on rocky Scottish coasts), and the Abingdon tortoise vs introduced goats in Gal'apagos.
6. Score-maximising format. Lay out the answer as a labelled list – (a), (b), (c), (d), (e) – with the definition and example on separate lines. This gives the examiner discrete bullet points to tick off and is the safest way to lock 5 of 5 marks.

Why this matters. The five terms cover four of the six rows of Table 11.1 (commensalism, parasitism, mutualism, competition) plus one prey-defence trait (camouflage). They commonly appear together in 5-mark ``define and give example'' questions on board papers.

(a) Orchid on mango (commensalism); (b) liver fluke (parasitism); (c) cryptically coloured frogs/insects (camouflage); (d) lichens (mutualism); (e) flamingoes vs fishes (interspecific competition).

Picture-first. The simplest way to remember the logistic curve is to draw the J-curve first (exponential), then bend it over horizontal when it hits the dashed K line. The result is the S-curve (sigmoid). This visual trick makes the description impossible to forget.

1. Build the equation by inspection. Start from the exponential law dN/dt = rN. Multiply the right-hand side by a bracket chosen to satisfy two boundary conditions: it must equal 1 at N = 0 (so logistic matches exponential at low density) and equal 0 at N = K (so growth stops at the carrying capacity). The simplest bracket satisfying both is (K - N)/K. Substituting, dNdt = rN(K-NK). That is the Verhulst–Pearl logistic equation. This construction explains why it has exactly the right asymptote.
2. Identify the four phases on the S-curve visible in Fig. 11.3 (Curve b). Lag → Acceleration → Deceleration → Stationary. Mark each on your hand-drawn S-curve before writing the description. Each phase scores 0.5 mark in board marking schemes, so all four must appear.
   • Lag phase: few individuals, slow gross increase.
   • Acceleration phase: density rises rapidly, (K - N)/K ≈ 1, growth ≈ exponential.
   • Deceleration phase: density approaches K, (K - N)/K → 0, growth slows.
   • Stationary phase: N = K, births balance deaths, dN/dt ≈ 0.
3. Find the inflection point. Differentiate dN/dt with respect to t and set the result to zero; this yields the inflection condition N = K/2. At that density the instantaneous growth rate is maximum: (dNdt)_max = r · K2 · K - K/2K = rK4. This N = K/2 point is the famous maximum sustainable yield of fisheries science – the best harvesting density a manager can target without driving the stock toward collapse.
4. Contrast with exponential (Curve a in Fig. 11.3). Exponential growth assumes unlimited resources: dN/dt = rN, N_t = N₀ e^rt, J-shaped curve, indefinite rise. Logistic assumes finite resources: dN/dt = rN(K-N)/K, N asymptotes at K, S-shaped sigmoid. NCERT calls logistic ``more realistic'' for natural populations because every real habitat eventually runs out of food, space or both.
5. NCERT-cited applications. The chapter encourages students to plot India's census data over the last 100 years and check the growth pattern – early decades approximate exponential growth, recent decades show deceleration as India's effective K becomes limiting (resources, agriculture, urban land).

Why this matters. The shape of a population's growth curve (J vs S) tells an ecologist whether the species is in early colonisation (J) or near its carrying capacity (S). The bend is what makes the logistic curve diagnostic, and the maximum-yield point at N = K/2 is the bridge from ecology to fisheries management, harvesting regulations and recovery targets in conservation biology.

Logistic (Verhulst–Pearl) growth: dNdt = rN(K-N)/K. S-shaped sigmoid curve with four phases – lag, acceleration, deceleration, stationary – plateauing at the carrying capacity K; maximum growth rate rK/4 at N = K/2.

Elimination angle. The fastest MCQ strategy for species-interaction questions is to convert each option into a (+/-/0) sign-pair and match it against NCERT Table 11.1. Among the six interactions in the table, exactly one row has the pair (+, -) that is offered here as an option and is named parasitism (predation also has (+, -), but predation is not among the choices).

1. Decode option (a). ``One organism is benefited.'' This sentence is incomplete – it does not say what happens to the other partner. The pair could be (+, +), (+, 0), or (+, -) – i.e. mutualism, commensalism, or parasitism/predation. Reject (a) for being underspecified.
2. Decode option (b). ``Both are benefited.'' This is the sign pair (+, +), which Table 11.1 names mutualism (lichens, mycorrhizae, fig–wasp). Wrong category for parasitism. Reject (b).
3. Decode option (c). ``One benefited, other unaffected.'' Sign pair (+, 0), which Table 11.1 names commensalism (orchid on mango, cattle egret with cattle). Wrong category for parasitism. Reject (c).
4. Decode option (d). ``One benefited, other affected (harmed).'' Sign pair (+, -). In Table 11.1 this matches two rows – parasitism and predation. Since predation is not offered and parasitism is, option (d) is the unique correct match. Accept (d).
5. Reality check with a textbook example. Plasmodium (the malarial parasite) in human RBCs: the parasite gains nutrition and a site for reproduction; the host suffers fever, anaemia and, in severe cases, organ failure. The sign pair (+, -) is unambiguous, confirming option (d). Other NCERT examples that fit the same pair: the human liver fluke; Cuscuta on hedge plants; lice on humans; ticks on dogs.
6. Common trap to avoid. Some students pick option (a) because it ``sounds parasitic''. Option (a) is not wrong, it is just insufficiently specific; a multiple-choice question rewards the option that contains the full diagnostic information, which is (d).

Why this matters. Sign-pair thinking turns every species-interaction MCQ into a one-step lookup; you do not need to memorise the names if you can read the +/-/0 pair off the option text. This is a particularly useful tactic in NEET, where the same distractor pattern (one correct option + one underspecified-but-tempting option) recurs.

(d) One organism is benefited, other is affected. The sign pair is (+, -), matching parasitism in NCERT Table 11.1.

Pick-and-explain angle. The exercise says ``list any three'', so pick the three with the cleanest one-line definitions, a memorable NCERT example, and a strong evolutionary or applied interpretation: density, natality + mortality, and age distribution. These three together also generate the rest of population ecology – the growth-rate equation dN/dt comes out of natality minus mortality applied to a density.

1. Population density N – the central quantity. Population size per unit area or volume at a given moment. Multiple metrics are available, with the choice depending on the species: total count (small mammals), biomass (microbes), per-cent cover (vegetation), or indirect indices (tiger census via pug marks and fecal pellets). NCERT example: stating banyan density by number underestimates its ecological role compared with 200 carrot grass plants, so per-cent cover or biomass is the better measure for trees. Density is the variable against which every ecological process is evaluated.
2. Natality and mortality – the rate engine. Per capita birth rate (b) and death rate (d) drive the change in density. NCERT lotus example: 8 new offspring among 20 plants in a year gives b = 820 = 0.4 offspring per lotus per year. Fruit fly example: 4 deaths out of 40 flies in a week gives d = 440 = 0.1 deaths per fly per week. Their difference is the intrinsic rate of natural increase r = b - d, which drives the exponential growth equation dN/dt = rN. Natality and mortality are therefore the two quantities ecology needs to predict population trajectories.
3. Age distribution and the age pyramid. The percentage of individuals in each age class (pre-reproductive, reproductive, post-reproductive). NCERT Fig. 11.1 shows three pyramid shapes:
   • Broad-based, tapering top → expanding / growing population (many young, high birth rate).
   • Columnar (roughly equal age classes) → stable population (births ≈ deaths).
   • Narrow-based, broad top → declining population (few young, low birth rate, ageing demographic).
   Demographers use the age pyramid to project future population trends and to inform policy decisions (housing, schools, pensions). For wildlife populations, the same pyramid predicts whether a species is recovering or heading for extinction.
4. Other valid picks if asked for more. Sex ratio (proportion of males to females), population growth rate dN/dt, immigration (I) and emigration (E) (the spatial flux terms in the balance equation N_t+1 = N_t + [(B + I) - (D + E)]). Any three of these are acceptable answers; the three chosen above are the highest-yield for the mark scheme because each has a calculable NCERT example.
5. Common application – wildlife census. Combining all three: tigers in a reserve are censused for density (pug marks), natality and mortality (camera-trap records of cubs and carcasses), and age distribution (scat-DNA aging). Together these tell managers whether to add anti-poaching patrols, expand the buffer zone, or relocate breeders.

Why this matters. These characteristics are the raw data that feed the exponential and logistic models. Without them, an ecologist has no way to predict whether a population is heading for extinction, plateau, or runaway growth – and therefore no basis on which to prescribe conservation, harvesting, or pest-control action.

Three key population characteristics: population density N; natality and mortality (per-capita birth and death rates, NCERT lotus example = 0.4/yr); and age distribution / age pyramid (expanding, stable, or declining).